Why doesn't the bidentate ligand form dative bonding three times? I was looking at this question where  it asked you to work out the final complex formed when an excess of 1,2-diaminoethane is added to and aqueous solution of copper (II) sulphate.

So the equation I worked out would have been:
$$\ce{[Cu(H2O)6]^2+ + 3(NH2CH2CH2NH2)-> [Cu(NH2CH2CH2NH2)3]^2+ + 6H2O}.$$

Therefore I worked out the complex to be: $$\ce{[Cu(NH2CH2CH2NH2)3]^2+}$$

However the answer was : $$\ce{[Cu(NH2CH2CH2NH2)2(H2O)2]^2+}$$

Why is this?

Orthocresol already pointed you towards the Jahn-Teller effect. The effect is there in $\ce{[Cu(H2O)6]^2+}$, the hexaaquacopper(II) cation, but more pronounced in the tetraammindiaquacopper(II) cation $\ce{[Cu(NH3)4(H2O)2]^2+}$. Compare an image of the complex as shown below:

Even the naked eye can see that the copper-oxygen bonds are much longer than the copper-nitrogen ones. Measuring the distances in crystal structures containing that cation gives $203~\mathrm{pm}$ for the $\ce{Cu-N}$ bonds and $251~\mathrm{pm}$ for $\ce{Cu-O}$.[1] Why is this the case? To answer that we need to refer to the molecular orbital scheme of a typical octahedral complex:[2]

The three lowest groups of orbitals containing six MOs in total are typically populated by the ligand lone pairs which form the coordinate bond. The next set of two groups — five orbitals in total — are the metal d-orbitals that can (or cannot, in case of $\mathrm{t_{2g}}$) interact with the ligands. For copper(II), we need to fill in nine metal d-electrons meaning that the $\mathrm{t_{2g}}$ level is fully populated while $\mathrm{e_g}$ has three electrons in two orbitals — a denatured state.

This denatured state is not energy-optimal. Energy could be released if we somehow manage to unequalise the $\mathrm{e_g}$ orbitals, give them different energies and have the lower-energetic one be fully populated. Remember that the $\mathrm{e_g}$ orbitals are $\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$. In a gedankenexperiment, remove the two ligands in $z$-direction symmetrically away from the metal centre. Any orbital that has a $z$ contribution should now be stabilised — most notably, that includes $\mathrm{d}_{z^2}$. The orbital scheme looks slightly different, the complex’ symmetry is no longer $O_\mathrm{h}$ but $D_\mathrm{4h}$ but we have gained energy by being able to differentiate between the $\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$ orbitals. This is the Jahn-Teller effect.

Consequences of the effect include that ammin ligands preferentially coordinate copper(II) equatorially, and that the tetraammindiaqua complex is the preferred one in high aquaeous ammonia concentrations. Hexaammincopper(II) can be prepared only in liquid ammonia to the best of my knowledge.

Now let’s consider using $\ce{en}$ (ethylenediamine) rather than $\ce{NH3}$. The bridge between the two amino groups is long enough to bridge across one edge of the octahedron but not any further. Replacing all the equatorial ligands with $\ce{en}$ makes sense because we will get another very symmetric complex $\ce{[Cu(en)2(H2O)2]^2+}$, and because these are the ones that are replaced first. If we now wanted to replace the two other water molecules, we would create a rather distorted $\ce{[Cu(en)3]^2+}$ complex, where two $\ce{en}$ molecules have one short, strong and one long, weak bond. This is an unfavourable condition, since the former shorter, stronger bonds have to be weakened to get there. In short, $\ce{en}$ behaves like ammonia and only occupies the equatorial positions, not the axial ones. A consequence again of Jahn-Teller.

[1]: Data taken from Professor Klüfers’ internet scriptum for his general and inorganic chemistry course at the LMU Munich (section 23.4).
[2]: First presented in this answer and originally taken from Professor Klüfers’ coordination chemistry course.


Why is diene more reactive than alkene? The chemical reduction of 1,3-butadiene with sodium in alcohol gives mainly the 1,4-addition product that is but-2-ene.
Under these condition the isolated double bond are not reduced. This suggest that dienes are much more reactive than simple alkenes.

Why is diene more reactive than alkene even though the conjugated double bond of diene (in the example of excerpts given above) is stabilized with resonance?
How is 1,4-addition product the major product?

It is not only the diene itself which is stabilised by resonance, but on electrophilic addition the carbocation formed is also stabilised by resonance if the diene is conjugated. Thus, forming the intermediate for a conjugated diene is easier than forming the intermediate for the alkene, even though the starting compound itself is more stable in case of conjugated dienes and therefore conjugated dienes react faster.


What are the consequences of having an element with an atomic number above 137? In other words, is an element such as Feynmanium the last chemical element that can physically exist on a planet such as Earth?

The critical value "137" occurs in calculations based upon relativistic theories that treat the nucleus as a point, such and Dirac theory or Sommerfeld theory.
However, when the nucleus is correctly treated as having a non-zero size, a significantly higher value is calculated.
The most recent calculations I see are in New method for solving the $ \bf Z > 137$ problem and determining hydrogen-like energy levels (2014) Phys.-Usp. 57, 189.  (alternative link to pre-print):

Following the development of the Dirac theory in 1928, Dirac [reference 2], Darwin [reference 3] and Gordon [reference 4] obtained expression (1) as a result of exact solution of the Dirac equation in the Coulomb field of a point charge (-Ze). ...  Formula (1) [results in a complex number] if Z [is greater than 137]...the complexity of energy levels in (1) is often called the “Z>137 catastrophe”.
In 1945, Pomeranchuk and Smorodinsky [6] considered
an atomic system with [the nucleus being a sphere of radius $0.8 \times 10^{-12}$cm and having uniform charge within the nucleus and found 175 to be the critical value].

The article goes on to explain that since the 1945 calculation, various calculations, with various nuclear charge distribution models, have calculated critical values in the 170-175 range.  The article's own calculations find 178.


Decomposition products of hexaurea-chromic-chloride-trihydrate I am sitting over a TG diagram of the decomposition steps of hexaurea-chromic-chloride-trihydrate ($\ce{[Cr(CO(NH2)2)6]Cl3 \cdot 3 H2O}$) and cannot really make sense of the products it might decompose into.

It is clear that water leaves first (-3.5%), and the next step might be (with a wide error margin) fumes of $\ce{CO2}$, $\ce{H2O}$ and $\ce{2 NH3}$. What is left is then $\ce{[Cr(CO(NH2)2)5]Cl3}$.

I have not found any resources on what might happen further - the four urea ligands could decompose into $$\ce{4 CO} + \ce{5 NH3} + \ce{3/2 N2} + \ce{1/2 H}$$ and I am left with $\ce{CrCl3}$ and urea, but those steps show up last (17.9% + 17.3% + 5.4% = 40.6%), leaving the 15.5% in the middle unexplained.

What other decomposition could take place after water and two urea ligands are gone? Is there a way to figure that out other than by guessing or experiment?

Using the article Klaus Warzecha linked to and some trial-and-error, a possible decomposition could look like this and is in accordance with the TG diagram in the question:

Decomposition of $\ce{3 H2O}$ and $\ce{2 CO(NH2)2}$:
$\ce{[Cr(CO(NH2)2)6]Cl3 $\cdot$ 3 H2O -> [Cr(CO(NH2)2)4]Cl3 + 3 H2O \uparrow + 2 CO(NH2)2 \uparrow}$
Decomposition of $\ce{2 CO(NH2)2}$ and $\ce{2 HCl}$:
$\ce{[Cr(CO(NH2)2)4]Cl3 -> CrH6C2N4ClO2 + 2 CO(NH2)2 \uparrow + 2 HCl \uparrow}$
Decomposition of $\ce{2 CO2}$, $\ce{N2}$, $\ce{3 H2}$ and $\ce{1/2 Cl2}$ and oxidation of $\ce{Cr}$ to form $\ce{CrN2}$:
$\ce{CrH6C2N4ClO2 -> CrN2 + 2 CO2 \uparrow + N2 \uparrow + 3 H2 \uparrow + 1/2 Cl2 \uparrow}$

Given the temperature range and pressure (probably standard pressure), the formation of $\ce{CrO2}$ is unlikely.


Wattage to produce 6L of air from Water in less than a second? So I understand the question is probably vague to a scientist, but this is all the specifics I could mustard up with my non-scientific background.

So I understand the basics of the process and have used it occationally in a garage environment to fiddle around, but never really thought about the applications until a particular Indiegogo campaign claimed to have created artificial gills small enough to hold in your mouth and generating 45 minutes of breathable air.
The campaign have gotten a wide spread of media attention and it got me thinking, how much wattage would be required to produce 6L of air over the course of a second which happens to be a rough average of a adults lung capacity.

I understand that submarines use this process to create breathable air, but as someone mentioned they have a nuclear reactor onboard and "it would not be wise to attach that to your face".

Would shortening a Lithium Ion/Polymer battery be sufficient to create such a quick reaction? Or are we talking megawatts in order to produce such a volume of air in that time span?

Update to the numbers used:

Pointed out by @DavePhD, you don't use 6L per breath, rather 6L on average per minute.
Taking into account that this would be a physical exercise under water, you'd normally use anything between 14-17L per minute, on average 16L.

Also noted is that I should have used P instead of the term wattage, thanks @Loong

I guess it would have to be the amount energy released by the reaction $$\ce{2H2 + O2 -> 2H2O}$$ where you have four liters of hydrogen gas and two liters of oxygen gas, all divided by one second (or less, increasing the wattage as you decrease the time).

You asked about "air" — but you can't get normal air from electrolysis of water.

So, the heat of combustion for hydrogen is $286\ \mathrm{kJ/mol}$. $$286\ \frac{\mathrm{kJ}}{\mathrm{mol}} \times \frac{4\ \mathrm L}{22.4\ \frac{\mathrm L}{\mathrm{mol}}} = 51.1\ \mathrm{kJ}$$ and that is assuming atmospheric pressure which is certainly low, but how low depends on your depth below the surface of the water. So with the "barely under the surface" energy of $51.1\ \mathrm{kJ}$, we can then divide that by seconds: $$\frac{51.1\ \mathrm{kJ}}{1\ \mathrm s} = 51.1\ \mathrm{kW}$$ so you would need a bare minimum of $51.1\ \mathrm{kW}$ of power to convert water into a 67:33 mixture of $\ce{H2:O2}$. That seems like a lot of energy in a small amount of time to me: think about a six-liter balloon with the perfect combustion ratio of hydrogen and oxygen. That reaction is a lot faster than one second, but it's still a lot of energy. Tough to get into a small, portable battery and have it fill a capacitor with that amount of e energy every few seconds.

I'll be honest, I'm just doing these calculations on-the-fly, and I may have made a mistake or five, but it seems reasonable to me. If anyone takes any issue with my answer, please, feel free to edit/give suggestions.

Edit: This Wikipedia page has some information on hydrogen-oxygen diving air. They use a 96:4 ratio. Not sure why, but if that's the best ratio, this would take significantly more energy per breath, since a lot of oxygen would just get dumped.


A book like Feynman's Lectures to Physics, which deals with Chemistry and provides insights I have been reading The Feynman Lectures, and it provides a lot of basic knowledge in physics. I would like to know if there exists a book for chemistry like this too.

About 20 years ago, I have got the German translation of Atoms, Electrons and Change by Peter W. Atkins. This short and simple book was inspired by the lectures on chemistry given by Michael Faraday (such as The Chemical History of a Candle).


Is there a broader nomenclature convention to explain why the ‘a’ in ‘chlorophyll a’ is italicized? By convention, the letter that specifies which side chain a chlorophyll molecule contains is italicized: for instance, chlorophyll a. Does this convention stem from any broader principle of chemical nomenclature, or is it more of a quirk unique to chlorophyll?

This is only a partial answer to your question.

The suffixes a and b to the name chlorophyll were most likely introduced by Richard Willstätter and Arthur Stoll. They are widely used in their book UNTERSUCHUNGEN ÜBER CHLOROPHYLL (Julius Springer, Berlin, 1913). Here, however, the suffixes are not yet italicized and appear as normal text.


The unit of energy, 1 dm³ atm, is equal to how many calories? I surfed through the internet and visited a number of chemistry websites but only in vain. I couldn't find an answer.
Please help! Your quick response will be appreciated.


Is it possible to find the ratio of isotopes only given the mean mass number? Three isotopes of an element have mass numbers $(M)$, $(M+1)$ and $(M+2)$. If the mean mass number is $(M+0.5)$, then the ratio of the amounts of the three isotopes is?

Let amount of $(M)$, $(M+1)$, and $(M+2)$ be $x$, $y$ and $z$ respectively.
$$\frac{(M)x + (M + 1)y + (M + 2)z}{x + y + z} = (M + 0.5)$$

After further simplification, I got $x-y=3z$.

I don't know how to proceed further. Can someone give me a hint, or reassure me that this question can't be solved with the given data?

The answer is given as $x:y:z=4:1:1$.

You have two equations and 3 unknowns, so you can't solve it with just that. Say a, b, c are the fractions (as a decimal) of each isotope...

$$ a(x) + b(x+1) + c(x+2) = (x+\frac{1}{2}) $$

$$a + b + c = 1 $$

The 4:1:1 solution works. Another that works is 3:0:1. Another is 7:4:1. There are infinitely many solutions.


How to convert cell diagram to a chemical equation? How would one convert on the right side of the cell diagram:
$$\ce{Zn(s) | Zn^2+(aq) || MnO4-(aq), Mn^2+(aq), H+(aq) | Pt(s)}$$
$$
\begin{align}
\ce{Zn(s) &-> Zn^2+(aq)} \tag{ox}\\
\ce{MnO4-(aq) &-> Mn^2+(aq)}\tag{red}
\end{align}
$$
Obviously, one would then balance and combine the half-equations. I just need to know if the unbalanced half-equations are correct.

Assuming you meant $\ce{MnO^-4}$ and not $\ce{MnO^{4-}}$, your reduction half reaction is incorrect. Because according to your equation $\ce{Mn^{2+}}$ is being oxidised to $\ce{MnO^-4}$ rather than being reduced.
The correct half cell reactions would be:
$$
\ce{Zn -> Zn^{2+} + 2e-}\\
\ce{MnO^-4 + 8H+ + 5e- -> Mn^{2+} + 4H2O}\\
$$
And so the overall balanced reaction would come out to be:
$$\ce{5Zn + 2MnO^-4 + 16H+ -> 5Zn^{2+} + 2Mn^{2+} + 8H2O}$$


Extracting Copper from Copper Carbonate In a crucible, we added copper carbonate and the total weight of both is 37.04g. After, we heated the crucible with the copper carbonate. It was easy to judge when the reaction had taken place as the colour of the substance had changed from the light green of copper carbonate to the black of copper oxide.After we observed this, we weighed the copper oxide and we notified the result which was 36.42g. We should have heated and weighed the crucible again, but we weren't able because it was broken(due to temperature change). But theoretically, which would be approximately the value of the second weighing?I'm curious because after searching on the Internet, I have seen both increase and reduction of the remaining mass.

The repeated heatings were just to ensure that you had oxidized all of the copper carbonate to copper oxide. If you heated it a second time and the mass decreased, that would just mean that some of the copper carbonated had not been oxidized by the first heating. Then you would just keep heating until the mass quit decreasing, meaning that you had definitely oxidized all of it.

So the answer to you question is simply that the mass theoretically should not have decreased at all with a second heating, unless some of the copper carbonate had not been oxidized the first time.


Does SiO compound exist? If carbon monoxide can exist then why can't we have silicon monoxide?
Both carbon and silicon belong to group 14 and have similar chemical properties.

Short answer: It does, but unstable and hard to detect reliably..

Long answer.

First of all, we should look at structure of carbon dioxide and silicon dioxide. It is pretty easy to see that silicon forms polymeric dioxide with no double bonds, while carbon forms molecules with two double bonds. It is not an anomaly, obtaining silicon compounds with double bonds $\ce{Si-E}$ is tricky at best. This is attributed to larger atom size and more diffuse $p$-orbitals for silicon, making formation of double bonds unfavoured in most circumstances.

Now, let's look at carbon monooxide. It is a molecule with bonding very similar to that in dinitrogen molecule. The carbon atom has formal negative charge and oxygen has formal positive charge. However, because of high electronegativity of oxygen, electron density is pulled back towards oxygen, so the resulting dipole moment is diminutive. The molecules contains not double, but triple bond, making formation of similar molecule for silicon highly unfavorable. Still, silicon monooxide can be detected in gas phase.

Now, why solid phase with composition $\ce{SiO}$ is not formed is a more interesting question. Unfortunately, such questions not always have an easy answer, as rules guarding stoichiometry of solid phases is often base on arcane concepts of atom packings and electron count. It is entirely possible, that such a phase would be obtained one day in some exotic conditions. For clarity, I would use sterical considerations. $\ce{Si-O-Si}$ fragment is roughly linear, while $\ce{Si}$ atom favors tetrahedral coordination. Going from here, only two simple phases with no dangling bonds can be formed, based on diamond structure. It is structures with either $\ce{Si}$ or $\ce{SiO4}$ units in the structure nodes. It IS an oversimplification, as silicon dioxide forms many rather esoteric crystal structures, because $\ce{Si-O-Si}$ fragment isn't perfectly linear, but is is close enough in my opinion.

Do not rely on oxidation states in prediction of existence of covalent compounds,   +2 oxidation state for silicon is a thing. It exists in $\ce{Si6Cl12}$, adopting structure similar to cyclohexane.


Why do we use Ḡ ("G bar") as a notation for partial molar free energy? For partial molar entropy there is a bar that is positioned above the $G$ with a subscript I that is sometimes added to the $G~(\bar{G}).$ The bar I read sometimes denote the "partial nature of the $G$".

However, is it not rhetorical. I am looking for the explanation as to what is so partial about this quantity with regard to it's occurrence in the equation.

Thermodynamic properties are functions of composition, which can be represented as $(\chi _1,...,\chi _N)$ where $\chi _i$ is the mole fraction of component $i$ present in the system. Now imagine a change in the composition such that the amount of component $i$ is increased by a very small quantity $dn _i$ (alternately, imagine the system to be infinitely large), and a thermodynamic property $X$ is changed by $dX$. Then (using the notation you describe) we define the partial molar property as $$\bar{X}=\left(\frac{\partial X}{\partial n_i}\right)_{p,T,n_j}$$ This notation means that we make a limiting, infinitely small change in $n_i$, such that the change in $X$ is for all practical purposes nil, ie $\Delta X=0$. The partial molar quantity describes the slope of the property $X$ (which can be represented as a surface in a multidimensional manifold) along the composition coordinate $n_i$, at a specific composition $(\chi _1,...,\chi _N)$.


What is the methodology to discover thousands of compounds from a complex mixture? So this is something that has always amazed me,

  More than 1,000 chemical compounds in coffee,[1] and their molecular and physiological effects, are areas of active research in food chemistry.

https://en.wikipedia.org/wiki/List_of_coffee_chemicals

How is it possible to identify so many separate compounds from just a single matter? What methods are used?

Most compounds can only be analytically quantified individually, so the first step in analyzing a complex mixture is typically to separated the different compounds from each other. For something like coffee, this would typically be done via different types of chromatography (i.e. gas (GC), liquid (HPLC), ion (IC)).

As the separated compounds exit the chromatography column, they can then be analyzed. This is usually done in real time using a number of techniques; the most relevant in this case is mass spectrometry (MS) following GC or HPLC.

In short, the primary techniques used to analyzed the compounds found in an organic sample like coffee would be GC-MS and HPLC-MS.


Is sugar necessary for the sweetening effect of food? A lot of food packages, nowadays, are mentioning "0 g sugar" or "sugar free". Then how do those still taste sweet?

There are plenty of sweet things that are not sugar(s)

People have a sweet tooth (or, to put it another way humans like sweet things). This is an evolutionary adaptation because fresh fruits are both good food (both nutritionally and in providing lots of energy) and rare in pre-modern food gathering cultures. So we are adapted to like things that are sweet as eating them is good for us (in moderation).

In the modern world we can have as much sweet stuff as we want but this is no longer a survival advantage as too much sugar provides more energy than we need to survive or thrive and eating it just makes us fat.

So the modern food industry has adapted to give us sweet-tasting things that don't overload us with unneeded energy. Many of those food ingredients mimic the sweetness of sugar without containing many calories, so allow us to eat sweet food without the same risk of consuming too many empty calories and getting fat.

There are many chemical that taste sweet but are not sugars. The successful ones tend to tase a great deal sweeter than the typical sugars found in fruits and related plants (glucose, sucrose, fructose and others). For example, aspartame is hundreds of time sweeter than sucrose but is a peptide not a sugar; Saccharin is an aromatic suphimide and is also hundreds of times sweeter than sucrose; acesulfame K is similar to saccharin.

There are also sugar-like sweetness that the body cannot digest but which taste sweet. Steria is a complex sugar with sweetness but few calories; Sucralose is a modified sugar the body can't digest so tastes sweet without containing the calories of sucrose.

The point of the claims on "sugar-free" foods is that they are sweet but non-calorific. Plenty of compounds can mimic the sweetness of real sugars like sucrose but are calorie free either because they are far sweeter (so much less stuff is required for the same level of sweetness) or because the body can't digest the compound and turn it into unneeded calories.


Hybridisation of similar energy orbitals Hybridisation is the mixing of orbitals of similar energies, so when we perform hybridisation, why don't we prefer $4\mathrm s$ in place of the $3\mathrm d$ orbital for hybridisation, even though $4\mathrm s$ has less and more comparable energy to $3\mathrm p$ and $3\mathrm s$? Why can't we make an outer orbital complex like $\mathrm{sp^3s}$ instead of $\mathrm{sp^3d}$?

I would suggest that this entire factory of bogus hybridization stories should be eliminated from general and organic chemistry curricula. The word mixing is worse, as it is a jug where one can add different orbitals and run a blender and here is the mixture. This stuff is good for rote memorization only.
Only after a thorough QM course and mathematical foundation, these ideas should be introduced with an example along with historical development. Neither the textbook author, nor the teacher know what it is. The only real experimental observable is only electron densities via X-ray diffraction or other techniques.


Why does a weak acid and a salt containing its conjugate base form an acidic buffer? I read this today:

  If you mix equal amounts of $1 M$ $\ce{H_3PO_4}$ and $1 M$ $\ce{NaH_2PO_4}$ you get a buffer at $pH<6$.

The reason given is that a weak acid plus a salt containing the conjugate base forms a acidic buffer.

  If you mix equal amounts of $1 M$ $\ce{NH_3}$ and $1 M$ $\ce{NH_4Cl}$, you get a buffer at $pH>8$.

The reason given is that a weak base plus a salt containing the conjugate acid forms a basic buffer.

I don't understand why this is true. Let's take the first statement for example. If you have equal amounts of a weak acid and its conjugate base, then by definition you have a buffer. Whether the buffer is acidic or basic shouldn't depend on whether the salt contains the conjugate acid or the conjugate base, right?

The statement "a weak acid plus a salt containing the conjugate base forms an acidic buffer." is not correct. More properly, a mixture of equal amounts of a weak acid (but not too weak: with a pKa < 7) with its conjugate base (salt) does indeed give an acidic buffer. To get a basic buffer you would use equal amounts of an even weaker acid (with a pKa > 7) and its conjugate base (salt). Similar considerations apply to formation of buffers from a weak base plus a conjugate acid salt.


Which one of displayed formula below does not represent an isomer of heptane? All of these appeared to be correct. I’ve already counted their hydrogen and carbon number. Please help!

[1]: https://i.stack.imgur.com/tnQvO.jpg

The structure in the question and option (D) are the same thing. Note that a branching as apparently seems is not a real branch on the principle chain that would give an isomer. Basically (D) is same as that given in the question.


Cahn–Ingold–Prelog priority of CN vs CH2NH2 Why does $\ce{CH2NH2}$ have lower priority than $\ce{CN}$? The carbon on the first molecule is bonded to (N,H,H) while the carbon on the second molecule is just bonded to (N). I can't seem to find this scenario anywhere online.

Each multiple bond is considered to be equivalent to an individual bond. So $\ce{-CN}$ has the same priority as (N,N,N), making it higher than (N,H,H). Read more about it here.


Why are equilibrium constants unitless? I haven’t quite reached the point where I can read a full-fledged text on chemical kinetics and thermodynamics yet, so bear with me, please.

I’m wondering why a value like $K_\text{eq} = \frac{[\ce{NO}]^2[\ce{O2}]}{[\ce{NO2}]^2}$ wouldn't have units of M?

This source explains it well. It looks like part of class material, but it clearly explains the dimensionlessness of $K_{eq}$.

The resolution of this apparent paradox is that the above equation, while perfectly satisfactory for everyday use, is not technically correct.  A more correct version is:
$$K_{eq} = \frac{\frac{\lvert B \rvert_{eq}}{\lvert B \rvert_{ss}} \frac{\lvert C \rvert_{eq}}{\lvert C \rvert_{ss}}}{\frac{\lvert A\rvert_{eq}}{\lvert A\rvert_{ss}}}$$
where the "ss" subscripts refer to the concentration of that species in the standard state.  (By this definition, Keq is always unitless.)

It then goes on to state:

Strictly speaking, division by the standard state concentrations is also necessary in every thermodynamics equation in which you take the log of a concentration product, otherwise the units don’t come out right.)  We will NEVER use this "correct" version of the equation in this class (well, never except in one problem on this week’s problem set...), and $K_{eq}$ for a reaction with unequal numbers of reactants and products is ALWAYS given with units,
even in published papers.

[1] http://www.bio.brandeis.edu/classes/archives/biochem102/102-02stand.pdf


Why there is a color fading ( to colorless) when Cerium Ammonium Nitrate (CAN) is mixed with ethanol or butanol? I have mixed CAN in butanol and also ethanol (separately), the solution became dark red immediately but after a couple of minutes the color totally faded and turned into a transparent solution, I am curious to understand the reason behind it.

Recall that Ce(IV) is a pretty strong oxidizing agent. So oxidation of alcohols and fading of the color is expected. The mystery lies in the immediate intense red color. It has been relatively recently studied in great detail and well known long time ago.
See these references:

V. Briois, D. Lützenkirchen-Hecht, F. Villain, E. Fonda, S. Belin, B. Griesebock, and R. Frahm, "Time-Resolved Study of the Oxidation of Ethanol by Cerium(IV) Using Combined Quick-XANES, UV−Vis, and Raman Spectroscopies," J. Phys. Chem. A 2005, 109(2), 320–329 (DOI: https://doi.org/10.1021/jp046691t).

Lewis Brewster Young and Walter S. Trahanovsky, "Oxidation of organic compounds with cerium(IV). VII. Formation constants for 1:1 cerium(IV)-alcohol complexes," J. Am. Chem. Soc. 1969, 91(18), 5060–5068 (DOI: https://doi.org/10.1021/ja01046a021).

Well, you can see Ce(IV), alcohols form transient 1:1 complexes with alcohols, but such complexes don't last long because of oxidation. In ref. 1, one can deduce that alcohol replaces one of monodentate ligand (nitrate), and thus absorption spectrum of this complex changes for a few seconds.


Is the 'sweat' on the side of bottles distilled water? Are the water droplets that form on the side of water bottles (or other drink containers) distilled water? Assume that the sides of the container are completely clean.

Yes, here is the definition for distilled.

purify (a liquid) by vaporizing it, then condensing it by cooling
the vapor, and collecting the resulting liquid.

The sweat on the water bottle is the process of distillation, but only the condensation part of the process is seen on the water bottle.


Why is the number 2 needed for 2-methylpropanal? I think there is just one possible position for the methyl group to 'fit in' to propanal, however, my textbook and wiki suggests the correct names should be '2-methylpropanal'. Why is the '2' needed?

Well I suppose you are correct in a way- but it does help us see the molecule quicker in our heads...

it should be written methyl-2-propan-1-al or methyl-2-propan-2-al
otherwise it would be a butan-1-al  or butan-2-al

let me explain further>
the new IUPAC nomenclature  advises the use of numbers instead of the iso epithet- in organic chemistryIn old naming -as with propanol and iso-propanol - so with propanal

Isopropyl alcohol

Preferred IUPAC name

Propan-2-ol

-
Isopropyl alcohol (IUPAC name propan-2-ol; commonly called isopropanol) is a compound with the chemical formula C3H8O. -It is a structural isomer of 1-propanol.

wiki  Chemical Names: ISOBUTYRALDEHYDE; 2-Methylpropanal; 78-84-2; Isobutanal; Isobutylaldehyde; Propanal, 2-methyl- More... Molecular Formula: C4H8O or (CH3)2CHCHO.


Reaction of nitrobenzene with sodium metal in ethanol The reactant is nitrobenzene(only) and reagent is Na metal in solvent liquid ammonia along with ethanol (which i am assuming is playing the role of proton donor please correct me if i am wrong).

The first thing that comes in mind is "alkyne" but there is not any alkyne in nitrobenzene( not even in its resonance structures).How this reaction will proceed?

In answer, the DBE is just reduced by one.I was also trying what if it was toluene?Can i have a explanation based on mechanism?

Benzene Rings or Substituted Phenyl rings undergoes reduction in presence of $\ce{Na}$ in liq. $\ce{NH3}$ in presence of $\ce{EtOH}$ solvent, and generally forms substituted Cyclohexa-$1,4$-dienes.

This reduction generally follows Single Electron Transfer (SET) Mechanism.The single unpaired electron of Sodium is transferred to the Benzene ring forming a radical which subsequently forms a Carbanion. This carbanion eventually takes the Proton from $\ce{EtOH}$ and becomes Cyclohexa-$1,4$-diene.

But in the first step, the transfer of electron by $\ce{Na}$ occurs in that place, where the intermidiate formed after this step gets maximum stabilisation. In case of Electron-withdrawing group, such as $\ce{-NO2, - COOH, -SO3H, -CHO, -COOR, -COR }$ etc., that transfer occurs para to that group, as the negative charge generated is highly stabilised by that electron-withdrawing group. The mechanism thus in case of Nitrobenzene is as follows,

But if the group is electron donating such as $\ce{-CH3, -OH, -OMe, -NMe2 }$ etc. The transfer will occur in such a way to stabilise the generated negative charge by keeping it in meta to the group (where the effect of electron donation by the group is least). For those compounds, the mechanism and corresponding product will be as follows,

Thus, in your case, the product will be $\text{3-nitrocyclohexa-1,4-diene}$ as described in the first mechanism.


The major product of the following reaction Here,1,2-dibromobutane is reacted with
i) alc. $\ce{KOH}$
ii) $\ce{NaNH2}$ in liq. $\ce{NH3}$

According to me, since alc. $\ce{KOH}$ causes dehydrohalogenation, using Saytzeff's rule the product obtained after reaction (i) should be one of the either:

And after reaction (ii), since $\ce{NH2-}$ is a strong base, an E2 reaction proceeds and again using Saytzeff's rule the final product should be:

However, according to the answer key the final product obtained is given as:

Can someone help by explaining where and how I went wrong in my attempt to solve the question?

When you form the allene, the methylene hydrogen has enhanced acidity because deprotonating it leads to a carbanion in which the negative charge is delocalized:

$\ce{CH3-CH=C=CH2} - \ce{H^+ -> CH3-CH=C=CH^- <-> CH3-C^-H-C#CH}$

The proton then reattaches to the opposite end of the allene chain to form a more stable product.  In effect this is a (strong) base catalyzed tautomerization.


heat of fusion in an equation You add $100.0\ \mathrm g$ of water at $55.0\ \mathrm{^\circ C}$ to $100.0\ \mathrm g$ of ice at $0.00\ \mathrm{^\circ C}$.  Some of the ice melts and cools the water to $0.00\ \mathrm{^\circ C}$.  When thermal equilibrium is established at $0.00\ \mathrm{^\circ C}$, what mass of ice has melted?
  $\Delta H$ of fusion for water is $333\ \mathrm{J/g}$.

My question is I got a negative for $q$.  Would this change to a positive because you need heat to melt the ice?  I no you can't have a negative. So I am not sure if it should be a neg or a positive

$q = m \times \text{specific heat} \times (T_\text{final} - T_\text{initial})$
specific heat of water is $4.186$ (should memorize or on a chart)
$q= 100\ \mathrm g \times 4.186\ \mathrm J \times (0 -55) = -23\,023\ \mathrm J$

mass ice × heat of fusion
$q= m \times H_\text{fusion}$
$-23\,023\ \mathrm J = m \times 333\ \mathrm{J/g} = 69.12\ \mathrm g$

If you go by thermodynamic conventions,

$$\begin{align}q > 0 &\implies \text{ heat is gained by system}\\
q < 0 &\implies \text{ heat is lost by system} \end{align}$$

You initially considered water as your system, and it lost heat, therefore  $q$ was negative.
Now when you do calculation for ice, it becomes your system, and the heat is gained by ice, $\implies$ positive $q$


Does a first order reaction really get completed only at infinity? For a $1^{\text{st}}$ order reaction $ A \rightarrow B$, the corresponding rate law is: $$\text{rate} = k[A]^{1}$$ and the integrated rate law for this is $$ \ln \left(\frac{A}{A-x}\right) = kt$$

Going by this equation, every first-order reaction must get completed only at infinity. Is this true? If not, then what is the correct explanation?

To add to the other answers that only address the mathematical behavior of the first order rate equation close to $t=\infty$, let me address what actually happens physically.

Before you reach $[A]_t=0$, the first order rate equation actually breaks down. It isn't valid anymore, because one of the assumptions on which it is based, namely that the reaction rate is deterministic, breaks down.

When you get to really low quantities of (one of) your reactants, reaction events, i.e. two molecules meeting, are blind luck. Actually, for high quantities this is also the case, but then the behavior is a physical version of the law of large numbers: a probabilistic system behaves deterministic on average.

You can simulate this basically by flipping coins whether a reaction occurs or not. If you do this for a simple $A$ to $B$ reaction you get graphs like the following:

Here the chance of reaction per timestep for each molecule $A$ is 10%. You can see that the curves at low number of molecules $N$, are jumpy (and actually would look different everytime I rerun my calculation), while for $N=1000$ you already see the fairly smooth and familiar first order behavior. Just note that close to t=10 also the $N=1000$ case is behaving jumpy again as you can see from the zoom below.


Nicotine from tobacco? I'm reading a couple of methods of extracting nicotine from tobacco. I have read that many people trying to commit suicide with nicotine. The method is simple:

Boil some tobacco with water
Leave it for a few hours
filter the solution
boil it till you get a thick black syrup thing

That's pure nicotine and they claim that 2-3 drops of that can kill you. Is that true? What do you think?

I am not thinking to commit suicide. I just asking because it is really  weird that so little a dose can be fatal.

MaxW's comment linked to the relevant Wikipedia page:

  Nicotine is regarded as a potentially lethal poison.[17] The LD50 of nicotine is 50 mg/kg for rats and 3 mg/kg for mice. 30–60 mg (0.5–1.0 mg/kg) can be a lethal dosage for adult humans.[11][56] However the widely used human LD50 estimate of 0.5–1.0 mg/kg was questioned in a 2013 review, in light of several documented cases of humans surviving much higher doses; the 2013 review suggests that the lower limit causing fatal outcomes is 500–1000 mg of ingested nicotine, corresponding to 6.5–13 mg/kg orally.[13] Nevertheless, nicotine has a relatively high toxicity in comparison to many other alkaloids such as caffeine, which has an LD50of 127 mg/kg when administered to mice.[57]

You ask:

  That's pure nicotine and they claim that 2-3 drops of that can kill you. Is that true?

"Drops" is an imprecise non-quantitative term, but let's suppose that a drop is about 20 μL in volume.  Then 3 drops is 60 μL or 0.06 mL.  The density of the nicotine "oil", if pure, is about the same as water's.  So 60 μL should equal about 60 mg of nicotine.  According to the paragraph from Wikipedia, this corresponds to a widely used figure for a lethal dose.

However, note the text later in the paragraph that a lethal dose is closer to 500-1000 mg.  That would require ten to twenty drops, not just two or three.

Undoubtedly what is a "lethal" dose varies widely from person to person and I have no doubt that consuming 40 or 60 mg of nicotine at a time is very dangerous.  That's the same amount of nicotine as smoking 50-70 cigarettes or chewing ten to fifteen pieces of nicotine gum at the same time.  That can't be good for you.

  Boil some tobacco with water

The word you haven't thought about is "some".  According to a different Wikipedia page:

  It constitutes approximately 0.6–3.0% of the dry weight of tobacco[6]

Since plants are probably 70% water, that means that nicotine is roughly 0.2% to 1% of the wet weight of tobacco leaves.  So to obtain three drops of pure nicotine (approximately 60 mg), we would need at a minimum 6 g of tobacco.  That isn't very much! But keep in mind this calculation doesn't allow for any losses during extraction, and also I'm not sure if the Wikipedia figure for the nicotine content of tobacco refers to "fermented" leaves or to freshly harvested leaf tissue.  So for a variety of reasons this calculation could be way off.


In the reaction between salicylic acid and acetic anhydride to produce aspirin, why is it necessary to quench the reaction? I understand that water is added to hydrolyse acetic anhydride into acetic acid, but what is the point of this?

Acetic anhydride is contaminating your final product. Quench simply means to destroy excess reagent before continuing. It might not matter too much in this particularly reaction since you're about to purify via recrystallization anyway, but imagine if the next step involved adding a very expensive nucleophile. That nucleophile would add to acetic anhydride and be wasted.


Green Tea and Caffeine Is the caffeine from green tea that is extracted acidic?
This goes for powder and liquid form.

If I add Alkaline water (pH: 7 or higher) will it change?

The caffeine is in a free base form unless prepared otherwise . Caffeine is an alkaloid and a base , slightly alkaline. If you add base like naoh to it the solution will become more alkaline.


What chemicals cause the smell of a septic tank? I am wondering what chemicals exactly contribute to the smell of a septic tank system. I read about $\ce{H2SO4}$, $\ce{NH3}$, but I am not sure. Is methane also a factor or is that odorless?

According to the Wikipedia article Septic tanks, as septic tanks are generally anaerobic environments, pungent smells from the formation of hydrogen sulfide $\ce{H2S}$, which is sometimes referred to as 'rotten egg gas'. As you stated, ammonia $\ce{NH3}$ can be produced - also with a distinct smell.

Also, according to the Wikipedia article, fermentation could occur, producing the odourless methane and carbon dioxide as well as phosphates.

Incidentally, $\ce{H2SO4}$is the chemical formula for sulfuric acid.


A contradiction regarding the reaction coefficient expression in the Nernst equation I came up with the following question in my NCERT textbook:

  Write the Nernst equation and the emf of the following cell at $\pu{298 K}$.

  $$\ce{Fe}|\ce{Fe^2+}(\pu{0.001 M})||\ce{H+}(\pu{1 M})|\ce{H2 (g)}(\pu{1 bar})|\ce{Pt (s)}$$

  Given: $E^\circ(\ce{Fe^2+}/\ce{Fe}) = \pu{-0.44 V}$.

As per my understanding, the data of activity coefficients must given because the reaction coefficient in terms of concentration is not dimensionless and its logarithm would be meaningless.
$$\ce{Fe + 2H+ -> Fe^2+ + H2}$$
$$Q=\ce{[Fe^2+]/[H+]^2}$$
Secondly, using a unit of concentration other than mol/litre would definitely yield a different value. So I decided to check its solution from a reference book. In the solution, they have used the above expression of molar concentrations.

So, my question is, is this expression correct?

This is why for equilibrium expressions we use activity and fugacity. Both of these values are dimensionless and are generally quite closely related to the concentration and partial pressure, respectively, for reasonable concentrations.


Rate of reaction graph An autocatalytic reaction is a reaction in which one of the products catalyses the reaction.
In general what will be the shape of curve if the rate of reaction was plotted against time for an autocatalytic reaction?

I think answer should be A (in image) as an autocatalyst should always increase rate of reaction until there is a limiting factor for example concentration but the answer is B.

As Curt F. has correctly showed that, in general, if the rate of reaction is plotted against time for an autocatalytic reaction, it would be the shape of curve B (in your question). Since Curt F. has already worked with the mathematical part, only I can do is show you some literature evidence. One of research on an epoxy resin composite hardening (Ref.1) state that:

  Composite structures for aerospace applications are mainly made by the well-known prepreg technology. In order to achieve adequate prepreg processing schedules, and consequently maximum fiber strength utilization, one has to know in deep the cure kinetics of matrix, which held the fibers together. This work describes a procedure to study the cure kinetic and has as example how aromatic amine hardeners influence the cure kinetics of an epoxy resin used in advanced composites. The investigation was carried out by using the DSC technique and it was found that depending on the system used the cure kinetics of the formulation obeys order $n$ or autocatalytic order.

For Comprehensive details for autocatalyzation and review, see Ref.2 & 3.

References:

Michelle L. Costa, Luiz C. Pardini, Mirabel C. Rezende, “Influence of aromatic amine hardeners in the cure kinetics of an epoxy resin used in advanced composites,” Mat. Res. 2005, 8(1), 65–70 (http://dx.doi.org/10.1590/S1516-14392005000100012).
Donna G. Blackmond, “Asymmetric autocatalysis and its implications for the origin of homochirality,” Proc. Natl. Acad. Sci. USA 2004, 101(16), 5732–5736 (https://doi.org/10.1073/pnas.0308363101).
Andrew J. Bissette, Stephen P. Fletcher, “Mechanisms of Autocatalysis,” Angew. Chem. Int. Ed. 2013, 52(49), 12800–12826 (https://doi.org/10.1002/anie.201303822).


How exactly do sulfites prevent oxidation of wine? I know that sulfites are used in wines to prevent oxidation, but how exactly does this happen, which reactions take place and where do the products go?

Sulphites work as anti-oxidants in wine because they are very easy to oxidise to sulphates. Schematically:

$$\ce{SO_3^{2-} + \text{oxidant}\to SO_4^{2-}}$$

This ease with which the sulphite ion is oxidised to the sulphate ion means that sulphite added to wine tends to 'mop up' any oxidisers (like air oxygen) that might get into the wine, rather than the alcohol (and maybe other substances in the wine) becoming oxidised.

The sulphite method has the added advantage that it, as well as its sulphate reaction product, is quite harmless to humans.


Why should a complexometric titration be kept neutral? In my next lab, I'll be doing a complexometric titration of a copper solution with EDTA and a murexide indicator. The lab guide advises to keep the pH level nearly neutral by adding a diluted $\ce{NH3}$ solution. Why is that necessary?

I understand that during the formation of the copper complex $\ce{[Cu(EDTA)]^{2-}}$, the solution becomes more acidic. Do we need to add the basic solution in order to make sure the murexide does not change color due to the decreasing pH level? Or is it simply so that we can revert the color change (of the formation of the copper complex) and continue the titration until the equivalence point is reached (and all copper ions are in complex form)?

Your last guess is exact. This is a classical method, the equivalent point is seen by our eyes, so in our interest is for the complex to be formed so we can recognize the colour, and we should add $\ce{NH3}$ so the colour won't change immediately after the equivalent point.


Ozone is above, nitrogen is below Ozone is heavier than nitrogen. Then why is ozone in stratosphere and nitrogen in troposphere?

The entirety of the Earth's atmosphere can be divided into two regions: the homosphere and the heterosphere. The former extends from the earth's surface up to about $\pu{100km}$.  Within this region, turbulence and mixing, even within the relatively calm, persistent temperature inversion known as the stratosphere, prevents any segregation of gases based on molecular/atomic weight.  With the exception of water vapor, the composition of the major constituent gases only varies by a few hundred ppm from any given region of the homosphere to another.

The heterosphere extends from about $\pu{100km}$ above the earth's surface upward until merging with the interplanetary medium at around $\pu{100,000km}$. Within this region, the mean free path of any gas is great enough that gravitational forces have a chance to segregate the different species based on their mass, rather than by bulk processes like turbulence.

The Encyclopaedia Britannica describes the heterosphere this way:

  In the heterosphere, the concentrations of the heavier constituents,
  such as nitrogen or oxygen, decrease more rapidly with increasing
  altitude than do the concentrations of lighter gases, such as hydrogen
  or helium; and eventually the atmosphere is dominated by the lighter
  gases. Under mean daytime conditions, helium and its ion become
  dominant around $\pu{1,000 km}$ ($\pu{620 miles}$) and hydrogen and protons above
  $\pu{2,500 km}$ ($\pu{1,555 miles}$).

At sufficient altitude, $\ce{H+}$ becomes the dominant species and hydrogen is the only element of significance that escapes the atmosphere based on its low mass.

So, to reiterate the answer to the question; the troposphere and stratosphere both lie well within the homosphere, and thus segregation based on mass plays no roll in their distribution throughout this region of the atmosphere.


Why does coke foam up when I add it to (cold) black coffee (no sugar, no milk)? Album of the different gifs detailing this situation: Album of coke-coffee related gifs
I thought that the answer would be similar to the ice-cream float situation, but I don't see what in the coffee would cause the soda to float.
Could it be some reaction with the coffee oils and the coke?

Above is a shot of my materials:

Can of Coke Zero (ingredients: Carbonated water, caramel color,
phosphoric acid, aspartame, potassium benzoate, natural flavors,
potassium citrate, acesulfame potatssium, caffeine)
Nespresso coffee    (ingredients: Roast and ground coffee, hot water)

Above is after the coffee cooled.
Experiments:
Here is a link to a gif of me pouring the coffee:
Gif of me pouring coke into the coffee
To address one of the comments, here is a gif of me pouring coke into an empty bottle: Gif of me pouring coke into an empty bottle, comparison
To address another comment, link to me pouring coke into a mug of tea (instead of coffee): Gif of me pouring coke into a mug of tea

There are two likely contributing factors here. What you're seeing is rapid release of carbon dioxide from solution, indicating that its solubility has decreased and that the rate of re-equilibrating to that new lower solubility is quite fast. The solubility of carbon dioxide is a factor of

1) the partial pressure of CO2 in the air above the liquid (higher pressure forces more into solution),

2) the pH of the liquid (because higher pH leads to formation of more $\ce{HCO3-}$ and $\ce{CO3^2-}$ thus increasing the total CO2-related species in solution), and

3) temperature of the solution (like most gases, CO2 decreases in solubility as temp increases).

In your case, (1) and (3) are probably most important. [EDIT: I overlooked that the coffee was cold, so (3) is actually not likely to be relevant] The can before opening is pressurized with a high CO2 atmosphere, so it is overloaded with CO2 relative to just about any ambient condition. In order to equilibrate with ambient conditions, it needs to lose some CO2. On the other hand, you've diluted it into the coffee, so there is a slight offsetting effect there. The coffee is slightly acidic, but is not likely to be very well buffered, and probably is not much different than diluting with water.

In addition to these thermodynamic effects that change the equilibrium dissolved CO2 concentration, there is also the kinetic issue of how quickly the CO2 escapes in order to return to equilibrium. The CO2 can only leave solution at the boundary points where the liquid meets air. The stirring effect of pouring the liquid (even into an empty glass) creates more surface area of the liquid/air boundary and also keeps the CO2 well distributed throughout the solution so that the areas near the boundary are continuously resupplied with CO2 as it is lost to the air.

To determine if coffee specifically has any effect, I would try the same thing with water instead of coffee. I suspect the result will be very similar.

UPDATE based on new data:
I was wrong about the coffee itself not being important. The coffee mixture clearly foams more than the tea. There are two possibilities I can think of (not mutually exclusive):
1) the rate of CO2 evolution is greater in coffee. As the other answer postulates, that could be due to faster nucleation by particles in the coffee or, if the pH goes up very quickly, buffering by acids in the coffee once the pH reaches 5 or so (from the Coke starting point ~ pH 2).
2) the same amount of CO2 is released, but the coffee mixture foams more. Studies on sparkling wines such as the one described here indicate that proteins, especially ones with carbohydrates attached, contribute to foaming by acting as surfactants (that means like a detergent molecule, essentially making soap bubbles). It is likely that coffee contains similar molecules. I now think this is the likely explanation, though I also like the idea that there are more nucleation sites from the dissolved solids.

The way to test some of these hypotheses would be to filter the coffee with varying molecular weight cut-off filters, but I suspect you don't have access to that sort of equipment?


Why are protic acids not used for the formation of thioacetals? The standard procedure for protecting a carbonyl group as an acetal uses dry acid, such as $\ce{TsOH}$, but when forming a thioacetal, the Lewis acid $\ce{BF3}$ seems to be the catalyst of choice. For example, see Corey and Seebach's original demonstration of dithiane acidity: Org. Synth. 1970, 50, 72.

Is there any particular theoretical or practical reason why this is the case? I am guessing that the boron trifluoride coordinates exclusively to the harder oxygen atom, making it a better leaving group and hence favouring the formation of products (since the reaction is under thermodynamic control). However, I don't see any reason why acid wouldn't be able to do the job equally well.

The pKa/leaving group ability between alcohols and thiols may have something to do with it.

  Alcohol pKa around 17, thiol around 13.

As you pointed out the whole reaction is in equilibrium. But in the case of the thioacetal, after the initial addition theres a strong driving force to kick back out the sulfur to reform the carbonyl rather than committing to the thioacetyl formation since the sulfur is a relatively stable leaving group.

Protic acids are almost always wet, and as such the water would prevent the reaction, whereas Lewis acids are generally either neat, or in ethereal solutions.


Yield of possible products from acid catalysed hydration of an alkene with a phenyl group Here, (d) may be the major product due to high temperature.
My questions are:

Will a kinetically favoured (minor) product (c) form?

What shall be the order of yield of products?

A basic principle in OChem says that stability of intermediate is the key factor in deciding the major product, in general. Another factor is stability of final product.

Will a kinetically favoured (minor) product (c) form?

It will form, as far as it's intermediate is known to exist (and it exists). However, that will be present in very minor amount, as the intermediate of it will not be as stable as that of (b) and (d). Additionally, the stable intermediate of (c) has 7 α-H, while that of (a) has 2 α-H, so product (c) will be major than product (a).

Here, (d) may be the major product due to high temperature.

The intermediates of (b) and (d) are equally stable. So, here we use the second factor. By looking at the structures of product (b) and product (d), we can say that (d) is major product due to extended conjugation and hyperconjugation with 9 α-H. Really?
The answer is YES. It is known that acid-catalyzed hydration reaction is highly exothermic$1$, which is why it is favored at low temperatures! (In accordance with the "Le Chatelier's Principle", a system will consume heat when temperature of surrounding is raised).
Now, I think, it's pretty clear what the order of yield of products should be :)

References
$1$: https://en.wikipedia.org/wiki/Hydration_reaction#Alkenes


Finding EMF of a galvanic cell without standard potentials For the galvanic cell
$$\ce{Ag | AgCl(s), KCl (\pu{0.2 M}) || KBr (\pu{0.001 M}), AgBr(s) | Ag}$$
find the EMF generated given $K_\mathrm{sp}(\ce{AgCl}) = \pu{2.8e-10},$ $K_\mathrm{sp}(\ce{AgBr}) = \pu{3.3e-13}.$

This is the question from JEE exam (1992).
How to start solving the problem since the $E^°$ of individual half reactions is not given? How to write the $E^°$ for the cell without it? Or is it not needed?

How to start solving the problem since the $E^\circ$ of individual half reactions is not given?

This is a concentration cell, i.e. the half reactions at the anode and at the cathode are the same (except for the direction).

AgCl(s) electrode
$$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}$$
$$\ce{Ag+(aq) + e- -> Ag(s)} $$

AgBr(s) electrode
$$\ce{Ag(s) -> Ag+(aq) + e-}$$

$$\ce{Ag+(aq) + Br-(aq) <=> AgBr(s)}$$

The standard reduction potentials will cancel out, i.e. $E^\circ (\mathrm{cell}) = 0$.

Further thoughts

  [Comment by EdV] The Ag|AgCl electrodes I have are commercial, but making them in the lab is just a matter of oxidizing Ag wire in a chloride solution, so the electrode is Ag wire with an adherent coating of AgCl... I have never seen one of these made by just sticking an Ag wire in the Ag halide, but I guess it would work.

I added more to the answer prompted by that thoughtful comment.

  [...my own comment] I pictured the silver electrode submerged in the solution, with the solid halide on the bottom. I am puzzled now too. Does it make a difference if the electrode touches the solid, the liquid, or both?

What are the actual reduction potentials?

$$\ce{AgCl(s) + e− <=> Ag(s) + Cl−}\ \ \ \ E^\circ_\mathrm{red} = \pu{+0.22233 V}\tag{1}$$
$$\ce{AgBr(s) + e− <=> Ag(s) + Br−}\ \ \ \ E^\circ_\mathrm{red} = \pu{  +0.07133 V}\tag{2}$$
$$\ce{Ag+ +  e−  <=> Ag(s)}\ \ \ \ E^\circ_\mathrm{red} = \pu{  +0.7996 V}\tag{3}$$

Are they related?

If you subtract (3) from (1), you get the dissolution reaction of AgCl, if you subtract (3) from (2), you get the dissolution reaction of AgBr. Thus, standard reduction potentials for (1) and (3) should be different by

$$ -\frac{RT}{zF} \ln K_\mathrm{sp}(\ce{AgCl})$$

and standard reduction potentials of (2) and (3) should be different by

$$ -\frac{RT}{zF} \ln K_\mathrm{sp}(\ce{AgBr})$$

Finally, standard reduction potentials (1) and (2) should be different by

$$ -\frac{RT}{zF} \ln \frac{K_\mathrm{sp}(\ce{AgBr})}{K_\mathrm{sp}(\ce{AgCl})}$$

Numerical answer using half reaction (1) and (2)

$$\ce{AgCl(s) + Br-(aq) <=> AgBr(s) + Cl-(aq)}$$

$$Q = \frac{[\ce{Cl-}]}{[\ce{Br-}]} = 200$$

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

$$= \pu{(0.22233 V− 0.07133 V) - 0.13612 V = 0.0149 V}$$

Numerical answer using half reaction (3) twice

$$\ce{Ag+(c) + Ag(b) <=> Ag(c) + Ag+(b)}$$

"c" stands for chloride side, and "b" stands for bromide side. For consistency, I am using the following values for the solubility products (derived from difference of standard reduction potentials of half reactions (1), (2) and (3)).

$$K_\mathrm{sp}(\ce{AgCl}) = \pu{1.74e−10}$$
$$K_\mathrm{sp}(\ce{AgBr}) = \pu{4.89e−13}$$

$$[\ce{Ag+}]_c = K_\mathrm{sp}(\ce{AgCl}) / [\ce{Cl-}] = \pu{8,27e−10}  $$
$$[\ce{Ag+}]_b = K_\mathrm{sp}(\ce{AgBr}) / [\ce{Br-}] = \pu{4.89e−10}  $$

$$ Q = \frac{[\ce{Ag+}]_c}{[\ce{Ag+}]_b} = 0.560 $$

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

$$\pu{= 0 - (-0.0149 V) = 0.0149 V}$$


Final concentration of stock solution upon dilution If we have a stock solution that is $\pu{1 mg/mL}$, and we have $\pu{0.1 mL}$ of that stock to which we add $\pu{0.9 mL}$ of water, what is the final concentration?

For this I assumed I had to use the concentration formula:

$$\mathrm{M_{1}V_{1} = M_{2}V_{2}}$$

I thought the initial concentration was $\pu{1 mg/mL}$, since that is my stock. I thought the initial volume was $\pu{0.1 mL}$, since that is what they needed from the initial stock. I assume the final volume is $\pu{1 mL}$, since that is the total volumes of the components.

$$\mathrm{(1\;mg/mL)\cdot(0.1\;mL) = (M2)\cdot(1\;mL)}$$

$$\mathrm{M_2 = \frac{0.1\;mg}{1\;mL}}$$

$$\mathrm{M_2 = 0.1\;mg/mL}$$

I feel I am right due to the units evaluating properly out, since the original concentration is giving in $\mathrm{mg/mL}$ , the final should be also in $\mathrm{mg/mL}$, right?

You are correct.  You might find it more intuitive to think of it this way:

In the end you will have $\pu{1 mL}$ of solution. In that end solution you have $\pu{0.1 mL} \times \frac{\pu{1mg}}{\pu{mL}}$ of stock, or (cancelling the $\pu{mL}$s) $\pu{0.1mg}$ of the chemical in question.  Therefore, final concentration is $\frac{\pu{0.1mg}}{\pu{1mL}} = 0.1 \frac{\pu{mg}}{\pu{mL}}$.


How do I determine the molecular vibrations of linear molecules? I know how to do it for just about every other point group, but the $D_{\infty \mathrm h}$ and $C_{\infty \mathrm v}$ character tables aren't as straightforward. In particular, I'm interested in the vibrational modes of carbon dioxide, $\ce{CO2}$.

$$\begin{array}{c|cccc|cc} \hline
C_{\infty\mathrm{v}}  & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} &  & \\ \hline
\mathrm{A_1} \equiv \Sigma^+ & 1 & 1 & \cdots & 1 & z & x^2 + y^2, z^2 \\
\mathrm{A_2} \equiv \Sigma^- & 1 & 1 & \cdots & -1 & R_z & \\
\mathrm{E_1} \equiv \Pi      & 2 & 2 \cos\phi & \cdots & 0 & (x,y), (R_x,R_y) & (xz,yz) \\
\mathrm{E_2} \equiv \Delta   & 2 & 2 \cos 2\phi & \cdots & 0 & & (x^2-y^2,xy) \\
\mathrm{E_3} \equiv \Phi     & 2 & 2 \cos 3\phi & \cdots & 0 & & \\
\vdots & \vdots & \vdots & \ddots & \vdots &  & \\ \hline
\end{array}$$

$\,$

$$\small \begin{array}{c|cccccccc|cc} \hline
D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline
\mathrm{A_{1g}} \equiv \Sigma^+_{\mathrm{g}} & 1 & 1 & \cdots & 1 & 1 & 1 & \cdots & 1 &  & x^2 + y^2, z^2 \\
\mathrm{A_{2g}} \equiv \Sigma^-_{\mathrm{g}} & 1 & 1 & \cdots & -1 & 1 & 1 & \cdots & -1 & R_z & \\
\mathrm{E_{1g}} \equiv \Pi_{\mathrm{g}} & 2 & 2\cos\phi & \cdots & 0 & 2 & -2\cos\phi & \cdots & 0 & (R_x,R_y) & (xz,yz) \\
\mathrm{E_{2g}} \equiv \Delta_{\mathrm{g}} & 2 & 2\cos 2\phi & \cdots & 0 & 2 & 2\cos 2\phi & \cdots & 0 & & (x^2-y^2,xy) \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots & & \\
\mathrm{A_{1u}} \equiv \Sigma^+_{\mathrm{u}} & 1 & 1 & \cdots & 1 & -1 & -1 & \cdots & -1 & z & \\
\mathrm{A_{2u}} \equiv \Sigma^-_{\mathrm{u}} & 1 & 1 & \cdots & -1 & -1 & -1 & \cdots & 1 & & \\
\mathrm{E_{1u}} \equiv \Pi_{\mathrm{u}} & 2 & 2\cos\phi & \cdots & 0 & -2 & 2\cos\phi & \cdots & 0 & (x,y) & \\
\mathrm{E_{2u}} \equiv \Delta_{\mathrm{u}} & 2 & 2\cos 2\phi & \cdots & 0 & -2 & -2\cos 2\phi & \cdots & 0 & & \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots & & \\ \hline
\end{array}$$

As Tyberius noted, the projection formula does not work for infinite order groups (this is because the Hermitian form on characters is defined to be G-invariant by averaging over all elements in a group. That is, it works because it is possible to hit all the elements of the group in some order).

How can one work in the infinite dimensional groups then? One property that still holds for these particular infinite groups (though not in general 1) is Maschke's Theorem, that every representation is the direct sum of irreducible representations. So we can still determine what irreps make up a rep by adding them up.

First we need to find the representation we wish to reduce. I find that the easiest way to do this is to start with just the atoms (instead of the vectors) and then include the vector contribution afterwards. For the sake of a new example, I will do the linear $\ce{I4^{2-}}$ anion instead of $\ce{CO2}$.

$$\small \begin{array}{c|cccccccc} \hline
D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline
\Gamma_\text{atoms} & 4 & 4 & \cdots & 4 & 0 & 0 & \cdots & 0 &\\
\end{array}$$

Now, we multiply by the vector's contributions to the character.
$$\begin{array}{c|c}E&3\\\hline C_2&-1\\\hline \sigma&1\\\hline i&-3\\\hline C_n&1+2\cos(\frac{2\pi}{n})=1+2\cos\theta\\\hline S_n&-1+2\cos(\frac{2\pi}{n})=-1+2\cos\theta\end{array}$$

These come from taking the trace of the matrix representations of each symmetry operation. For example, a rotation about the z-axis by an angle $\theta$ is given by:
$$\left[\begin{array}{ccc}\cos\theta&\sin\theta&0\\-\sin\theta&\cos\theta&0\\0&0&1\end{array}\right]$$

$$\small \begin{array}{c|cccccccc} \hline
D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline
\Gamma_\text{xyz} & 12 & 4+8\cos\theta & \cdots & 4 & 0 & 0 & \cdots & 0 &\\
\end{array}$$

Now, we are only interested in the vibrational modes so we must subtract off the translational and rotational degrees of freedom. We can check the character table for these (they are the ones which correspond to $z$, $(x,y)$ and $(R_x, R_y)$, $R_z$ doesn't count because we define the molecular axis to be the $z$-axis.
$$\small \begin{array}{c|cccccccc} \hline
D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline
\Gamma_\text{xyz} & 12 & 4+8\cos\theta & \cdots & 4 & 0 & 0 & \cdots & 0 &\\
-\Gamma_\text{rot} & 2 & 2\cos\theta&\cdots&0 & 2 & -2\cos\theta&\cdots&0 \\
-\Gamma_\text{trans} & 3 & 1+2\cos\theta&\cdots& 1 & -3 & -1+2\cos\theta&\cdots&-1 \\
\hline\Gamma_\text{vib} & 7 & 3+4\cos\theta&\cdots& 3 & 1 & 1&\cdots&1 \\
\end{array}$$

So now we have the actual representation of the vibrations. How do we go about decomposing it? We want to eventually subtract off all of the irreps until we have the zero vector so we consider what effect each irrep has and try to move our remaining rep closer to zero. The first thing to note is that only $\Pi_g$ and $\Pi_u$ irreps have $2\cos\theta$. $\Pi_u$ adds $2\cos\theta$ under both proper and improper rotations, while $\Pi_g$ adds to proper and subtracts from improper. We note that in our rep, we only have $\cos\theta$ terms under proper rotations and so we need to subtract off enough $\Pi_u$'s until there are the same number of $\cos\theta$ (with different signs) under proper and improper rotations, at which point we should subtract off $\Pi_g$s until there are no $\cos\theta$ terms.

$$\small \begin{array}{c|cccccccc} \hline
D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline
\Gamma_\text{vib} & 7 & 3+4\cos\theta&\cdots& 3 & 1 & 1&\cdots&1 \\
-\Pi_u & 2 & 2\cos\theta&\cdots&0 & -2 & 2\cos\theta&\cdots&1 \\
-\Pi_g & 2 & 2\cos\theta&\cdots&0 & 2 & -2\cos\theta&\cdots&1 \\
\hline\Gamma & 3 & 3 &\cdots& 3 & 1 & 1&\cdots&1 \\
\end{array}$$

Similar to the $\Pi_g/\Pi_u$, the $\Sigma_g^+$ shift adds 1 to all positions, while $\Sigma_u^+$ adds 1 to all positions before $i$ and subtracts from all after. We subtract enough $\Sigma_g^+$ until all the positions before and after $i$ have equal magnitude.
$$\small \begin{array}{c|cccccccc} \hline
D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline
\Gamma & 3 & 3 &\cdots& 3 & 1 & 1&\cdots&1 \\
-2\Sigma_g^+ & 2 & 2 &\cdots&2 & 2 & 2&\cdots&2 \\
-\Sigma_u^+ & 1 & 1 &\cdots&1 & -1 & -1&\cdots&-1 \\
\hline \Gamma& 0 & 0 &\cdots&0 & 0 & 0&\cdots&0
\end{array}$$

Thus, we have found that the vibrational modes of $\ce{I_4^{2-}}$ to be $2\Sigma_g^+ +\Sigma_u^+ + \Pi_u + \Pi_g$.

We can check this with the formulas given in Group Theory And Chemistry by David M. Bishop as noted by Feodoran. One can also follow Tyberius's suggestion and work in $D_{2h}$ which will give the decomposition $2A_g+B_{1u}+B_{2u}+B_{3u}+B_{2g}+B_{3g}$.

It is important to note that correlation from the $D_{2h}$ point group to $D_{\infty h}$ works fine for vibrations because you cannot have terms higher than $\Pi$ and so there is no ambiguity. For things like electronic states, though, in which direct products must be decomposed, you will run into some problems. If the $D_{2h}$ result includes $A_g+B_{1g}$ does that correlate with a $\Sigma_g^+ +\Sigma_g^-$ or a $\Delta_g$ state in $D_{\infty h}$? Working in the parent infinite group may be more difficult without the reduction formula, but these problems are avoided.

The reason going to lower symmetry works is because all characters under the infinitely many proper and improper rotations are the same for each irrep (namely $2\cos n\theta$) and so we can represent the behavior of all of them by the behavior of one of them. Hence, we pick $\theta=\pi$ and ignore all the others and this produces the $D_{2h}$ point group (the two improper rotation axes are equivalent to $i$ and the two proper axes are equivalent to one $C_2$ about $z$). This is origin of the ambiguity. In $D_{\infty h}$, $\Sigma_g^+ + \Sigma_g^-$ are equal to $\Delta_g$ at $\theta=\pi$, but not in general.

Having now identified the origin of the ambiguity, one can tune how fast these ambiguities occur though. The problem occurs when the $\theta$ chosen to represent all $\theta$ has the property $1=\cos n\theta$ where $n$ represents the period of the ambiguity. So for $D_{2h}$, $\theta=\pi$ and $1=\cos2\theta$ and the period of the ambiguity is 2. That is, every other term will be ambiguously identical ($\Sigma, \Delta, \Gamma, \ldots$ are the same and $\Pi,\Phi,\ldots$ are the same). By dropping to $D_{4h}$ instead, $\theta=\pi/2$ and $1=\cos4\theta$. The ambiguity period is longer and we can distinguish $\Delta$ from $\Sigma$ by correlation from the $D_{4h}$ point group. We can thus select the point group to descend to based on how much resolving power we expect to need in the parent point group (though we should stick to even numbers to maintain inversion symmetry for $D_{\infty h}$). Of course we could also stay in the parent point group and never have a problem, but also have no reduction formula.


Can neutralization ever be complete? Suppose we have 1 eq of acetic acid and we want neutralize it using sodium hydroxide. After adding 1 eq of sodium hydroxide 1 eq of sodium acetate should form and neutralization should be complete but because of hydrolysis little bit of acetic acid will form again so we would have to add more sodium hydroxide and this process would go on forever so how will neutralization ever be complete?

My guess is that after adding 1 eq of NaOH the conc. of acetic acid becomes so low that it is neglected and the neutralization is assumed to be over.

The $\ce{K_a}$ expression for acetic acid, in aqueous solution, is given by:

$\ce{K_a} =\dfrac{\ce{[CH3COO-][H+]}}{\ce{[CH3COOH]}}$

The expression can be rearranged to give

$\ce{[CH3COOH]} = \dfrac{\ce{[CH3COO-][H+]}}{\ce{K_a}}$

so no matter what the pH, all the acetic acid will never be reacted.

So in a titration let's assume that we start out with 0.1000 moles of whatever acid $\ce{[HA]_{initial}}$.

After base has added and $0.1000 \gg \ce{[HA]_{final}}$ and thus $0.0001 \gt \ce{[HA]_{final}}$, then the reaction is  quantitatively complete.

This is strictly a significant figures argument. You can't do a titration to 10 significant figures. 3 is decent, if you get 4 you're a magician.
When 0.1000 moles of base have been added to the solution the equivalent point has been reached.

The key point is that the moles of base is equal to moles of acid, regardless of the pH.
When ever you get the "signal" that the titration is over you have reached the end point.

The signal could be, for example, a color change of an indicator, the swing of the needle on a pH meter, or the formation of a precipitate.


Why do the reactions of Magnesium in steam and cold water differ? "Magnesium has a very slight reaction with cold water, but burns in steam.

A very clean coil of magnesium dropped into cold water eventually gets covered in small bubbles of hydrogen which float it to the surface. Magnesium hydroxide is formed as a very thin layer on the magnesium and this tends to stop the reaction.

Magnesium burns in steam with its typical white flame to produce white magnesium oxide and hydrogen." - From Chemguide.

May I know why the reactions of Magnesium in steam and cold water differ?

Magnesium is a very active metal- highest activity of all engineering metals. Your question gives you part of the answer. In cold water it forms a decent protective layer. Given greater activation energy- simple as hot water, that layers is simply damaged the oxidation of the magnesium proceeds unimpeded at a greater rate. Once its in steam the oxidation is extremely rapid.


Which thermodynamic ensemble does Gaussian's BOMD functionality sample? Someone sent me a Born-Oppenheimer molecular dynamics output with the following settings line:

# b3lyp/sto-3g BOMD=(MaxPoints=35, Ntraj=1, Update=15, RTemp=298) guess=read geom=allcheck

Based on my understanding of the BOMD keyword documentation, it sounds like the temperature is only used for putting energy into the vibrational modes, which are then used to initialize the nuclear velocities. Since there is no mention of a thermostat in the output or the documentation, this implies that calculations can only be run in the NVE ensemble.

Are my assumptions correct?

@pentavalentcarbon: As of now (Gaussian16), I believe that BOMD calculations can only be run in the NVE ensemble.

However, you can do ADMP (also available with Gaussian and uses an extended Lagrangian scheme), and offers equivalent functionality to BOMD. Thermostatting is available here, but the caveat is it implements only the simple velocity rescaling algorithm. On performing simple tests this seems to be "nearly" canonical on some systems I have worked on. i.e small ion-water clusters.


Measuring a high Michaelis constant using fluorescence We have the task of measuring kinetic parameters of an oxidase reaction that has a $K_M$ of about $2 \,\text{mM}$. For that, we want to use a fluorescence assay based on Amplex Red. The latter is a fluorogenic substance of an upper detection limit of $5 \, \mu \text{M}$. At higher concentrations, the inner filter effect causes non-linear behaviour.

Usually, we would use substrate and fluorophore concentrations of a similar range of the $K_M$ value. Obviously, we cannot do that here, as the fluorophore concentration would be way above the upper detection limit. This led us to the idea of reducing the substrate, fluorophore, and enzyme concentration by the same factor to do measurements in a micromolar range. Later we could simply scale up the results.

We doubt that this method is possible as the Michaelis-Menthen kinetics never look at the ratio between substrates and enzyme, but only regard absolute substrate concentrations.

So our question is if the proposed method is in fact not applicable and if so, what your suggestions would be?

Based on your description, it sounds like you are measuring the production of hydrogen peroxide by an oxidase enzyme. In that case, it is not necessary for the Amplex Red to be at the same concentration as the substrate. You only need enough reagent to determine the initial rate of reaction at each substrate concentration, ie when only a small portion of the substrate has been consumed.
If $\pu{5 μM}$ is not sufficient for that purpose, you can determine the resorufin (produced by reaction of Amplex Red with $\ce{H2O2})$ by absorbance rather than fluorescence for lower sensitivity and higher maximum. According to the Thermo Fisher assay kit info, resorufin has an extinction coefficient of $\pu{58000 M^-1 cm^-1},$ so with a standard $\pu{1 cm}$ path length cuvette, your linear range would probably go up to around $\pu{50 μM}$ depending on your spectrophotometer. With a short path cuvette (these are available with $\pu{0.2 cm}$ and $\pu{0.1 cm}$ path lengths), you could go quite a bit higher, certainly more than sufficient for your needs.
As for why you cannot just scale the concentrations down, consider the limiting case where the Michaelis constant is equal to the $K_\mathrm{d}$ of the substrate. We have that
$$K_\mathrm{d} = \frac{[\ce{E}][\ce{S}]}{[\ce{ES}]},$$
so $[\ce{S}] \approx K_\mathrm{d}$ when $[\ce{E}] = [\ce{ES}]$ (assuming $[\ce{S}]_\mathrm{tot} \gg [\ce{ES}]).$ Thus, the absolute concentration of $\ce{E}$ is not a factor, as long as it is small enough that the amount of $\ce{S}$ bound does not substantially change the concentration of free $\ce{S}.$ You should get the same result for $K_\mathrm{M}$ regardless of what concentration of enzyme you use.


Phase Diagrams and Equilibrium In this link

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch14/phase.php

It says (in the latter part ) that for all combinations of Pressure and Temperature along line BC

the rate of boiling of liquid to form a gas= the rate of condensation of gas to form a liquid.

And C is the Critical Point.  Similar thing has been said  for other lines also

Now 100℃ (or boiling point of the liquid) will lie somewhere on BC

So according to the link rate of boiling of water to form gas is the same as the rate of condensation of gas to form water.

But I have read that at boiling point all the liquid changes to gas or there is no equilibrium at that point.

So why then it says- the rate of boiling of liquid to form a gas= the rate of condensation of gas to form a liquid.

I am not good at chemistry so this might not be a good question. But I was trying to understand phase Diagrams for the first time !

In the phase diagram of a pure substance there are regions of a single phase e.g. solid, liquid or vapour. The boundary lines you refer to are where the substance exits as $2$ two phases, say solid and vapour in equilibrium with one another.

The Gibbs phase rule states that the number of degrees of freedom, F,  which is the number of (intensive) variables that can be varied is given by
$$F = C-P+2$$
where C is the number of components, $1$ for a pure substance, and P the number of phases, which is $1$ anywhere in the diagram except on a boundary line. Therefore the number of degrees of freedom is $2$, (except on a boundary line) which means that pressure and temperature can both be varied independently.

The boundary lines indicates the condition of P and T where there are two phases in equilibrium, say solid and vapour. As there are $2$ phases there is now only $1$ degree of freedom so that temperature and pressure are no longer able to be varied independently. Thus on the liquid /vapour curve, if the pressure is increased some vapour is condensed (heat is given out) and the temperature increases to a new value but one that that is still on the boundary line, i.e. vapour is still in equilibrium with the liquid. Along any boundary curve the two phases are in equilibrium and so, e.g., the rate of melting is equal to rate of freezing. (Any point along the liquid/vapour boundary line can be considered to be boiling, but we define the normal boiling point to be the boiling temperature at $1$ atm pressure.)

At the triple point, where the three boundary curves meet, there are no degrees of freedom and this point is fixed. At and above the critical point the density of the liquid and vapour are the same and the liquid's meniscus  disappears. The super-critical fluid looks very much like the swirling foam that one sees for a short while on pouring a bottle of Guinness into a glass.

The boundary lines can be calculated for a substance using the Clausius and Clausius-Clapeyron equations.


Ionic Product of Water Why is the ionic product for water ($K_\text{w}$) temperature dependent? The concentration of $\ce{H+}$ and $\ce{OH-}$ ions in pure water is $[\ce{H+}] = [\ce{OH-}] = 1.0\times 10^{-7}\ \mathrm{mol/l}$ at $25\ \mathrm{^\circ C}$, and therefore the ionic product of water at $25\ \mathrm{^\circ C}$ is $1.0\times 10^{-14}$. Why it is so? What is the effect of temperature on this value?

I think this link should really help you out in understanding how the equilibrium constant depends on temperature.

Now, considering the case of water dissociation:
$$\ce{H2O <=> H+ + HO-}$$

We can calculate the free energy of the reaction ($\Delta_\text{r}G$), using the Gibbs energy for the reactant and the products (in aqueous solution):

$$\Delta G(\ce{H2O}) = -237.13\ \mathrm{kJ/mol}$$
$$\Delta G(\ce{H+}) = 0\ \mathrm{kJ/mol}$$
$$\Delta G(\ce{HO-}) = -157.24\ \mathrm{kJ/mol}$$

$$\Delta_\text{r}G = \Delta G(\ce{H+}) + \Delta G(\ce{HO-}) - \Delta G(\ce{H2O}) = -157.24\ \mathrm{kJ/mol} + 237.13\ \mathrm{kJ/mol} = 79.89\ \mathrm{kJ/mol}$$
The values of the Gibbs energies were taken from the 8th edition of Atkins' Physical Chemistry.
Now, applying the formula below (which can be obtained from $\Delta_\text{r}G = -RT\ln K_\text{eq}$) for $T = 298\ \mathrm{K}$ (or $25\ \mathrm{^\circ C}$):
$$K_\text{eq} = \operatorname{e}^{-\frac{\Delta_\text{r}G}{RT}}$$
we get something like this :
$$K_\text{eq} = 2.71^{-\frac{79890\ \mathrm{J\ mol^{-1}}}{8.3145\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 298\ \mathrm{K}}} = 9.93 \times 10^{-15}$$
This value can be approximated to ${10^{-14}}$.


Noyori hydrogenation and the Curtin–Hammett principle My question is relating to how the Curtin–Hammett principle applies to homogeneous hydrogenation using chiral ligands.

Figure 0: An example of a quadrant diagram used as a mnemonic to predict the facial selectivity of reduction (in this case of a 3-hydroxyketone)

We commonly draw mnemonics (i.e. quadrant diagrams) to 'predict' the most likely products, however this prediction surely works on the assumption that the energy of the intermediate complexes is responsible for the overall outcome of the reaction since in using these mnemonics we're assuming that steric clashes are what determines the 'lowest energy' binding mode (which is supposedly not the case in the example in figure 1).

Does the mnemonic take into account this kinetic preference, or are we only able to predict the product for hydrogenations not taking place via Curtin–Hammett kinetics? (and if so, can one generalise if the Curtin–Hammett principle is applicable for a given reduction? I seem to recall reading something along the lines of Ru(II) and Ru(I) systems operating by different kinetics, for example).

Ruthenium-BINAP complexes (and related complexes using modified ligand systems) are commonly employed for asymmetric homogeneous hydrogenation/reduction of a wide range of substrates (the general caveat is the need for 2 point binding in order to get suitable control).

An example is below in figure 1 using a diphosphine ligand to selectively reduce the enamine, which introduces a new stereocentre and essentially furnishes an amino acid derivative.

Figure 1: Example of a Ru-catalysed asymmetric hydrogenation (from Organotransition metal chemistry, from bonding to catalysis, Hartwig).

When thinking about these reactions, we often use 'quadrant diagrams' and other mnemonics in order to predict how the substrate will bind (based on minimising steric interactions), but essentially binding of the substrate to the complex yields two diastereomeric transition states which are able to interconvert (figure 2).

Figure 2: Proposed 'mechanism' of a Ru-catalysed asymmetric hydrogenation  (from Organotransition metal chemistry, from bonding to catalysis, Hartwig).

What is unusual in this reaction is that the energy (and hence population) of the diastereomeric intermediates does not have any influence on the ratio of products formed, with the higher energy/less populated intermediate being the one going on to form the product. This is an example of Curtin–Hammett kinetics, with the oxidative addition step, in this case, being the thing that determines which product forms.

Figure 3: Reaction profile (from Organotransition metal chemistry, from bonding to catalysis, Hartwig).

There's no way to predict this.

Mnemonics are just rules, and rules are just rules. There will be exceptions, and you won't know when you should apply them a priori. If you could, it would be part of the rule by definition.

The Curtin-Hammett conditions apply when the rate of interconversion between intermediates is significantly faster than the rate of reaction from the intermediates onto their respective products. If the rate of interconversion is slow, then the ratio of products is simply governed by the ratio of intermediates.

The problem comes about from determining if the interconversion is fast. If you're looking at proton transfers or some kinds of conformational changes, it's usually pretty straightforward. Otherwise, only intuition can guide you ahead of time. Once you have experimental results, you might be able to rationalize this fact and determine that the reaction is governed by Curtin-Hammett kinetics, but there's no way to know for sure. This is just due to the complexity of binding entropy, solvent effects, etc, coupled with the intrinsic barrier for the forwards reaction.

Chemistry is hard.


Resonance stabilization and size of ligand atoms I am told that for these two molecules, one of them is not as resonance stabilized as the other. Apparently it's the chlorine one, and it's because of the mismatch in the size of chlorine and carbon. This makes sense prima facie, but can someone confirm this is true? I can see how carbon and nitrogen have similarly sized p orbitals and thus have good overlap, and how chlorine and carbon might have poor p orbital overlap, like a BB pellet with a basketball.

Yes, this is true. The $\pi$ overlap between a valence $\ce{p}$ orbital of a second row element and one of a higher row element is much smaller than the $\pi$ overlap between two second row elements. One reason is that the bond length is generally higher when higher row elements are involved (due to their size). Another reason is, that the $3 \ce{p}$ or $4 \ce{p}$ orbitals are much more diffuse than the $2 \ce{p}$ orbitals. If you want to read deeper into the topic have a look at:

[1] W. Kutzelnigg, Angew. Chem. Int. Ed. Engl., 23, 272 (1984).

[2] N. C. Norman, Polyhedron, 12, 2431 (1993).


Is pitchblende the same as Uraninite? Different websites say different things on the matter.

 For example:

 This Website   Holds that pitchblende is also known by the name Uraninite.

This Website   Holds that pitchblende is a type of Uraninite.

My question is, which website is correct? I have always known the first site to be a reputable source, so I assume that this is a contentious topic.

If you search online for Uraninite then you find various websites saying that uraninite is called pitchblende.

On wikipedia you find :

  Uraninite is a radioactive, uranium-rich mineral and ore with a
  chemical composition that is largely UO2, but also contains UO3 and
  oxides of lead, thorium, and rare earth elements. It is most commonly
  known as pitchblende

even the word pitchblende will automatically redirect you to the uraninite wikipedia page .

Another trustworthy website is mindat.org : uraninite information can be found on this link:http://www.mindat.org/min-4102.html (It seems that there are various names associated to the mineral):

  Re-named in 1845 by Wilhelm Karl von Haidinger for its composition
  containing uranium. Originally recognized before the element uranium
  was known and called "schwarz beck-erz" by Franz Ernst Brückmann in
  1727. Subsequently described with several names including "pseudogalena" and "pitch-like zinc-blende" by Johan Gottschalk
  Wallerius (Vallerius) in 1747. Named pechblende by Axel Cronstedt in
  1758. Named "uranerz" by Martin Klaproth after he discovered the element uranium in this mineral. Many spelling variations of the above
  names known.


Why does the mole-fraction equilibrium constant depend on pressure and volume? The ratio of rate constants for the forward and backward reactions gives us the equilibrium constant, $K_c$. For a sample reaction:

$$
\ce{A \underset{k_{-1}}{\overset{k_1}{<=>}}  P + Q}\quad \Longrightarrow \quad K_c = {k_1\over k_{-1}} = {C_P C_Q\over C_A}
$$

Per kinetic theory, rate constants depend only on temperature and the presence of any catalysts:

$$
k_i = A_i e^{-E_{a,i}\over RT}
$$

Thus, their ratio (i.e., the equilibrium constant, $K_c$) also must depend only on temperature:

So: Why does $K_x$, the mole-fraction equilibrium constant, depend on pressure and volume?

TLDR; Because only the concentration equilibrium constant is directly related to the rate constants themselves.

Given that pressure has a minimal effect on equilibria in solution, I'm going to answer assuming you're interested in gas-phase equilibria.  The first step in showing why pressure/volume matter to $K_x$ is to move to the partial-pressure equilibrium constant, $K_p$. For a system composed entirely of ideal gases, the partial pressure of each species is related to its concentration as:

$$
{P_i\over RT} = {n_i\over V} = C_i
$$

Substituting this into the expression for $K_c$ for your example reaction gives:

$$
K_c = {C_P C_Q\over C_A} = {1\over RT}{P_P P_Q\over P_A} = {1\over RT}K_p
$$

Thus, for this reaction:

$$
K_p = RT\,K_c \tag{1}\label{Kp}
$$

For other reactions, the power of $RT$ on the RHS of Eq. $\eqref{Kp}$ can differ, depending on the change in the sum of the stoichiometric coefficients between the reactants and the products.

Since $K_c = f(T)$  and $K_p = (RT)^nK_c$, then $K_p$ is also dependent only on temperature (with ideal gases only!), not pressure or volume.

Where pressure/volume enter the picture is when one goes from $K_p$ to $K_x$.  Again assuming ideal behavior, the mole fraction of a species is defined as:

$$
P_i = x_i P
$$

Substituting this into the $K_P$ expression:

$$
K_p = {P_P P_Q\over P_A} = P {x_P x_Q\over x_A} = PK_x
$$

Thus, for this reaction:

$$
K_x = {1\over P}K_p \tag{2}\label{Kx}
$$

Similar to the above, the power of $1\over P$ in Eq. $\eqref{Kx}$ can vary, depending on the change in the sums of the stoichiometric coefficients between the reactants and the products.

In any event, this is where the dependence on $P$ enters the expression for $K_x$.

This dependence on pressure implies a related dependence on volume, since by the ideal gas law:

$$
V = {nRT \over P}
$$

and, equivalently:

$$
P = {nRT \over V}\tag{3}\label{igl-P}
$$

If one is working with a system where the volume and temperature are the control variables but the pressure is not, then in general the pressure will vary implicitly according to Eq. $\eqref{igl-P}$. If the temperature and/or volume change in any way other than proportionally $($i.e., ${T\over V} = \text{constant})$, then the pressure dependence of Eq. $\eqref{Kx}$ will manifest itself in a fashion that looks like a volume dependence.


Mixing 1 cubic meter of HCl and NaOH: Temperature rise of a metal container? Say I (perfectly) mix $1\ \mathrm{m^3}$ of 1 M $\ce{HCl}$ with $1\ \mathrm{m^3}$ of 1 M $\ce{NaOH}$ at standard air pressure and room temperature within a metal vessel. Heat of neutralization for strong acids and bases is $-55.8\ \mathrm{kJ/mol}$ (is that actually just the bond enthalpy of an $\ce{H-OH}$ bond?).

So the $2\ \mathrm{m^3}$ of $\ce{H2O}$ amount to around $2 \cdot 10^6\ \mathrm{mol}$ of water ($M(\ce{H2O}) \overset{\wedge}{=} 20\ \mathrm{g\ mol^{-1}}$). Therefore, $\Delta_\mathrm n H$ is around $2 \cdot 10^6 \times 55.8\ \mathrm{kJ/mol}$, around $110 \cdot 10^6\ \mathrm{kJ}$.

I would now like to calculate the temperature rise.

The heat released is $Q = m c_p \Delta T$, where $m$ is the mass of the solution ($2000\ \mathrm{kg}$), $c_p$ is the specific heat capacity of the solution and $\Delta T$ ist the change in temperature. But that's exactly what I'm trying to find. So unless I know $Q$, I can't find $\Delta T$. Therfore, how do I find $Q$?

You have 1 mole/liter of each reactant.  How many liters of each solution do you have?  That gives you the number of moles of reactants. They will create way less than $2\cdot 10^6$ moles of water-you were counting the water that was already there.  Once you have the right number of moles of reactant, that is the number of moles of water produced.  Multiplying by your heat per mole gives $Q$.  Then divide by $mc_p$ to get $\Delta T$


Is there an actual specific heat for aluminum in its gaseous state? I've scourged the internet and I can't seem to find it anywhere although I've found the specific heat values for aluminum in both solid and liquid states. I'm trying to construct a heating curve and I need the specific heat capacity of aluminum in its gaseous state in order to calculate the amount of energy that would be necessary to bring aluminum from its liquid state to a gaseous state.

By "bring aluminum from its liquid state to a gaseous state", I assume you mean to a gaseous state at a temperature above aluminum's boiling point—since to merely convert it from a liquid to a gas, you only need the enthalpy of vaporization.
According to https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node18.html,
most metallic vapors have molar heat capacities very close to those of ideal gases, i.e., $C_v = 3/2 R = 12.47$ J/(mol K) and $C_p = 5/2 R = 20.79$ J/(mol K).  Thus you can obtain aluminum's specific heat capacity by dividing by the atomic mass, giving $C_v = 0.46$ J/(g K)  and $C_p = 0.77$ J/(g K)
The reason for this is that metals don't form metallic bonds in the gaseous state (metallic bonding is a collective phenomenon that requires bulk metal), and are thus monatomic*.  And the heat capacities of monatomic gases are simple, since they have only translational degrees of freedom.
[*As real gases, they may not be exactly monatomic, since they can still transiently associate via van der Waals forces.  But treating them as monatomic is a good approximation.]
The exception to the above would occur when the temperature is sufficiently high to cause the metal to ionize, in which case you would have a higher heat capacity because you would now have two (or more) particles for every ionized aluminum atom. But since aluminum's first ionization energy is 5.985 eV/atom = $577$ kJ/mol, and $RT$ at $3000$ K is $25$ kJ/mol, I don't expect there to be significant ionization close to its $2792$ K boiling point. [Also, some might argue that, once it starts ionizing, it is no longer a gas, but a gas-plasma mixture.]


Does knowing the ratio of masses of the elements in a compound lead to the unique chemical identity of the compound? I was posed this question from a friend, however I am not necessarily sure. I think: yes, because a molecule of any compound is composed of a whole-number ratio of atoms of its elements, per the law of definite proportions. However, I feel as if I am wrong.

No, knowing the mass ratios is not sufficient by itself. In the absence of additional information (for example, molar mass) that would only be enough information to determine the empirical formula, which is the formula that contains the smallest integral ratios of atoms. (E.g., the empirical formula of $\ce{C2H6}$ would be $\ce{CH3}$.) There are (theoretically) infinitely many molecules with the same atomic and mass ratios, yet differing in the actual molecular formula.

Further, even given the additional information of molar mass (so that the molecular formula can be determined), the existence of isomers precludes identification of a singular, specific compound without more sophisticated analysis. For example, $\ce{C3H6}$ could be either cyclopropane or propene. $\ce{C3H6O}$ could be as innocuous as acetone, or it could be highly carcinogenic epoxypropane. With more atoms, the number of possible isomers grows very rapidly. There's a nice table giving the number of isomers of alkanes of different lengths in this article. By the time you get to $\ce{C30H62}$, the number is $4111846763$. (Note, this is only considering structural isomers; accounting for stereoisomerism, the number would be even more astronomical.)