How are chemistry and materials science related? I'm currently a chemistry major and now I am considering doing a minor in materials science.

Before making a decision, I would like to know how the two disciplines are related and what are their differences such as techniques, approaches, emphasis, etc.

How will my knowledge in chemistry be useful in my understanding of materials science, and how useful is a knowledge of materials science to a chemist?

An undergraduate chemistry course here in the UK typically has a materials component to it in the form of a module or two, so you bet they're related!

It's sometimes not that fruitful to view materials science and chemistry as a different subject—much like physic vs chemistry—however we can associate certain motivations to either topic as we see fit. Please note that this is quite subjective; this view does indeed vary from person to person.

First of all, materials is more interested in the solid state than chemistry in general. We might like to look at crystal structures via x-ray crystallography (see Bravais lattices, space groups etc.) as well as the rather intriguing quasi-crystals (for which Dan Shectman won the Nobel in Chemistry in 2011). The skills gained in a materials crystallography course would no doubt help with chemistry. The converse is also true since many substances in chemistry are amenable to study via crystallography, such as proteins. You therefore may also have a crystallography module in your chemistry course—looking at molecules in the solid state not typically viewed as a 'material'. If it has appreciable electron density, you can study it by x-ray crystallography!

Another aspect of materials science looks at properties of materials, such as stress/strain curves etc.I've had such a lecture series specifically dealing with polymers in a chemistry course; one of the challenges of modern day chemistry is to invent and modify polymer architectures for specific industrial applications. It's clear then that chemistry and materials science complement each other here. Whilst materials looks at interesting properties of given materials, chemistry constructs those polymers.

Materials science can differ from chemistry in that it focuses (or at least did in my case) on the band-structure of solids more than chemistry does. So it's possible that a materials course would give you a more nuanced appreciation of semi-conductors and the like. Again, this would help overwhelmingly if you decided to go into the field of conducting polymers—a currently very active field in R&D, especially if it involves replacing rare earths.

To summarise: materials science and chemistry go hand in hand in my quick overview, however some aspects of chemistry (such as organic) view the solid-state as the 'graveyard of chemistry'—you'd be hard pressed to find reactions which occur in the solid state! If you want to find out a little more before delving into materials, then finding a quick Youtube course might help to sway you either way. In my opinion there are far more ridiculous pairings in the world than chemistry and materials!


Determine reaction rate from reaction time? These are the results of my experiment (solution A and B were reacted and the reaction time was measured).

Data Table Edit:

How would I determine the rate of reaction (so I can find the rate law) from the reaction time?

Solution A: $\pu{4.3 g}$ of potassium iodate per $\pu{1 L}$ of water

Solution B: $\pu{2 g}$ of sodium bisulphite, $\rm5~mL~1.0~M~\ce{H2SO4}$, 4 g soluble starch per $\pu{1 L}$ of water

I found the concentrations of $\ce{[KIO3]}$ by dividing mass of $\ce{KIO3}$ by molar mass of $\ce{KIO3} (\rm214.001\frac{~g}{mol})$ and then dividing that by volume of solution A. I also divided the mass of $\ce{NaHSO3}$ by molar mass of $\ce{NaHSO3} (\rm104.061\frac{g}{mol}$) and then dividing that by volume of solution B to find the concentrations of $\ce{[NaHSO3]}$.

I then tried to find the reaction rate by $$\frac{1}{3}\times\frac{\ce{[NaHSO3]}}{t}$$ and $$\frac{\ce{[KIO3]}}{t}$$ as the reaction is $$\ce{IO3- + 3 HSO3- -> I- + 3 HSO4-}$$

Not sure if I'm on the right track or not.

Source:

Reaction Rate was found by $[\ce{NaHSO3}]$/Reaction Time and $[\ce{KIO3}]$/Reaction Time.

Rate order was found by log(Reaction Rate)/log([$\ce{NaHSO3}$]) and log(Reaction Rate)/log([$\ce{KIO3}$]).

Would appreciate it if anyone could tell me if I did this correctly.

The reactant orders in iodine clock reactions generally turn out to be first order.  Your calculated orders are suggestive of first order dependences, but there is a systematic error evident in the data that possibly stems from the experimental design.  In addition, I don't believe that the manner in which you calculated the orders is correct.  The approach in this experiment is to use the method of initial rates.  Briefly, a differential rate law (which gives instantaneous rate as a function of concentrations) can be written as
rate $=k[\ce{A}]^n [\ce{B}]^m$ for a reaction $\ce{A + B -> products}$.  To find the order of reactant A, for example, one varies the initial concentration of A and keeps the concentration of B constant. This generates a pseudo-rate law in which the concentration for B is treated as a constant(k'):  rate = $kk'[\ce{A}]^n$.  This relation can be linearized as follows:

$$
\begin{align}
\log \text{rate} &= \log kk'[\ce{A}]^n\\
\log \text{rate} &= \log [\ce{A}]^n + \log kk'\\
\log \text{rate} &= n \log[\ce{A}] + \log kk' \tag{I}
\end{align}
$$

Trials are conducted in which [A] is varied and the initial rate is determined.  A plot of log rate vs. log[A] should be linear and its slope should equal n, the order of reactant A. If, on the other hand, we solve EQ I for n we get
$$
\begin{align}
n &= \frac{\log \text{rate} - \log kk'}{\log[\ce{A}]} \\
n &= \frac{\log \frac{\text{rate}}{kk'}}{\log[\ce{A}]}
\end{align}
$$

Thus calculating n as $\log \text{rate}/\log[\ce{A}]$ (your approach) is not fully correct because the pseudo-rate constant $kk'$ is not accounted for.  I plotted your $[\ce{HSO3-}]$ data in the form of $\log \text{rate}$ vs. $\log [\ce{HSO3-}]$ and obtained a value for n (the slope) of 0.84 with $r = 0.997$ (note: the rates must be treated as negative values because reactant is being consumed).  I advise you to attempt such a plot for the data where $\ce{[IO3-]}$ was varied.  Regarding the systematic error, you should think about the fact that you determined average rates experimentally, meaning reactant concentration was actually decreasing during the timed intervals and, therefore, the rates you calculated reflected an average reactant concentration, not an initial concentration.


How to find the balanced reaction equation for the decomposition of potassium dichromate in non aqueous conditions? I can't find a method to balance a reaction: the heating of potassium dichromate, which forms potassium dichromate (III), dichromate oxide (III) and oxygen.
I can't add water products as $\ce{H+}$, $\ce{{}^{-}OH}$ because we are not in solution.
So, how can I balance it?

Who oxidises in this redox? The oxygen inside the salt? The oxygen of $\ce{H2O}$ which is in the air?

I think that part of your difficulty lies in incorrect naming of the products which are simply potassium chromate(VI) and chromium(III) oxide together with oxygen. The equation then becomes:

$$\ce {4K2Cr2O7 -> 4K2CrO4 + 2Cr2O3 + 3O2}$$

As for what is oxidized and reduced; the chromium is reduced from chromium(VI) to chromium(III) in going from $\ce{K2Cr2O7}$ to $\ce{Cr2O3}$ and some of the oxygen is oxidized from −2 in the dichromate to 0 in the oxygen gas.


Why is tert-Butyl (tBu) often not listed as protecting group for alcohol? I am not seing tBu listed as alcohol protection group in all of these places:

https://en.wikipedia.org/wiki/Protecting_group#Alcohol_protecting_groups
http://www.synarchive.com/protecting-group/
http://www.organic-reaction.com/organic-synthesis/protecting-groups/ether/

On the other hand it seems to be quite common in peptide synthesis to protect the alcohol of serine, threonine and tyrosine with tBu.

It is listed on the German Wikipedia page: https://de.wikipedia.org/wiki/Schutzgruppe#Alkohole

Can I conclude from this that tBu is very rarely used in non-peptide syntheses? Or is it just plain missing from the English Wikipedia article?

Protecting group selection is a balancing act of ease of preparation and deprotection. In common practice, use of a protecting group adds two steps to a synthetic route. Installation and removing a tert-butyl group requires stronger (acidic) conditions. Of course, it's a rock while in place and inert to nearly all conditions short of strong acid.

Peptide synthesis is particularly amenable to use of tert-butyl protection of hydroxyl groups because (1) all of the suitably protected residues are commercially available and (2) the protecting groups have been designed to be globally deprotected with trifluoroacetic acid. These factors minimize the inherent inefficiency to protecting groups.

The workhorse protecting groups for alcohols are silyl ethers (e.g.trimethylsilyl, triethylsilyl, tertbutyldimethylsilyl, triisopropylsilyl, tertbutyldiphenylsilyl). The silyl ethers are easier to form and cleave than standard ethers and have excellent compatibility with reactions that require alcohol protection.

Specialized (carbon) ethers can be useful as well, including unsubstituted or substituted benzyl ether (removed by oxidative or reductive conditions) and acetals (removed by mild acid treatment).

For a more complete description of protecting groups (including tert-butyl), see the books by Greene and Wuts or Kocienski.


Can Fe and Ca react and become unusable in some foods? In Turkish cuisine there is a popular food called "tahin pekmez" (tahini with grape molasses).
According to a 2017 article on some Turkish news site, mixing these is supposedly counter-productive due to the calcium in the tahini bonding with the iron in the molasses and becoming unusable to the human body.[1]  From what I can tell, the article doesn't cite any study for this claim.
Obviously I'm inclined to believe that the claim is click-bait, but I thought let's give it a chance and ask people who know their chemistry.
I hope this isn't too basic or otherwise off-topic for Chemistry StackExchange.
[1] https://www.yenisafak.com/gundem/tahin-ve-pekmezi-karistirmayin-2804300 (in Turkish)

Studies have indicated that calcium inhibits iron absorption in humans. However, numerous dietary factors influence iron absorption, and the effect of calcium may not be as pronounced when calcium is served as part of a whole diet. Calcium does not have any significant long-term effect on iron absorption. In addition, milk and milk products have not been found to affect iron absorption.
References:

Lönnerdal B. Calcium and iron absorption--mechanisms and public health relevance. Int J Vitam Nutr Res. 2010 Oct; 80(4-5):293-9. doi: 10.1024/0300-9831/a000036.
Grinder-Pedersen et.al., Calcium from milk or calcium-fortified foods does not inhibit nonheme-iron absorption from a whole diet consumed over a 4-d period, The American Journal of Clinical Nutrition, Volume 80, Issue 2, August 2004, Pages 404–409, doi: 10.1093/ajcn/80.2.404
https://www.dairynutrition.ca/nutrients-in-milk-products/calcium/calcium-and-iron-absorption-is-there-an-interaction


How are the toxic components in a mixture identified? I was doing research on buckwheat sprouts (as to whether I should include them in my diet) and I came across a website that said eating too much can cause fagopyrism, which is caused by compounds called fagopyrins:

I was just wondering how they came to the conclusion that it was these compounds that were dangerous? If someone reports side effects, with the number of different chemical compounds in a plant, how would they narrow it down to that one?
Do they just toxicity test all of the possible compounds? Or do they look for certain functional groups and structures to narrow down their search? Ie is there a way you can predict from the structure that a compound could be potentially toxic?
Thanks!

You have many questions nested together. Whenever you have a mixture + identification of components in your mind, bring chromatography to your mind. You must be familiar with paper chromatography. Same idea, but applied to more sophisticated instruments.
When you have a completely unknown mixture, you start a so-called screening procedure with a high-performance liquid chromatograph and a mass spectrometer. The HPLC separates the components, and the mass spectrometer identifies the molecules from the fragmentation pattern and molecular mass.
You start to see how many components are there in the HPLC-MS chromatogram. Let us say you had natural products and you see 100 peaks (which is not unusual at all).
Later you scale up the process, go at a preparative scale and start collecting fractions of mixtures. You perform biological tests on them, and the relevant fraction is further studied by more advanced tools such as recycling the HPLC-NMR system.
The question you ask is equivalent to a decent Ph.D. project in natural product isolation and characterization followed by investigating their biological activity. In pharmaceutical companies, HPLCs run day and night with a typical HPLC may have 10-15 columns for screening as to which column under which mobile phase will produce the best separation.
I did not check how fagopyrins were discovered, but the process is general. Wiki gave a hint on the first German reference Fagopyrin, ein photodynamischer Farbstoff aus Buchweizen (Fagopyrum esculentum) = Fagopyrin, a photodynamic dye out of buckwheat, locked behind paywall. This procedure is from 1950s, when HPLCs did not exist but the technique of chromatography was very well known.

Wir haben aus einem Methanolextrakt von getrockneten Buchweizenblüten
durch fraktionierte Verteilung und chromatographische Adsorption einen
photodynamisch wirksamen Farbstoff in Form eines dunkelroten, fast
schwarzen mikrokrystallinen Pulvers isoliert, der im folgenden als
Fagopyrin bezeichnet ist.
We have isolated from a methanol extract of dried buckwheat flowers by
fractional partitioning and chromatographic adsorption a
photodynamically active dye in the form of a dark red, almost black
microcrystalline powder, hereafter referred to as fagopyrin.


Comparison of rate of dehydration of primary, secondary & tertiary alcohols on cyclohexane Rate of dehydration when given compounds are treated with conc. $\ce{H2SO4}$ is

According to me the answer should be $R > Q > S > P$ but my textbook solution gives the answer as R > Q > P > S (Sadly, there is no explanation).

Dehydration of secondary & tertiary alcohols in the presence of conc. $\ce{H2SO4}$ proceed through E1 mechanism whereas most primary alcohols proceed through E2 mechanism.

In the molecular R, the carbocation formed after the alcohol group leaves the molecule after protonation is a tertiary carbocation and not just that, it is resonance stabilized as well.

In the molecule Q, you get a secondary carbocation and this is resonance stabilized as well.

In molecule S, you get a secondary carbocation but there is no resonance stabilization.

In molecule P, the dehydration goes via E2 mechanism and there is a bulky cyclohexene group attached to the $\alpha$-carbon which should be contributing to steric hindrance.

If the question did not have molecule P, then its pretty much obvious that the answer is R > Q > S but with molecule P, I am not quite sure where to place it.

I thought that P will be least reactive since it proceeds via E2 mechanism and there is one bulky group attached to the carbon.

So how do I come up with a concrete answer to justify why molecule P should react faster than molecule S but slower than molecule Q.

The rate of dehydration is related to the ease of formation of the carbocation and the energy of the carbocation intermediate.

The ease of formation of carbocation is tertiary>secondary>primary. That is the ease of abstraction of $\ce{OH2+}$ COULD BE be R>Q/S>P.

Let's look at a comparison of the energy of the carbocation intermediates in all these cases.

  A carbocation may be stabilized by resonance by a carbon-carbon double bond next to the ionized carbon. Such cations as allyl cation $\ce{CH2=CH–CH2+}$ and benzyl cation $\ce{C6H5–CH2+}$ are more stable than most other carbocations. Molecules that can form allyl or benzyl carbocations are especially reactive. These carbocations where the C+ is adjacent to another carbon atom that has a double or triple bond have extra stability because of the overlap of the empty p orbital of the carbocation with the p orbitals of the $\pi$ bond. This overlap of the orbitals allows the charge to be shared between multiple atoms – delocalization of the charge - and, therefore, stabilizes the carbocation.

Let's look at them individually,

P - allylic cation, primary alcohol
Q - allylic cation, secondary alcohol
R - allylic cation, tertiary alcohol
S - NOT an allylic cation, secondary alcohol

$\ce{R}$ is clearly the winner.

Now we may conclude that the charge dispersal is better in resonance than it is in hyperconjugation (explanation)

Here are the 3D diagrams of the P carbocation. Hydrogen cation doesn't look like it has to face significant steric hindrance to perform protonation of the alcohol.


How can we combine all the three gas laws into a single ideal gas equation? In all texts that I have read, it has been stated that the combined gas law for ideal gases was derived from the individual gas laws proposed by Boyle, Charles and Avogadro.

My confusion is this is that, in each individual law, some variables of the system's state are to be kept constant. However, none of these conditions are considered in the combined gas equation. Also, how is it possible for us to suggest that for two states with variables $P$, $V$ and $T$:

$$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$$

It is indeed the case that the rearrangement of the different gas laws is not very satisfactory as it employs the different variables without any constraints. This issue has been addressed by S. Levine in J. Chem. Educ., 1985, 62 (5), p 399, but in the case you do not have access to this reference, I'll shortly repeat his arguments here.

As $P$, $V$ and $T$ are thermodynamic sate functions, we may write the total differential of $T$ as

$$\mathrm{d}T=\left (\frac{\partial T}{\partial V} \right )_P \mathrm{d}V + \left (\frac{\partial T}{\partial P} \right )_V\mathrm{d}P \equiv X\mathrm{d}V+ Y\mathrm{d}P$$

since this differential must be exact, the following relation must hold

$$\left (\frac{\partial X}{\partial P} \right )_V=\left (\frac{\partial Y}{\partial V} \right )_P$$

We can find simple functions satisfying this condition
$$X=cP\text{ and }Y=cV$$
where $c$ is a constant that we may choose to be $c=1/R$ for later convenience.  Substitution of these functions in the total differential gives
$$\mathrm{d}T=\frac{1}{R}\left (P\mathrm{d}V +  V\mathrm{d}P\right )$$
and integrating on both sides from 0K to T yields
$$T-T_0=\frac{1}{R}\left [(PV)_T-(PV)_{T_0} \right ]$$
and since $T_0=0$ and $PV=0$ at $\pu{0K}$ we arrive at the ideal gas law.


Why does H2O2 have a gauche structure? The gauche structure of $\ce{H_2O2}$ is quite well known. It has the two $\ce{OH}$ groups in different planes at right angles. Why does it adopt this structure?

I can understand that it might have something to do with the repulsions between the lone pairs on the oxygens. But in that case a staggered structure would make more sense since that minimises the repulsion even further. Maybe another factor is at play here that I am missing.

EDIT: My main question in this post is why the gauche structure arises in the first place. However, the other post is mainly focussed on the variation of the dihedral angle in different phases. Although the OP asks why hydrogen peroxide adopts a gauche structure, a satisfactory answer has not been provided. (Note that the OP has accepted the provided answer. Since that does not answer my main question, I think a new question is justified in this regard)

In liquid or gas and even in solid hydrogen peroxide has a gauche structure. One reason for this is to reduce electronic repulsion and the other is to give a well packed structure in solid phase.also the interactions between different molecules leads to a gauche structure in liquid and gas.


How to dock aggregated structures comprised of "elementary" protein units like LEGO pieces? There are proteins with intrinsic symmetries. For example:

I was wondering how to use transformations such as: rotations, translation, replications to construct possible structures using 2 types of proteins.

Assume I found a binding site of the 2 proteins using Molecular Dynamics simulators.

For example the above one and this one:

Can I use group theory way to find possible configurations of the 2 proteins?
How can I use the tensor product of the symmetry groups ($G_1 \times G_2$) to find possible structures?
What graphic/framework would you recommend do perform all the manipulations on the proteins?
Is it possible to use the transfer to create some algebraic structure that will help to construct a mechanical configurations comprised of proteins?
Is there any tool from Baker's Lab I can use for this task?
Amy recommendation for a tool I could use to replicate, rotate and shift structure would be appreciated us well.

TL;DR:

My goal is to use group theory or Monte Carlo simulation driven by group theory insight to create macro molecular structures using multiple (dozens) instances of the two proteins (the LEGO building blocks).
Any idea that relates symmetry, transformations, group theory to create structures is much welcome.

Clarification:

Given a binding point between 2 proteins, which software can I use to create the dimers into macro-structures like in LEGO namely only by replicating, translating and rotating the Lego pieces?

Try DNA origami instead

While protein assemblies are well known, rationally using proteins to assemble larger supramolecular structures is still a difficult problem:

Predicting protein folding is an unsolved problem. Sometimes even small modifications to a sequence can change the folding propensity.
Predicting protein-protein interactions is often difficult, due to challenges in understanding protein dynamics, intermolecular interactions, and accurate predictions of protein electrostatics.

There has been some progress in terms of forcing particular protein-protein assembly through intentional side-chain modification and cross-linking, and reactive end-groups (e.g protein-based metal-organic frameworks (MOFs).

That said, use of proteins for supramolecular assemblies is far harder and less prevalent than using DNA assemblies. At the moment, DNA origami is far superior for these purposes.

Edit: If you're really looking to just copy and displace atoms, I don't think you need any particular software. I'd write a Python (or other script) that reads in the PDB files, and writes out duplicate atoms displaced in the XYZ directions as needed. Group theory can help predict the angles, but you'll still need to know the protein size to work out the geometry.


IUPAC Nomenclature of Cumene Why is cumene referred as isopropyl benzene? And not sec propyl benzene, since the carbon of that chain is the middle one?

Let me abstract your question to “Why is the $\style{font-weight:bold}{\ce{-CH(CH3)2}}$ group sometimes named ‘isopropyl’ but almost never ‘sec-propyl’?”

iso alkyls

‘Iso’, obviously as in ‘isomer’, in IUPAC organic nomenclature (last one is from 2013) evolved to non-preferred, retained names for several smaller alkyls, that are linear but specific single branching at the carbon next to the terminal carbon, by additional methyl.

These are (intentionally skipping our dour discussed $\ce{C3}$ branched alkyl):

isobutyl    $\ce{-CH2-CH(CH3)2}$
isopentyl   $\ce{-CH2-CH2-CH(CH3)2}$
(isohexyl    $\ce{-CH2-CH2-CH2-CH(CH3)2}$  , only in old IUPAC 1979 nomenclature, see here)

(However, in general naming of bulk chemical products, larger alkyls with different branching are named isoalkyl, e.g. ‘isooctyl’ $\ce{-CH2-CH(CH2CH3)-CH2-CH2-CH2-CH3}$.)

sec- alkyls

Obviously ‘sec-’ is for 'secondary’, meaning there's a branching into 2 carbons at the first carbon atom. The list of retained names has only one item:

sec-butyl $\ce{-CH(CH3)-CH2-CH3}$

but our candidate would fit in here as well.

Why only sec-butyl? A major reason could be that the next “sec-pentyl” would represent two different structures:

$\ce{-CH(CH3)-CH2-CH2-CH3}$
or $\ce{-CH(CH2CH3)2}$ ?

(Now one might ask, why there is not some naming system for alkyls with methyl branch at the first atom, similar to iso alkyls with the branching at the next-to-last carbon. I don't know.)

Why not sec-propyl? The $\ce{-CH(CH3)2}$ is a special case that can fall into both isoalkyl and sec-alkyl class. The better defined one, also resulting with simpler spoken name and simpler typography (no dash and italic) – isoalkyl, was once chosen. That's my humble hypothesis.

(And please note that the preferred IUPAC name for this group is propan-2-yl. Alternate name is 1-methylethyl. Isopropyl is still retained.)


Given the mass of precipitate, how can I calculate the relative amount of the original dissolved compounds? A simple mixture of $\ce{NaCl}$ and $\ce{NaBr}$ weighing $0.180~ \mathrm{g}$ is treated with $\ce{AgNO3}$ solution to give $0.3715~ \mathrm{g}$ of precipitate. Calculate the content of $\ce{NaCl}$ and $\ce{NaBr}$ in the mixture.

The answer given in book are $\ce{NaCl}$ = $0.0682~ \mathrm{g}$ and $\ce{NaBr}$ = $0.1118~ \mathrm{g}$.

The answers I got are $0.1519~ \mathrm{g}$ $\ce{NaCl}$ and $0.0281~ \mathrm{g}$ $\ce{NaBr}$. Please tell me where I made mistake.

Working:

$$\ce{NaCl + AgNO3 -> NaNO3 + AgCl}$$
$58.5~ \mathrm{g}$ $\ce{NaCl}$ give $143.5~ \mathrm{g}$ $\ce{AgCl}$

$0.3715~ \mathrm{g}$ $\ce{AgCl}$ give $58.5/143.5 \cdot 0.3715~ \mathrm{g}$ $\ce{NaCl}$ = $0.1519~ \mathrm{g}$ $\ce{NaCl}$

$\ce{NaBr}$ = $0.180 ~ \mathrm{g}$ - $0.1519~ \mathrm{g}$ = $0.0281~ \mathrm{g}$ $\ce{NaBr}$

x = number of moles of chloride
y = number of moles of bromide

Since you know the mass of the original $\ce{NaCl}$/$\ce{NaBr }$mixture and the molar masses of $\ce{NaCl}$ and $\ce{NaBr}$, you can write an equation of the form:

$58.44 \pu{\frac{g}{mol}}~ x + 102.89 \pu{\frac{g}{mol}}~ y = 0.180~ \pu{g}$

Based on this idea, you can write a similar equation for the final products $\ce{AgCl}$ and $\ce{AgBr}$. With that in hand you will then have a system of equations you can solve (for x and y) that can be converted back into the masses of those salts.


Why is Electrochemistry often the method of choice for determining Ksp values? I don't understand the relationship they have with one another. Please, someone explain.

Perhaps a simple example will illustrate how a solubility product constant, $\ce{K_{sp}}$, may be determined via measurement of an appropriate voltaic cell potential. Consider the estimation of the $\ce{K_{sp}}$ value for the following equilibrium:

$$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq) \quad K = K_{sp}} \tag{1}$$

This equilibrium lies far to the left because AgCl has very poor solubility in water and, as a direct consequence, the concentrations of silver ions and chloride ions are very low. Therefore, the $\ce{K_{sp}}$ value is small, i.e., much less than unity, and a direct measurement of the $\ce{K_{sp}}$ value is difficult due to the low solubility of AgCl.

However, we have the following two half-reactions, with their standard reduction potentials (from standard tables):

$$\ce{Ag+ +  e−  <=> Ag(s)} \quad E^\circ_\mathrm{red} = \pu{+0.7996 V} \tag{2}$$

$$\ce{AgCl(s) + e− <=> Ag(s) + Cl−} \quad  E^\circ_\mathrm{red} = \pu{+0.22233 V} \tag{3}$$

Half-reaction (2) has the more positive standard reduction potential, hence more negative $\ce{\Delta G}$, so it will be the spontaneous reduction half-reaction at the cathode in a voltaic cell that can be constructed. Reversing half-reaction (3) yields the oxidation half-reaction at the anode in the constructed voltaic cell:

$$\ce{Ag(s) + Cl− <=> AgCl(s) + e−}  \tag{4}$$

So the voltaic cell notation is

$$\ce{Ag(s)|Cl^-(aq;\pu{1 M})||Ag^+(aq;\pu{1 M})|Ag(s)}$$

and the net cell reaction is obtained by adding half-reactions (2) and (4), resulting in

$$\ce{Ag+(aq) + Cl-(aq) <=> AgCl(s) \quad K = 1/K_{sp}} \tag{5}$$

This equilibrium lies far to the right because it is simply the reverse of half-reaction (1). Therefore, the $\ce{K}$ value is large. Indeed, it is simply the reciprocal of half-reaction (1)'s $\ce{K_{sp}}$ value. This is consistent with the fact that mixing aqueous solutions containing silver ions and chloride ions immediately results in spontaneous formation of AgCl solid that precipitates out of solution.

The Nernst equation, where $\ce{Q}$ is the reaction quotient, is

$$ {E_\mathrm{cell} = E^\circ_\mathrm{cell} - \frac{R T}{n F} \ln(Q) } \tag{6}$$

The natural logarithm term is basically a correction term that takes into account the fact that voltaic cells are usually not at standard state, i.e., they are not dead yet. This is accomplished using $\ce{Q}$. At standard state, $\ce{Q} = 1$, and at equilibrium, $\ce{Q} = K$, where $\ce{K}$ is the equilibrium constant. In the present case, $\ce{K}$ pertains to equation (5).

Now, at equilibrium (and ignoring units), $\ce{\Delta G_{cell} = 0, E_{cell} = 0 and Q = K}$, so

$$E^\circ_\mathrm{cell} = \frac{R T}{n F} \ln(K) = {E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} = \pu{+0.7996 V} -(\pu{+0.22233 V}) = \pu{+0.5773 V}} \tag{7}$$

Note that the standard cell potential is defined as

$$E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} \tag{8}$$

With $\pu{T = 298.15 K}$ and $\pu{n = 1}$, $\pu{RT/nF \approx 0.02569 V}$, so equation (7) can be solved for $\ce{K}$:

$$\mathrm{K = e^{nfE^\circ _{cell} /RT} \approx e^{+0.5773V/+0.02569V} = 5.74 \times 10^{9} \tag{9}}$$

Hence $$\pu{K_{sp} = 1/K = 1.74 \times 10^{-10}} \tag{10}$$
which is close to typically quoted values in the literature.


Magnesium powder mixed with CuSO4 solution $\ce{Mg}$ powder is stirred into $\ce{CuSO4}$, what is the balanced molecular equation for this reaction ? I have $\ce{Mg + CuSO4 -> Mg2SO4 + Cu}$. Also, is the ionic equation the same as the net ionic equation? I believe it is not, because Copper would act as a spectator ion, however, I am not too sure.

Your equation is very close, but you it is not $\ce{Mg2SO4}$. $\ce{Mg^{2+}}$ has a charge of $2+$, and $\ce{SO4^{2-}}$ has a charge of $2-$. They combine in a ratio of 1:1, so the result is

$$\ce{Mg + CuSO4 -> MgSO4 + Cu} ~ .$$

The ionic equation does not include the $\ce{SO4^{2-}}$, and in the case of this equation the net ionic equation is the same as the ionic equation. Ionic equations look at what happens to electrons.
In this case, electrons are being transferred from the magnesium to the copper (since the copper is bonded to $\ce{SO4^{2-}}$ it is in the form $\ce{Cu^2+}$).

$$\ce{Mg + Cu^{2+} -> Mg^{2+} + Cu} ~ .$$

Since the charge on the $\ce{SO4^{2-}}$ doesn't change, it's just a spectator ion.


Why is the total number of electrons in a neutral carbon atom different from the number of dots in a Lewis Dot structure for carbon? Lewis Structure has 4 but the neutral atom has 6. why?

Only the valence electrons are represented in a Lewis structure. The inner electrons are not important for the bonding, therefore not displayed.


How to create liquid carbon, and what kinds of properties does it have? Carbon sublimes at about 3915K. In a high pressure environment, shouldn't it be possible to create liquid carbon at high temperatures? Has this ever been done? What does it look like and what properties does it have?

It's possible to create liquid carbon, but it's extremely hard. As you can see on the Carbon phase diagram  if you want to obtain "easely" liquid carbon you must do it at $\ce{1 GPa}$ and $\ce{5000 K}$.

And for the properties I really don't know but carbon follows the octet rule and it might be very reactive as an electrophile because it will just have four electrons around it.

I hope it can help you a bit, i'm seeking in some data-bases but there are not really things to answer your question.

Have a nice day.


Why does hybridization produce a more stable configuration? My text, and the other sources I've checked, include information on the effects, and kinds, of hybridized orbitals; however, they do not explain what properties of hybridized bonds conduce greater stability.

Why does hybridization produce a (lower energy) more stable configuration?

What causes orbital hybridization?

Hybridization only occurs in response to bond formation, that is an atom will not just spontaneously hybridize on its own.  Further, an atom will only hybridize if hybridization leads to the formation of a more stable (lower energy) molecule.

Nature is very efficient at exploring the energy surface and finding whatever energy minimum is available.  For example, ammonia ($\ce{NH3}$) is roughly a tetrahedron with $\ce{H-N-H}$ bond angles around 109.5° (the tetrahedral angle) using (roughly) $\mathrm{sp^3}$ hybridized orbitals. For ammonia, mixing (hybridizing) it's $\mathrm{2s}$ and three $\mathrm{2p}$ orbitals to form four $\mathrm{sp^3}$ hybridized orbitals that are used to form three $\ce{N-H}$ bonds and one lone pair orbital produces a more stable molecule than if it were not hybridized.  On the other hand, the analogue phosphine ($\ce{PH3}$) chooses to remain roughly unhybridized, using p-orbitals to form its $\ce{P-H}$ bonds with a resultant $\ce{H-P-H}$ bond angle around 90°.  Bond strengths, steric factors, etc. all play into determining whether a hybridized or unhybridized structure will be the more stable.


Negative gradient graphs So basically I have a graph of Absorbance on y-axis against Time on x-axis and it is a negative curve. So this will give me a negative slope gradient. My question is how to reason the negative gradient. Do I write down the rate of concentration as a negative value and just reason it by saying that by being negative it means that the concentration is decreasing?

Yes. Beer's law tells us that the absorption is the product of path length, extinction and concentration. Thus concentration is absorption divided by the product of path and extinction. The rate of change of concentration (first derivative with respect to time of concentration is the rate of change of absorption divided by the product of path and extinction. If the rate of change of absorption is negative the rate of change of concentration is negative. Concentration is decreasing.


What are the units that measure free energy? Is it $\pu{kJ/mol}$? I have done problems and I always end up with these units, but is this correct?

What are the units that measure free energy?

There are several different free energies, but I'll tackle Gibbs and Helmholtz only here:

Gibbs free energy: $G=H-TS$, for $H=U+PV$. Hence, by dimensional analysis, Gibbs free energy has the SI units of Joule.
Helmholtz free energy: $A=U-TS$. Again, by dimensional analysis, Helmholtz free energy also has the SI units of Joule.

Since both are only "free energies", it makes sense to assign them the SI unit Joule.

That said, however, almost everywhere they are specified in terms of $\pu{kJ/mol}$ because that makes the comparison across different substances easier. Since $G$ is an extensive function, its value can vary even with the amount of substance. Your calculated $\Delta G_\mathrm f^\circ$ can differ from mine even if we take the same substance, just because we had the amounts different. Hence, to ease comparison and computations, they are "specified" in terms of $\pu{kJ/mol}$, although their correct SI unit is Joule.


Predicting reaction among compounds Suppose I have a group of organic compounds say A, B, C, D, E, F, G, H. I know their chemical formula and their names. Now there exists a reaction among these compounds but we do not know which compounds would react to give which products. It may be that A and B react to give C and D or B and D react to give F and H. If we are given these compounds, is it possible to identify a reaction that will occur among any number of these compounds to give the product which would belong to this set of compounds?

Note: If you have not understood any part of this question, please ask me before down-voting, I will edit the question and try to re-explain. Thank you. ALso I am a computer science student, so I dont have too deep knowledge in chemistry, so do help me.

I think your question relates to the field of combinatorial chemistry. This is an important area especially for pharmaceutical companies that want to try out lots of different structures to see if they have therapeutic potential.

Lets talk specifics. In the case of known chemistry - e.g. formation of esters from alcohols and carboxylic acids, you can identify the possible reactions. Start with a system where A, B, C, and D are different alcohols and E, F, G, and H are different carboxylic acids. Under the right conditions (acid catalyzed dehydration), the alcohols can combine with the acids all possible ways to give 16 different esters (AE, AF, AG, ... etc.). But A (an alcohol) will not combine with the other alcohols B, C, or D -- and acid E will not combine with acids F, G, or H. Under more harsh conditions, however, they could be induced to react within each group (to form ethers or acid anhydrides).

If you are talking about a random selection of compounds - especially ones with more complex structures and more functional groups, and a variety of reaction conditions, the outcome would be much less predictable.


How to calculate the charge distribution at the ortho/para or meta sites of a substituted benzene ring I've modelled a few substituted benzene rings in avogadro and retrieved the partial charges of each atom; I've tried to sum the partial charges of the carbon and hydrogen atoms at the ortho/para and meta sites to see the overall partial charge at that site but results don't make sense as the para site always ends up being 0.

How can I calculate the charge at the sites?

Summary: The method used does not handle π-conjugation

The partial charges in Avogadro (and many other software programs) are assigned by default by the Gasteiger-Marsili scheme†

Importantly, while the Gastieger charges are fairly good for atomic partial charges of organic-ish molecules, they're unlikely to reproduce effects in conjugated systems very well.

  A method is presented for the rapid calculation of atomic charges in σ-bonded and nonconjugated π-systems. Atoms are characterized by their orbital electronegativities. In the calculation only the connectivities of the atoms are considered. Thus only the topology of a molecule is of importance.

My suggestion is to use some type of quantum chemical calculation, e.g.:

NWChem
OpenMOPAC
ORCA
(etc.)

Even semiempirical quantum chemical methods (e.g. PM7) will allow you to assign charges based on the quantum electrostatic potential around the molecule. They'll be much more accurate at the trends you want.

† "Iterative Partial Equalization of Orbital Electronegativity - A Rapid Access to Atomic Charges," Tetrahedron, Vol. 36, pp. 3219-3228, 1980.


Liquid which neither dissolves in water nor oil I am looking for a liquid which is not miscible (or very close) with water or vegetable oil.  Ideally, I want to have a container which shows three layers (the three liquids) with clear boundary layers.

I was thinking that I should find a non-polar compound of very high molecular weight - such that difference in the density of that compound and oil is enough to draw it down out of the oil.  However, I haven't found it yet.

I would prefer a compound which is non-toxic and non-flammable if possible.

It feels like such a simple question, I must be missing something.

Fluorocarbons are compounds that are hydrophobe and lipophobe. It's a special property of perfluorinated compounds: They are non-polar and thus hydrophobic but in addition as the high electronegativity of fluorine reduces the polarizability of the atom, fluorocarbons are only weakly susceptible to the fleeting dipoles that form the basis of the London dispersion force. As a result, fluorocarbons have low intermolecular attractive forces and are lipophobic.
I'm not sure about their toxicity, but this seems to be an issue, e.g. here.


How do I determine the amount of Calcium in a substance using EDTA titration? I need to determine the amount of calcium in a substance (that also contains Magnesium) using EDTA titration. From what I understand, this can be done by adding Eriochrome T indicator to the solution, then titrating it with EDTA. The solution will start out Red then turn blue as the Ca ions bond with the EDTA. Many of the labs I search up online seem to follow these procedures. However, doesn't Magnesium also bond with EDTA? Wouldn't that affect my calculations since I'd be measuring Ca AND Mg as they both bind with both Eri T and EDTA?

Thanks

Yes, EDTA will chelate to both calcium and magnesium. However, since the complex formation constant for $\ce{Ca^2+}$ is significantly greater than that for $\ce{Mg^2+}$ ($\log \beta_{f,\,\ce{Ca^2+}} = 10.65$; $\log \beta_{f,\,\ce{Mg^2+}} = 8.79$), we are justified in assuming that EDTA preferentially chelates to $\ce{Ca^2+}$ and only chelates to $\ce{Mg^2+}$ when no $\ce{Ca^2+}$ is available.

Remark. How significant is "significant"? In principle we could avoid the approximation and solve for the system analytically, but then we would still have error from (i) imprecision in visual determination of endpoint, (ii) imprecision in determination of formation constants, and so on.


Can compounds found in concrete leach into soil against gravity? I spent a while trying to figure out which stackexchange site in which to post this question, forgive me if I got it wrong.

A number of scholarly articles indicate that concrete can leach calcium compounds under the right conditions. Could they ever wick upwards against gravity through dry or water-saturated soil?

An example would be a raised bed of soil with wooden sides and a concrete floor.

P.S. I also wonder about plastics leaching upwards, but this is admittedly most likely a different question.

Gravity is a small force on ions when compared to water flow and diffusion; lead ions won't sink in a graduated cylinder. But rather than ponder the drift upwards of ions from the concrete, consider the drift downwards of ions from the soil. For instance, the soil may be acidic, and will almost always have more water in it than the concrete has, so the acids in the soil will be much more mobile than the ions in the concrete, and will fairly easily drift into the concrete, neutralizing $Ca(OH)_2$ or even dissolving $CaCO_3$. Whether these calcium ions will travel upwards... well, yes, but  calcium compounds may adsorb on soil particles and travel even slower. So the soil may be as much a barrier as the concrete is a poor source of ions.
Concrete may lose some ions (especially $Ca(OH)_2$) in contact with flowing water, but even more, it tends to pick up $CO_2$ and the $Ca(OH)_2$ becomes $CaCO_3$, which is less soluble. So the soil right next to the concrete may be a little more alkaline than the soil at the top surface.


Reduction of chlorate with phosphorus Given
$$\ce{10KClO3 + 3P4 -> + 3P4O10 + 10KCl}$$

If 52.9 grams of $\ce{10KClO3}$ reacts with excess $\ce{P4}$, what mass of $\ce{3P4O10}$ could be produced? How many moles of $\ce{KCl}$ could be produced?

I am not sure if I have to calculate the excess mass of $\ce{P4}$ and then find the mass of $\ce{3P4O10}$ or if it is just telling me that $\ce{P4}$ is excess reactant. If it is calculating the excess mass, how would I do it if I do not know the original mass of  $\ce{P4}$?

The number of moles produced for each product will be the number of moles times of the limiting reagent times the reaction stoichiometry (the coefficients).

So to solve this you determine the number of moles of $\ce{KClO3}$ and use that as the base for determining the amount of $\ce{P4O10}$ and the moles of $\ce{KCl}$ produced.  Then multiple by the molecular weight per usual to get your answer.


Which it is the oxidation state of indium and phosphorus? I would like to know what the oxidation state of indium and phosphorus in a structure type rock salt. It could be +1 and -1? Yes or not? and why?

Thanks

I've estimated net atomic charges in rocksalt type InP by simple HF calculation in CRYSTAL06 in random basis.
Lattice parameter of 5.54 A is taken from:

K Kabita et al. First principle study on pressure-induced electronic structure and elastic properties of indium phosphide (InP). Indian J. Phys. (2015), DOI 10.1007/s12648-015-0701-0.

Results: http://pastebin.com/zdb8js8Z

Total atomic charges are 47.2861139, 16.7138861, so partial charges are:

In: +1.71
 P: -1.71

Given that oxidation state is an hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, the answer is probably better estimated as +2/-2, not 1.


What is the order of acidic strength among phenol, catechol, resorcinol and hydroquinone? This question is from a book that gives the answer as

$$
\text{hydroquinone} < \text{resorcinol} < \text{phenol} < \text{catechol}
$$

but does not provide any explanation for the answer. If I go by the stability of the conjugate base due to resonance, phenol seems to have the most number of resonance intermediates and seems to be most stable. Considerations about number of resonance intermediates of conjugate base gives an order that is nowhere close to that provided as an answers.

Please provide an explanation as to the provided answer or, if the answer is wrong, please supply the correct order with explanation.

A quick search on says that resorcinol is more acidic that catechol, followed by phenol.

Hydroquinone - strong +M effect and weak -I effect

Phenol - the normal resonance you'd expect in all phenols

Resorcinol - -I effect

Catechol - +M effect, stronger -I effect and hydrogen bonding

Taking these factors into account, it is easy to see why hydroquinone is least acidic. Hydrogen bonding and the vicinity of the $\ce{-OH}$ to the anion (for the stronger -I effect) make cactechol more aciic than the phenol despite its +M contribution. Resorcinol has only -I effect and no counteracting electron-donating effects, which makes it the most acidic.


Why does the reduction of NAD+ to NADH change the absorbance so much, and cause fluorescence? This is an updated question:
I don't have a very strong background in biophysics, but I want to understand the theory behind $\ce{NAD+}$ and $\ce{NADH}$ absorbance and fluorescence.

Background: I understand that the additional hydrogen that $\ce{NAD+}$ gains to form the reduced $\ce{NADH}$ results in the molecule being able to absorb light at $\pu{340nm}$. And that only the reduced $\ce{NADH}$ is able to produce fluorescence. This is also the property which is explored when studying, e.g., enzyme kinetics - if the enzyme binds $\ce{NADH}$.
Question: The oxidized form, $\ce{NAD+}$, still has an aromatic ring - why is it that it is not able to fluoresce? Why, in a nutshell, is the transfer of the hydrogens causing fluorescent in $\ce{NADH}$ and such a big change in absorbance spectrum?
Also, perhaps a bit naive question, but why is it fluorescence better to measure than absorbance?
I have tried to look for articles (the old and original) that explores these properties, but I can't seem to find any that purely speak about the spectroscopic properties.

Essentially NADH has an additional pair of electrons. These additional electrons which are oftein drawn as a lone pair of electrons alters the molecule in a similar way to the way in which many pH indicators change upon protonation.

Here are diagrams of the reduced and oxidized forms of NADH / NAD+

A typical indicator when protonated appears to have a colour, however upon deprotonation the indicator loses a hydrogen and gains a lone pair of electrons now appearing to have a different colour. The addition of the lone pair of electrons actually lowers the transition energy between the pi bonding and the pi anti-bonding states.

For example for methyl red it can be predicted that the unprotonated form has a HOMO/LUMO gap of 2.175 eV (according to extended Huckel calculations). For the protonated form of methyl red has a HOMO/LUMO gap of 3.275 eV.

This lowered energy correlates with a decreased absorbance energy, ultimately reflecting a higher energy wave of light (Frequency of Blue > Red). NADH's absorbance wavelength his greater than NAD+'s absorbance wavelength, because NADH is absorbing lower energy light. This, like pH indicators, is due to the lone pair lowering pi and pi anti-bonding energy state differences.


Are such high oxidation states as reported by Vojovodic et al. possible? I am a student in theoretical chemistry and I am confused about the paper: Trends in adsorption of electrocatalytic water splitting intermediates on cubic $\ce{ABO3}$ oxides (Montoya, J. H.; Doyle, A. D.; Nørskov, J. K.; Vojvodic, A.; Phys. Chem. Chem. Phys. 2018, 20 (5), 3813–3818. DOI: 10.1039/C7CP06539F), where the authors report DFT calculations on the oxides:

$\ce{MgBaO3}$ ($\ce{Mg^3+}$ and $\ce{Ba^3+}$)
$\ce{NaLaO3}$ ($\ce{La^5+}$)
$\ce{CaBO3}$ and $\ce{SrBO3}$ ($\ce{B^4+}$)
$\ce{ZrNaO3}$ ($\ce{Zr^5+}$)
$\ce{ScSrO3}$ ($\ce{Sc^4+}$)
$\ce{CaCuO3}$ ($\ce{Cu^4+}$)

... and many more. Are these oxidation states even possible?

Oxidation states are just numbers, a bookkeeping tool for chemistry. These hardly ever correspond to anything observable. Point in case: hypofluorous acid $\ce{HOF}$, see for example the oxidation state of oxygen and of fluorine in the confines of the definition. Maybe also of interst in that regard is the introduction to "oxidation state"/"oxidation number" in general.

Therefore instead of asking whether the oxidation states are stable, one should ask if the bulk/ molecular structure is stable.

When it comes to computationally aided catalyst design, you can pretty much use the entire periodic table to play with. It is a matter of interpreting the results that count, and obviously if experiment can reproduce such predictions. Where you start becomes a matter of taste and starting with perovskites is as good a guess as any.

The authors of the publication make quite an effort, calculating hundreds of potential catalysts. However, they also indicate problems within their methodology. Quoting from the supporting information:

  3 Oxygen evolution data
  [...]
  \begin{array}{lcl}
\text{formula} & \text{values of }\Delta G, \eta, \dots & \text{warnings}\\
\hline
\ce{MgBaO3} & [\cdots] & \text{a,b}\\
\ce{NaLaO3} & [\cdots] & \text{a,b}\\
\ce{CaBO3}  & [\cdots] & \text{b}\\
&\vdots&\\
\hline
\end{array}
  a: One or more runs did not converge or failed in Quantum Espresso, missing adsorbates are reconstructed from successful runs using scaling
  b: Changes in atomic positions of greater than 5.5 angstrom found in the slab or adsorbate during optimization, typically indicates structural instability
  [...]

And there are a lot more unstable species.

TL;DR: Basically, in these extreme cases, where you would observe abnormal oxidation states, the calculation also predicted that the bulk/ molecular structure is not stable.


Reaction of an alkyl halide with silver nitrite I want to know that whether both alkyl nitrite and nito alkane are produced or if nitroalkane is the only product. There are many different (contradicting) things written in different books that is confusing me.

On treating ethanolic solution of haloalkane with silver nitrite ($\ce{Ag-O-N=O}$), nitroalkane is formed because since the bond between $\ce{Ag-O}$ is covalent, the lone pair on nitrogen acts as an attacking site for nucleophilic substitution.
$$\ce{R-X + AgNO2 -> R-NO2 + AgX}$$
But on the other hand if haloalkane is treated with potassium nitrite ($\ce{KNO2}$), alkyl nitrite is formed as major product because since the bond between $\ce{K-O}$ is ionic in nature, the negative charge on oxygen serves as an attacking site.
$$\ce{R-X + KNO2 -> R-O-N=O + KX}$$


Reactivity comparison of phenol and toluene towards electrophilic substitution Which aromatic compound is more reactive towards electrophilic substitution?

  benzene
  chlorobenzene
  phenol
  toluene

  Correct answer: phenol

But, according to me, the answer should be toluene.

My reason: Toluene is an ortho-para directing group as well as an electron-donating group which increases the electron density at ortho-para position and activates the aromatic ring towards electrophilic substitution.

Where I am wrong?

You have to make a comparison between phenol and toluene. You have identified what toluene does, but you haven't described what phenol does.

The hydroxyl group in phenol increases electron density in the ring by resonance (+M effect). Note that the methyl group in toluene increases electron density only by hyperconjugation and inductive effect.

Resonance effects are generally far more superior to hyperconjugation. Hence, without doubt, phenol is more activated towards EAS as compared to toluene.


Is the kinetic energy turning into activation energy when the reaction takes place? So I want maybe a simple question that I had on a test today. For a reaction to take place it needs a minimum energy called activation energy. At school I heard that this energy is coming from the kinetic energy of the particles that collide. When the reaction takes place, is the kinetic energy small because it turned into activation energy?
If you need any clarification just ask me and sorry for any misspell or grammar error.

You got it right: when the reaction takes place, the kinetic energy of the reacting particles turns smaller because some of it is (sort of) spent on activation energy. It is just that the "when" in question is incredibly short.

Imagine two tennis balls flying at great speed and hitting each other head on. Sure enough, they will bounce off each other and fly in the opposite directions with the same speed. Some time into the collision, there must be an instant when they are not flying anywhere; their speed is 0, and so is kinetic energy. All energy they possessed has converted into the potential energy of elastic deformation, only to return into kinetic energy again really soon.

Same thing with molecules, except they are way smaller and fly way faster. Also, there might or might not be a reaction, and if it happens, then the kinetic energy of the molecules after the collision will not be the same as before. It may be greater, or it may be smaller; that's how exothermic and endothermic reactions happen.

So it goes.


What would happen if someone consumed a Silica Gel Packet? Most packaged goods have the tiny Silica Gel Packets packed along with a clear warning on every single packet.

My question is:
What hazards of desiccant silica gel are they concerned about that it is necessary to post the warning "THROW AWAY DO NOT EAT" on the packaging?

Essentially you need to understand that the silica gel packet is just $\ce{SiO2}$. These packets are used to absorb moisture (water vapor) to protect goods from spoiling.

To be honest with you here, nothing ever happens when you eat $\ce{SiO2}$. You consume it all the time from food and water. But, the reason these packets are so dangerous is because they might contain impurities such as $\ce{CoCl2}$ (blue color) and $methyl~violet$. These colored products are used to indicate moisture. These products are carcinogenic and are mutatic poisons. In most cases, it will not be fatal immediately, but given time, yes.

Generally, if the silica gel is colored, it is highly likely that they might be very dangerous. So don't eat them. :)

On top of this the silica beads tend to fracture so you have essentially consumed broken glass which will be quite painful as it tears up your digestive tract causing you internal bleeding.

Just to add a couple references: 1, 2, 3.
I don't know how far the the second and third ones are true but the first one is the most accurate.


Mercury, amalgams and mercury compounds Mercury is toxic. It forms amalgams and inorganic mercury compounds/salts are available.

My question is on toxicity. I am interested in its toxicity, not the mode of action (as it will be part of biology SE)

Is amalgam toxic? (sort of yes / no)
Mercury salts are toxic? (sort of Yes / no)
If mercury salts / amalgams are non-toxic, what makes them non-toxic. Is it because of the ionic nature / other metal or any other reason?

Toxicities are often greatly influenced by how you are exposed and in what dose.
Mercury is a fascinating metal. You can find it as the pure metal, a salt or in an organic form. In its standard state it is a liquid and is rather volatile. Large amounts are naturally released into the atmosphere, but humans contribute as well, for example: burning fossil fuels, such as coal, will release it into the atmosphere.
The forms of mercury are not set in stone it is quite able to switch forms from the free metal, to the salt, to the organic by processes in nature.
Orally consuming some liquid metal isn't the worst possible route of exposure; mercury was at one time even used as a laxative. The only absorption in the GI tact is thought to be due to mercury vapor. Skin exposure is undesirable, but we know that inhaling the vapor is more toxic because it is soluble in lipids. Upon entering the bloodstream, some makes it way past the blood/brain barrier (hence its neurotoxicity) and some is converted to Hg(II) by catalase.
Mercury's inorganic salts have a long useful history where some were used to treat syphilis, some for lights and some for hat making and so on. "Mad as a hatter" won its name due to chronic exposure to Hg(NO$_3$)$_2$, which was used in hat making. This salt isn't very lipid soluble, but more absorption does occur when swallowed than the free metal. These salts tend to build up in the kidneys after absorption and high exposure will cause renal failure. Not much will pass through the blood/brain barrier, but chronic exposure will eventually allow some across and cause neurotoxicity. They also tend to bind with proteins containing sulfur and this can cause some cell death. As a general rule, Hg(I) salts are less toxic than Hg(II), presumably due to the decreased solubility of Hg(I). This does not mean they cannot cause harm: mercuric chloride was used in a factory and the wastes were discharged into a nearby sea, anaerobic bacteria methylated the mercury, this methylmercury was absorbed by fish that were later eaten and hurt many people.
As an organic, mercury compounds tend to be considerably more lipid soluble and therefore can readily be absorbed or cross the blood/brain barrier, in comparison to the free metal. A terrible example is dimethylmercury, which claimed the life of an inorganic chemist who was accidentally exposed to some she was using in an NMR experiment.
I do not know much about its amalgams aside from gold/silver. I believe most of the toxicity was seen in gold refineries due to the fact that the amalgam is heated to remove the mercury, which leaves behind gold, and results in the release of mercury vapor.


Proton NMR splitting in 2-substituted furan I have recently run proton NMR on 1-(furan-2-yl)ethan-1-ol I synthesised in the lab.

My 400 MHz spectrum consists of the following peaks:

7.36 ppm (dd, J = 1.9, 0.9 Hz, 1H)
6.31 ppm (dd, J = 3.3, 1.9 Hz, 1H)
6.22 ppm (dt, J = 3.2, 0.9 Hz, 1H)
4.87 ppm (q, J = 6.6 Hz, 1H)
1.53 ppm (d, J = 6.6 Hz, 4H).

For the peak at 6.22 ppm, I am wondering why a doublet of triplets is observed rather than a doublet of doublets. I understand the long-range splitting that results in the dd, but am just a little surprised by the dt.
I attach a picture of the doublet of triplets.

I agree with Buck Thorn's comment. I suspect the coupling in the quartet is just not fully resolved. I'm not 100% sure why, but one possible reason is because there is further unresolved coupling with the OH peak: instead of being resolved as a splitting in the spectrum this just broadens the individual lines. Or it could just be that that proton relaxes faster.
You may be able to play around with window functions (or apodisation as it's referred to in Mestrenova) to try and tease out this small coupling. My suggestion would be: try enabling/selecting both "Gaussian" and "Exponential", set the exponential value to -2 Hz and the gaussian value to 0.2 GF, or something similar. See if the quartet is still a quartet. [From the comments, it seems that this has worked and that the quartet is actually a quartet of doublets under the hood.]
The window functions are basically mathematical functions that you multiply the FID by prior to Fourier transformation. Peaks in the FID generally have a damped sinusoidal form. The faster a signal decays to zero, the larger the corresponding linewidth will be, i.e. the broader the peak will be.
The two "big" window functions for 1D processing are exponential and Gaussian window functions. I won't go into too much detail because these are very well-described elsewhere. See e.g. Chapter 5 of Keeler's Understanding NMR Spectroscopy (2nd ed.), or Glenn Facey's blog posts on Exponential Line Broadening and Resolution Enhancement (which also include some helpful pictures). Googling for either "window function nmr" or "apodisation nmr"  will also throw up a huge number of documents.
The exponential one basically multiplies the FID by another decaying exponential. This is largest at the start and decreases as time passes, so effectively causes peaks to look as if they're decaying even faster. This leads to broader lines, but because the decaying exponential is largest at small $t$ where the signal is large, it "emphasises" the parts of the FID that have more signal. Effectively, you lose spectral resolution but gain some signal-to-noise.
A Gaussian window function (or a combination of Gaussian + exponential, which is more common) starts small, reaches a maximum in the middle, and tails off again towards the end. In a way it has an opposite effect to the exponential: by trying to "compensate" for the natural signal decay, it makes the peaks look as if they are decaying more slowly, i.e. you get narrower lineshapes. On the other hand, you lose signal-to-noise because you de-emphasise the start of the FID which has the greatest signal.


What is so special about positive hydrogen ions that they are used in pH calculation? I just learned about pH. If I understand it correctly, all it really tells is the concentration of positive hydrogen ions (protons) there are in a given liquid.

Why is this pH scale entirely dedicated to positive hydrogen ions? What is so special about protons/hydrogen ions? I would expect the scale to would count all positive ions, even if they are of atoms different from hydrogen. I don't understand why is the proton ion a "superstar", being so special that people made a scale dedicated just to it.

What we're really talking about with the acidity of a solution is the proportion of water which is dissociated into hydroxide $OH^-$ or hydronium ions $H_3O^+$ (hydronium is usually a more accurate representation than a free proton).

Other ions do affect the acidity of a solution, but by inducing water to dissociate. Check out the example on the wiki page. Pure water alone has a $pH$, because of spontaneous dissociation of some water molecules into ions. The $pH$ of pure water is 7, because the $pK_a$ of water as a weak acid is 14.

Basically, water is the special thing, and its constituent ions happen to be hydroxide and hydrogen / hydronium. Those ions are what we measure with a $pH$ electrode, so the scales are named after them. Water has a concentration around 55 M, vastly predominating in aqueous solutions.


Software for visualization of 3D models of chemical structures with atomic orbitals A number of programs are available that let you build, visualize, and analyze 3D models of chemical structures (I use Chem3D/ChemBio3D). Current versions usually include various possibilities for the calculation and graphical representation of molecular orbitals.

However, the resulting images can be overloaded or unclear. Introductory explanations using such images might be overdoing it. Therefore, many textbooks on organic chemistry use simple chemical structure diagrams based on atomic orbitals.

In particular for didactic purposes, I am looking for an intermediate solution. I would like to build, optimize, and visualize a 3D model of a chemical structure and show a graphical representation of the original atomic orbitals in the 3D model. Unfortunately, this option doesn’t seem to be available in Chem3D/ChemBio3D. Does any software exist that lets you build a 3D model of a chemical structure and visualize atomic orbitals?

As porphyrin mentions in the comments, have a look at Jmol.

It indeed does the rendering as in your original version, but with the many options (see this mo command documentation) and particularly interactivity provided by the scripting, I think you may pedagogically achieve what you are looking for. Proteopedia is a nice website where the efficiently use Jmol's scripting feature for educational purposes (try the green links).


Number of electrons in pseudo wavefunction and all electron wavefunction in pseudopotentials In pseudo potential theory most pseudo wavefunctions are specified to be the valence pseudo wavefunction. These pseudo wavefunctions are often compared with an all-electron wavefunction. My question is whether the all-electron wavefunction is the wavefunction for a valence electron as well. Also is the all-electron wavefunction a one electron wavefunction? Is the valence pseudo wavefunction a one electron wavefunction? Part of my confusion stems from the fact that some journals state that the charge of the pseudo wavefunction and all electron wavefunction have to be the same.

In pseudopotential approach the core electrons are replaced by the pseudopotential. In case of all electron calculation we obtain wavefunction for both core and valence electrons. In case of pseudopotential only the valence electron remains. So when we compare the all- electron and pseudo wavefunction we compare only the valence ones. All the wavefunctions are one electron wavefunction. The charge is basically the norm of the square of the
wavefunction. The condition of charge equivalence between the pseudo and all electron wavefunction is imposed to get correct scattering properties. This condition is known as norm conserving pseudopotential.


Is the speed of sound the fastest rate that a chemical reaction can progress through a substance? I read that a martensitic transformation can travel at the speed of sound in that material. Is that confirmed as being the fastest speed of a chemical reaction (transformation)?

Certainly, chemical reactions can propagate far faster than the speed of sound. If the reaction emits electromagnetic radiation, e.g. light, and is itself initiated by light, then the speed of propagation is the speed of light in that medium. See this article from Sandia National Labs, or from the Royal Society.


Accelerated Oxidation Of Iron When Coating Breaks I was presented with the following information:

If iron is painted, then oxygen and water are prevented from contacting the metal, and corrosion is avoided.
Sometimes the iron is coated with a thin layer of another metal, such as Tin (the layer of Tin keeps oxygen and water away from the iron)
If the layer of Tin is worn away or scratched, the presence of Tin will actually accelerate the corrosion of the underlying iron. (http://www.mikeblaber.org/oldwine/chm1046/notes/Electro/Corrode/Corrode.htm)

I understand why Iron would be oxidized when the coating breaks, as an observation of the reduction (oxidation) potential would show that it is more favorable to oxidize iron, over tin.

However, the portion I am confused on is:

Why would corrosion be accelerated?

The passage I was reading suggested that it was because Tin acts as a cathode and thus would accelerate the corrosion, but that does not make a lot of sense to me. As I would see it, there would be little difference whether the Tin is there or not, and thus the reaction (corrosion) should not be accelerated, but simply as quick as it would have been without the coating.

Any insight would be greatly appreciated! Thanks!

The main factor for this phenomena is the the values of the standard electrode potentials for of the metal ions.

The UC Davis ChemWiki page Corrosion, uses this scenario as an example:

  the presence of the second metal can actually increase the rate of corrosion. The values of the standard electrode potentials for $\ce{Sn^2+}$ (E° = −0.14 V) and $\ce{Fe^2+}$ (E° = −0.45 V) show that $\ce{Fe}$ is more easily oxidized than $\ce{Sn}$. As a result, the more corrosion-resistant metal (in this case, tin) accelerates the corrosion of iron by acting as the cathode and providing a large surface area for the reduction of oxygen

Tin is therefore more of a noble metal than iron - a galvanic corrosion chart is shown below, showing that tin is more 'noble' than iron (Image source):

The significance of tin being more noble than iron is that, according to the UC Davis ChemWiki page Electrochemical Corrosion, it can act as a depolariser, removing more electrons than the water/air alone - hence, accelerate the corrosion process.

Instead of the copper used in the example, it would be tin in your example.

A somewhat simplied illustration of the process is shown in the diagram below:


How does hydrogen bonding affect the boiling points of chemical compounds? My main question is why is the boiling point of methanol so much different from that of water?

I understand that both compound are able to develop hydrogen bonding, and obviously water can develop one more hydrogen bond than methanol.

But on the other hand:

  Due to presence of an electron-releasing alkyl group, the oxygen atom of an alcohol molecule is less-electron withdrawing on the remaining hydrogen atom. Therefore the electron deficiency of the hydrogen atom is less than that of a water molecule resulting in the formation of a weaker hydrogen bond.

What does this actually mean?

Both hydrogen, in methanol and water, are connected to oxygen, which has the same electronegativity and takes electrons off the hydrogen making it have a partial positive charge.

Why does a "less electron withdrawing oxygen atom" lead to a weaker hydrogen bond?

I thought that a less electron withdrawing oxygen atom would mean that the H is more negatively charged, meaning it would be stronger for hydrogen bonding, since F, O, N want more electrons as they are more electronegative?

A hydrogen bond is a particularly strong dipole:dipole interaction with some covalent character. If we ignore the covalent character for a moment, we can say the higher the partial positive charge on the hydrogen, the stronger the hydrogen bond (keeping the O-H bond distance constant). If you relate "electron deficient" to positive partial charge, it might help you to make sense of the textbook answer.

You write:

  Both hydrogen, in methanol and water, are connected to oxygen, which has the same electronegativity and takes electrons off the hydrogen making it have a partial positive charge.

Yes, that is the primary reason compounds like water, methanol and - say - hydrogen peroxide and acetic acid are hydrogen bond donors. But the chemistry beyond the OH group also plays a role. To illustrate, just compare the acidity of the proton connected to the oxygen in these four cases. They are very different. In a similar manner, the hydrogen bonding will be different (not as different, though).


How do I round a calculation involving both addition/subtraction and multiplication/division to the correct number of significant figures? I'm just learning about significant figures (sig figs) in my chemistry class, and I'm confused about the rules. An example to my problem: if I had the following expression:
$$\left ( \frac{2.378 - 1.2}{1.03} \right )$$
I've thought of 2 ways to approach this. Which one is correct (or is neither correct)?

Approach 1:

Round each intermediate operation as you go. This is what my textbook does for its examples, but it itself suggests "Approach 2" below. I'm just listing this approach here as an option since addition/subtraction and multiplication/division have their own rules for rounding (hence, maybe do intermediate-operation rounding).

1) Round the subtraction part.

$$\left ( \frac{1.2}{1.03} \right )$$

2) Do the division and round.

$$1.2\,\,\, (2\ \mathrm{sig\ figs})$$

Approach 2:

In actual practice, the book says:

  In a series of calculations, carry the extra digits through to the final result, THEN round.

I think it means this:

1) Do the subtraction, but keep the *exact* number while noting that the rounded number was supposed to have 2 sig figs.

$$\left ( \frac{1.178}{1.03} \right )$$

2) Do the division with the exact number and the denominator, and since this step produces the final result, NOW you round, rounding to 2 sig figs per the note in Step 1.

$$1.1\,\,\,(\text{2 sig figs})$$

The problem here is that these two approaches produce 2 different results. So, to repeat, which one is correct (or is it neither)? The textbook doesn't provide an example like this. I found two different websites, but they too conflict with each other.

http://www.occc.edu/kmbailey/chem1115tutorials/Sig_Fig_and_Math_Answers.htm
(From the solution to problem #3, I should use "Approach 1".)

http://fabice.com/misc/significant_figures.html
(It quotes, "In a composed operation, intermediate results are not rounded." So that means I should use "Approach 2?".)

Compared to the other questions on this forum, this question seems silly, but I find the concept of sig figs important if I want to understand chemistry at all.

Approach 2 is better than approach 1.

This approach corresponds with methodology employed in both web links you provided.

The basic 'rule' is to do your calculation keeping all digits as you go right until to the very end (without rounding parts as you suggest in approach 1). This ensures your calculation retains maximum accuracy. You leave the 'rounding' til the very end, that is when you report your answer, only 'so many' digits are 'significant'.

To use your example, when calculating the subtraction part: 2.378-1.2 = 1.178, in approach 1, by 'rounding' this down to 2 significant digits (1.2), you are reducing the accuracy of your calculation before you finish.

The correct (most accurate) way to calculate the result is to keep all significant digits in your calculation until the very end, that is, 1.178 / 1.03 = 1.14368932 and then report only the 'significant' digits. In this case, since the number 1.2 only has 2 significant digits, your final answer can only have 2 significant digits. That is, the final answer is only as precise as the least precise measurement (in your case, 1.2), so in reporting the final answer for your calculation, you can only rely on 2 significant digits, so you report 1.1 as your calculated answer.

If you re-examine the answer provided in your link (http://www.occc.edu/kmbailey/chem1115tutorials/Sig_Fig_and_Math_Answers.htm) in which you state problem 3 follows approach 1, you will find that the solution provided in that link actually follows approach 2, similar to what I described in the above paragraph. That is, all decimal places are retained in the calculation until the very, when the final answer is reported to the number of significant figures corresponding to the least significant measurement.


Is a person warmed by fire an example of endothermic reaction? My chemistry textbook gives the example of a man being warmed by a fire as an endothermic reaction. If the man's body is the system and it is increasing in temperature, how can this be endothermic?

I wouldn't call it an endothermic chemical reaction. The fire is an example of an exothermic reaction. The man constitutes part of the surroundings for the fire (the system), and heat is being transferred between the system and its surroundings. Looking at it the opposite way, the fire could be considered the surroundings (a heat reservoir) for the man (the system), and there is heat being transferred from the surroundings to the system. But the man is not undergoing a chemical reaction.


Rate and order of a reaction The rate of reaction can be measured from a concentration vs time graph of a particular reactant in a reaction.

But, if in your reaction you have different reactants of different orders surely they will each give a different concentration vs time graph and so different rate at any time t. This implies that the rate of reaction can have multiple values depending on which reactant you follow.

The rate of reaction is defined in IUPAC's Gold Book as:

$$r = - \frac{1}{a} \frac{\mathrm d[\ce{A}]}{\mathrm dt} = - \frac{1}{b} \frac{\mathrm d[\ce{B}]}{\mathrm dt} = \frac{1}{p} \frac{\mathrm d[\ce{P}]}{\mathrm dt} = \frac{1}{q} \frac{\mathrm d[\ce{Q}]}{\mathrm dt}$$

where $[\ce{X}]$ denotes the concentration of the substance $\ce{X}$, and the lower case letters represent the coefficients of those substances in the balanced equation. The rates for $\ce{P}$ and $\ce{Q}$ are positive, as these are products in the reaction.

You are right in that you can determine the reaction rate by studying the change in concentration of a particular reactant (or product) over time. But, to determine the rate r for the reaction, not just for a particular species, you must take into consideration the coefficients in the balanced equation. The stoichiometric numbers are included in the definition of the rate, $r$, so that it is independent of which reaction species you study. For example if $a = 1$ and $b = 3$ then $\ce{B}$ is consumed three times more rapidly than $\ce{A}$, but $r = -\mathrm d[\ce{A}]/\mathrm dt = -(1/3)\mathrm d[\ce{B}]/\mathrm dt$ is uniquely defined.

Source: https://en.wikipedia.org/wiki/Reaction_rate


Will Proving or Disproving of any of the following have effects on Chemistry in general? I am working on a project relating to-
https://en.wikipedia.org/wiki/Millennium_Prize_Problems

I wanted to list the effects of them being proved or disproved in different aspects of science and maths. It was fairly easy to find such results on physics and maths as the questions primarily come up from them however I couldn't find any reference to effects it would have on our understanding of chemistry.

Is it because there will be no effect at all and that they are the least concern to chemists (in terms of applicability) or if there are then what are they?

All right, let's sum it up.

P vs NP, if solved the way that pretty much everybody expects (that is, $\rm P\ne NP$), will have no far-reaching consequences, because that's what we were thinking for quite a while now. If it would miraculously happen to be otherwise (that is, $\rm P=NP$), that would be quite a shock to many fields, especially to computer science, and by extension, to computational chemistry. But I don't believe in miracles.
Navier–Stokes (no matter how it will turn out) will have consequences in fluid dynamics, and by extension, maybe in some relatively narrow areas of chemical technology.
Yang–Mills will cause repercussions in the areas of physics which are the most distant from chemistry.
The rest of the problems, as far as I understand, have no bearing whatsoever on our material world.


Isomers with molecular formula C₄H₇Cl What are the possible isomeric structures of chlorobutene $(\ce{C4H7Cl})?$

I've come up with seven:

cis-1-chlorobut-1-ene
trans-1-chlorobut-1-ene
2-chlorobut-1-ene
3-chlorobut-1-ene
1-chlorobut-2-ene
cis-2-chlorobut-2-ene
trans-2-chlorobut-2-ene

There are 12 different constitutional isomers:

There is also the further possibility of stereoisomerism in some of the compounds above.

3: Two diastereomers (cis- and trans-) are possible.
7, 8, 9: Geometrical isomerism is possible in these alkenes.
11: C-3 (the carbon with the chlorine) is a chiral centre, which leads to two possible enantiomers.

That makes a total of 17 possible isomers.


Optical isomerism in Co-ordination Compounds Optical isomerism in normal organic molecules can be found out by looking for chiral carbon, and the number of isomers can be found out by drawing different spacial orientations of the same.

But, I don't know how to check whether a given coordination compound is optically active or not, and how many optical isomers does it have. I don't know how to draw different spacial orientations with all those polydentate ligands.

I would like to learn the general guidelines for finding optical isomerism in co-ordination compounds. I would like to solve questions like how many no. of optical isomers are there for $\ce{[CoCl2(en)2]^+}$

Any 3rd-year inorganic chemistry textbook worth its weight in salt will provide some information about isomers in coordination compounds.  An optically active compound contains no rotation-reflection symmetry element.  Typically, this means that if a molecule has an inversion center or a mirror plane then it is not chiral.

It takes a systematic brute force and ignorance approach to determining the number of isomers a coordination compound has, and then a bit more visual thinking to determine if any of the isomers are enantiomers.  The following figure is taken from Miessler, Fischer, Tarr 5th edition and is useful for identifying isomers of 6-coordinate complexes.

Note that a 6-coordinate complex with 6 unique ligands in an octahedral field would be enantiomers. In fact, there are 15 different isomers, each of which is chiral resulting in a total of 30 isomers!

The method of determining isomers this way is to formulate a structure of the type M where  indicates two ligands that are trans to each other.  Then, you start varying a, b, c, ... until you've come up with as many unique combinations as possible.  For your case, you have 2 instances of a bidentate ligand.  Bidentate ligands are typically designated with capital letters in this notation, so the formula for your molecule would be $\ce{M(AA)_2b_2}$.  Further, since bidentate ligands are too small to coordinate in a trans conformation, you will decrease the number of possible isomers.  One isomer, for example, would be M, where I use A and A' to designate the different $\ce{(en)}$ ligands.

The answer to your question, then, is that there are two isomers, and they are enantiomers, since you can't rearrange the A, A' and b in the formula above to come up with another unique configuration.


Thermodynamics : Work Done by a Gas Suppose in a piston in which there is a gas, the gas exerts pressure P on surroundings, whereas the surrounding exerts a pressure $P[ext]$ on the gas.

In order for the gas to expand, P must be greater than P[ext]. So far I understand. Suppose initially that $P>P[ext]$ and the gas expands until $ P=P[ext]$ This is what I don't understand is that when you calculate the work done by the gas, you use the external pressure and not the gas pressure. This makes no sense to me. After all, if you calculate the work done on something, you should use the force or pressure applied on that object, or not the other way around, correct? So why do we use external pressure instead of internal pressure when calculating the work done by the gas?

Assuming an adiabatic wall between gas and surroundings then the expansion of the gas will do work on the surroundings.  What is that work?  It is the pdV work using the external pressure acting against the piston from the surroundings.   You are asking why not use the pressure of the gas.  You could, but only if the process was reversible.  That is, the gas is at equilibrium throughout the expansion process.

But what if the gas is not at equilibrium throughout the expansion?   The pressure may not be uniform inside the gas throughout the expansion.  If the gas is expanding irreversibly then the pressure just behind the piston will be lower than the other parts of the gas.   It's easier to calculate the work on the surroundings since you know the external pressure is uniformly applied.


"NBO diagrams" versus MO diagrams Molecular orbital diagrams are well known by chemists.
According to Wikipedia (emphasis mine),

  A molecular orbital diagram, or MO diagram, is a qualitative descriptive tool explaining chemical bonding in molecules in terms of molecular orbital theory in general and the linear combination of atomic orbitals (LCAO) molecular orbital method in particular.

An example is given below.

(Image by CCoil (talk) - own work, CC BY-SA 3.0, Link)

By the use of computational chemistry methodologies we can (1.) produce a molecular wavefunction of a diatomic molecule as a linear combination of basis functions (assume they are centred on each nuclei).
This would already satisfy the definition above and generate a suitable MO diagram.

On the other hand, there are special methodologies for obtaining the starting orbitals centred on atoms:

We could express molecular orbitals as a linear combination of optimised atomic orbitals of each atom, which would imply an extra calculation for each element in a molecule.
A change of basis for the MOs, from the basis set to the atomic orbitals, does the rest.
The natural bonding orbitals (NBOs) method allows one to produce natural atomic orbitals as well, which could also be used as initial orbitals centred on the atoms.

So my question is:

Are the three approaches above really arbitrary?
What makes one option better than the other?

A molecular orbital diagram is a schematic representation of how we interpret bonding in certain species. It is as much an accurate representation for a specific bonding situation as a Lewis structure for the the molecular geometry.
When we construct a molecular orbital diagram, usually we resort to shapes of hydrogen like orbitals. We most of the times do not even use mathematical descriptions, accurate energy levels, without changing the descriptive value much. Oftentimes we completely omit core orbitals and don't draw nodal planes/spheres for orbitals with $n=2$.
Such diagrams, however, provide qualitatively good descriptions of what otherwise is almost impossible to fathom. They even provide a solid starting point for quantum chemical calculations.
As such, the choice of atomic orbitals does not matter at all; a picture will never contain enough detail to work out the subtle differences of these choices.

By the use of computational chemistry methodologies we can (1.) produce a molecular wavefunction of a diatomic molecule as a linear combination of basis functions (assume they are centred on each nuclei). This would already satisfy the definition above and generate a suitable MO diagram.

Most basis sets indeed contain basis functions which are centred at the respective nucleus and they resemble atomic orbitals (they don't have to). Through a linear combination of such functions we can obtain suitable molecular orbitals within the respective approximations. The wave function is then obtained as a set of orthogonal molecular orbitals. This in principle is the essence of molecular orbital calculations, which is in turn essentially the same as calculations on the Hartree-Fock level of theory. We know that this level of theory doesn't contain nearly enough precision to accurately describe the most simple compounds because of the neglecting most of the correlation energy. Yet, it is impossible to represent the result of such a calculation as a diagram.

We could express molecular orbitals as a linear combination of optimised atomic orbitals of each atom, which would imply an extra calculation for each element in a molecule. A change of basis for the MOs, from the basis set to the atomic orbitals, does the rest.

Yes you can do that, but apart from adding significant computational cost, this would not really increase the accuracy of the calculation. First and foremost you need to account for a larger portion of the correlation energy, for example with density functional approximations, post-HF methods like perturbations theory MPn, coupled cluster, or configuration interaction. Instead of optimising the basis set for each atom in the molecule, it is much more effective adding more functions to the basis set until you reach the complete basis set limit. There are several types of basis set libraries in use now, which have been designed to effectively handle such calculations. I think it was John Pople who laid out the advantages of this system. And while his basis sets have significant shortcomings, they are still used today.
In any case, what you will not get is a better or more descriptive MO diagram.

The natural bonding orbitals (NBOs) method allows one to produce natural atomic orbitals as well, which could also be used as initial orbitals centred on the atoms.

Indeed natural bond orbital theory uses natural atomic orbitals (and other types), however, these are representations of the atomic orbital within the molecular environment.
The theory itself is a localisation technique of canonical orbitals that are already expressed in the chosen basis set. As a result we get a bonding picture which closely resembles a Lewis structure. While they are different from the actual basis set they still have the same functions and would still lead to the same solution.
It is best to treat MO diagrams as what they are: qualitative tools to describe bonding.


Does the potential energy increase when temperature is raised? The potential energy of a material is its ability to do work. So I would suspect the potential energy to increase during phase changes that increase entropy (solid $\rightarrow$ liquid and liquid $\rightarrow$ gas), and decrease in the opposite direction.

However, what about when the temperature is raised for a solid, up until the melting point - will the potential energy increase? The heat from a 1500 K gold plate could do work on liquid water (evaporate) to a higher extent than a plate at 300 K, so I guess the potential energy has increased. So would the potential energy as a function of temperature, of an arbitrary material, look something like this ($K$ should be $T$)?

Yes, potential energy increases with increasing temperature for at least the following three reasons:

At a higher temperature, more atoms/molecules are in excited electronic states.  Higher electronic states correspond to greater potential energy. Potential Energy is -2 times Kinetic Energy. So actually, at higher temperature, when more atoms are in higher electronic states, there is more potential energy and less kinetic energy (just considering electronic energy).
At higher temperature, more molecules are in excited vibrational states.  Higher vibrational states correspond to greater potential energy. Half the energy of each vibrational state is kinetic, half potential.
At higher temperature, more molecules are in excited rotational states.  While for a rigid rotor, this would not involve increased potential energy, real molecules are not rigid.  The rotation of a diatomic molecule about an axis perpendicular to the bond stretches the bond as a centrifugal distortion, which represents an increase in potential energy.

Additionally, for intermolecular forces, to the extent that they involve potentials of the form r^x, the Virial Theorem requires that total energy is divided into potential and kinetic energy in a specific proportion, so total energy can not be increased without increasing potential energy.

Free translational motion is not subject to a potential, nor is rigid rotation.


Why does an mercury ore exist as HgS? Cinnabar($\ce{HgS}$)  is one of the most common ores of mercury.
Why does it not exist at $\ce{HgO}$ or some other such compound?
Is there any reason for why it is so prevalent?
Does it react less with other compounds, or maybe due to some solid structure of HgS having some characteristic property giving it more stability?
I'm seeking an explanation without the mention of free energy, because it doesn't properly explain the phenomenon for me. I understand that cinnabar is abundant in the crust, and my question is why is $\ce{HgS}$ so common?

Why $\ce{HgS}$ and not $\ce{HgO}$?

Because of HSAB. F'x has already answered a previous question of why mercury has propensity towards thiols(or in general sulfur). Quoting the relevant information:

In Pearson's HSAB theory (hard and soft acids and bases), the reason
the $\ce{S-Hg}$ bond is be stronger than the $\ce{O-Hg}$ can be
explained because $\ce{S^2-}$ is a soft basis and $\ce{Hg(II)}$ is a
soft acid, making a good fit while $\ce{O^2-}$ is harder.

If this answer helps, please make sure to upvote F'xs answer.


How can I figure out whether a reaction is endothermic or exothermic without its enthalpy value? I already know the way to determine whether if the given reaction is exothermic or endothermic by the enthalpy values. But is it possible without enthalpy values, and just by looking at the reaction?

So as far as I know there are two types of chemical reactions, namely one way reactions and reversible reactions.

So how can I determine this factor in one way reactions and reversible reactions?

I guess that for reversible reactions we can use equilibrium constants such as $K_p$, $K_c$, but I am not quite sure whether it is correct. On the other hand, I feel that there is no such way to determine what I want in one way reactions. So how can I decide it?

First of all, the equilibrium constant $K$ in general doesn't tell too much about the enthalpy change of the reaction as it is related to the (Gibbs) free energy change $\Delta G = \Delta H - T\Delta S$. It is true that in some simple cases (such as gas phase reactions) the  entropy change can be approximated by looking at the stoichiometry of the reaction but in general it's usually more complicated.

Regarding your distinction between one-way and reversible reactions I think you may as well assume all reactions are reversible, but some have incredibly huge $K$ so the equilibrium is shifted totally to one side of the reaction - this is the result of a very negative $\Delta G$.

In case of a "reversible" reaction you can always look at the change of the equilibrium constant with temperature. The constant is

$K = \exp\left(-\frac{\Delta G}{RT}\right)=\exp\left(-\frac{\Delta H}{RT}\right)\exp\left(\frac{\Delta S}{R}\right)$

so from the temperature dependence you can deduce the sign of $\Delta H$. Naturally, this is assuming that $\Delta H$ and $\Delta S$ doesn't depend on the temperature in the temperature range you're looking at.

The figure shows the change of equilibrium constant with temperature (arbitrary units) for an exothermic (yellow line) and an endothermic (blue line) reaction.

Just by looking at the reaction I don't think you can tell the enthalpy change. Normally you have bonds breaking on the reactant side and others forming on the product side and normally there is a delicate balance between the two. People have come up with bond strength values (which could be determined from known $\Delta H$ values) and you can use those in your reaction.


Does using same calibration curve for chemical analysis introduce correlation in sample error/uncertainty values? I am using a hypothetical case to explain my question
Let us consider a chemical analysis method A (using a certain instrument), which is used to quantify the value of a particular species b. According the instrument manufacturer's notes, the instrument used needs to be calibrated once a month. I am analyzing several samples of b in the duration of a month, and using the same calibration curve to obtain the quantity of b along with the uncertainty bound for the value.
I want to understand the following,

If number of samples (over the period of a month) of b are using the same calibration curve to quantify it value, are their errors/uncertainties correlated (by any means, the calibration curve introduces it, however small it may be) ?
Additionally, method A has the ability to quantify several species (b, c, d,$\dots $), does errors/uncertainty of species from the same sample (containing different species like b, c,$\dots$) correlated in any way?
If there exists even a very small correlation, does it make sense to consider it for any further analysis?

I am new to the realm of chemistry (analytical chemistry to be precise), I request your help.
Thank you.
PS: Apologies if the tags are inaccurate

Please note that calibrating an instrument once a month is another story as compared to constructing the calibration curve from that instrument. The former is meant to check the accuracy of the instrument's output and the latter serves to determine the concentration of a substance by noting the response of the instrument to 6-7 standards.
Also note that all instrumental analysis techniques are relative with very very few exceptions (gravimetry, microwave spectroscopy, coulometry which are absolute etc.). The rest all need standards for comparison.
In analytical chemistry, it will be considered a bad practice to make a calibration curve one day and use it for the entire month. The good practice is to make a fresh calibration curve for every batch of samples and there should be a "reference" check which keeps on validating the calibration curve after a certain number of samples have been analyzed.
The main reason is that most instruments will show what is called as the drift i.e., a low frequency and hence long term noise in the baseline. The baseline response can rise and decline slowly.
Also the word correlation is perhaps not suitable in your situation. Correlation has a very specific meaning in signal processing/instrumental analysis. What you imply is the influence of interferences and something called as the matric effect. Suppose, you have b, along with c,d,e,...,f. Yes, the presence of the additional components can indeed influence the signal of b. In most cases, the signal of b is decreased causing a lower result than expected.
Beyond your hypothetical situation, one can say nothing. Which instrument you had mind?