Gibbs free energy of chemical compounds How the Gibbs free energy of a chemical compound is theoretically calculated from scratch?

When saying that the Gibbs free energy of NaCl is -384.1 kJ/mol, this value is related to the chemical bonds of Na-Cl.

We should be able to calculate this value from the sum of energies in chemical bonds of one mole of NaCl? Right? How?

The value you have reported is the standard Gibbs free energy change of formation, which has a specialized definition. A standard free energy change of formation is the free energy change that occurs when 1 mole of substance is formed by reaction of its constituent elements in their standard states at 1 atmosphere of pressure and 25 degrees Celsius.

Thus, for sodium chloride:

$$\ce{Na(s) +\frac{1}{2} Cl2(g) -> NaCl(s)}\ \ \ \Delta_f G^\circ =-384.1 \text{ kJ/mol}$$

Now, all we need to be able to do is measure $\Delta G$ for this reaction at standard temperature and pressure.

In practice, we will separately detemrine $\Delta_f H^\circ$ and $\Delta_f S^\circ$ and plug them into $\Delta_f G^\circ = \Delta_f H^\circ -T\Delta_f S^\circ$.

At constant pressure (which we are by definition operating at),  $\Delta H$ is equal to the heat transfer  $\Delta H = q = -C_p\Delta T$, which is equal to the heat capacity of the surrounding (in practice the calorimeter, which is reasonably isolated from the universe) multiplied by the change in temperature.

$\Delta S$ is still tricky to measure. At equilibrium, for a reversible process at constant temperature $\Delta S = \frac{q}{T} = \frac{\Delta H}{T}$. If the process is not reversible or you are not at equilibrium, then... then it is challenging.


Why can't we rotate a carbon while checking for chirality? If I were to rotate the front carbon clockwise by $60^\circ$, it would become achiral. But my text book says that it's chiral. Is it that rotating an individual atom about a sigma bond changes its conformation is that why we can't rotate it?

It simply depends on what is the subject. The book asks for the displayed conformation, which we agree is chiral.

The question would have been more tricky to answer if it were about optical activity of a sample.

In this case we know the conformer can not be isolated at room conditions but this is not a general rule.

And one can always cool down, at least in principle.


Why not an Avogadro's principle for solid matter? Avogadro stated that for any ideal gas, 1 mole of particles will occupy a specific volume at a specific ratio of pressure and temperature. So why doesn't this apply to a liquid or solid?

Avogadro's law, which can be written as $V \propto n$, where $V$ is the volume of the gas and $n$ is the amount of substance of the gas (measured in moles), can be thought of as just another manifestation of the ideal gas law rewritten as follows,
$$
  V = (RT/p) n \, .
$$
Consequently, strictly speaking, Avogadro's law is applicable only for a hypothetical ideal gas, but since real gases under usual conditions (such as standard temperature and pressure) are not far away in their properties from an ideal gas, Avogadro's law usually works well enough for them also. Substances in condensed phase on the other hand are milage away from an ideal gas in their properties, and thus, Avogadro's law is inapplicable for such cases.

To clarify in which sense Avogadro's law is inapplicable for substances in condensed phases, note that for solids and liquids, as well as for gases, the volume is proportional to the amount of substance. But Avogadro's law states not just the proportionality. It is also insists that the coefficient of the proportionality is the same for each and every gas, and this is the thing which can not be generalised to substances in condensed phases. For real gases, the ideal gas description which depends on the following assumptions (too name just a few):

molecules are almost point-like (effectively zero-sized) particles;
molecules do not exhibit any intermolecular forcers;

usually works as a good approximation. But for liquids and solids such a description a priori is physically horrible: molecules here are relatively bigger and more tightly packed so that their size and intermolecular interaction between them couldn't be ignored anymore.


Hybridisation of carbon atom in cyclopentadienyl anion How do I figure out the hybridisation of the carbon with the negative charge?

One logic is that it is an $\mathrm{sp^3}$ carbon because there are 3 sigma bonds and 1 lone pair around it.

Another reasoning, though, is that it is $\mathrm{sp^2}$; we don't count the lone pair, because it is in conjugation with the double bonds.

What is the correct explanation?

It entirely depends on what that electron pair is doing.

If the lone pair is hanging out by itself in its own orbital, then the carbon is probably sp3 hybridized, as greater hybridization is typically more stable. This is what you'll see on an alcohol or an amine lone pair. Those lone pairs aren't really interacting with anything else in the molecule, hence the "standard" logic of the rule you invoke.

However, in certain compounds you're able to stabilize the system further by delocalizing the electron pair. In this sort of case, the sp3 hybridized lone pair orbital is not able to participate in the delocalization, and thus will be higher in energy than an electron pair that is in the remaining p orbital of an sp2 hybridized carbon. This is the sort of thing you see with amide groups, where the "lone pair" on the nitrogen is sp2-like, which allows the electron pair to delocalize into the carbonyl, allowing for resonance stabilization.

Note that delocalization is not always going to happen. You can potentially draw up examples where delocalization would decrease the stability of the compound. For example, 1,4-Dihydropyrazines would be anti-aromatic if both lone pairs from each nitrogen were to participate in the ring system. This is highly disfavorable, and the molecule contorts itself to avoid it, putting the nitrogens into sp3 (pyramidal) hybridization, except where it needs to be planar to conjugate with an exocyclic double bond.

In this particular case - the one of the cyclopentadienyl anion - you would be able to form an aromatic system if the lone pair were to delocalize into the ring - much like pyrrole does. The lowered energy obtained from this aromatic delocalization more than makes up for the small amount of energy gained by going from sp2 to sp3, so the carbon should be sp2 hybridized.

The important thing here is not simply to invoke rote-memorized rules to determine hybridization states, but think about what the electrons are doing and what the comparative stability of the various options are. The hybridization will change to be the most stable (lowest energy) state. -- To be perfectly honest, neither sp2 nor sp3 is the "true" answer. The actual   answer to these sorts of questions lies more toward Molecular Orbital Theory, where the molecule is taken as a whole. Per-atom hybridization is a convenient shorthand, but still one which ultimately rests on the basis of the true molecular stability.


What makes teflon such a strong material? I don't know much about polymers but recently teflon made me get addicted to study their properties and nature.

As far as I know, the teflon polymer contains fluorine which is one of the most reactive elements but in the polymer state it is totally stable and moreover, it is applied as a coating to kitchen-wares and many utensils which are exposed to very high temperatures.

How did a reactive element like fluorine got such an exciting stability?

Teflon (PTFE, polytetrafluoroethylene) is prepared by the polymerization of tetrafluoroethylene as shown below.  If we took polyethylene and replaced all of the hydrogens with fluorines, we would have PTFE.

Here is a Table that compares the strengths of some single bonds involving a methyl group.

\begin{array}\hline
Experimental ~Bond~ Enthalpies for~ CH_3 - X~ (kcal/mole)\\ \hline
\end{array}
\begin{array}{|c|c|c|}\hline
I & 57.6    \\ \hline
Br & 72.1    \\ \hline
Cl & 83.7    \\ \hline
CH_{3} & 90.1    \\ \hline
H & 104.9    \\ \hline
F & 115.0    \\ \hline
\end{array}

The $\ce{C-F}$ bond is one of the strongest bonds to carbon known.  As you add additional fluorine atoms to the carbon, the bond strength increases even further.  While the $\ce{C-F}$ bond is a covalent bond, it has considerable ionic character making the carbon positively polarized and the fluorine negatively polarized.  This bond polarization results in a strong electrostatic attraction between carbon and fluorine which accounts for the high bond strength observed.  It is this very high bond strength that makes fluorocarbons so stable (unreactive).  There are not many conditions that can supply enough energy to break this very strong bond.


"Number of moles of electrons involved in the reaction" for hydrogen cell I understand we can calculate the maximum work done by the electromotive force (driving electrons through a cell) by multiplying the potential of the cell with the charge of one mole of electrons (Faraday's constant) and with "the number of moles of electrons involved in the reaction". I thought that in the case of a hydrogen cell: the reaction $\ce{H2 - 2e- -> 2H+}$ involves two electrons but I can see on wikipedia that in the Nernst equation for the standard hydrogen electrode $z$ (the number of moles involved) is considered to be 1.

So what does this $z$, or expression "electrons involved" actually mean?

Nernst equation is used for finding the cell potential when the various reactants, i.e. anodic and cathodic electrolytes are not in their standard states ($298\text{K}, 1\text{mol l}^{-1}$). Thus, in such cases, you first write down any balanced form of equation, which for our hydrogen electrode, I choose to write down $\ce{3H2 -> 6H+ +6e-}$ and these electrons are used in a standard $\ce{H2}$ cell on the other electrode. Hence the Nernst equation would give the cell potential for this concentration cell as ($E^\circ=0$ since the the standard potentials of both sides is the same since the materials are same, and the non-standard hydrogen cell is the anode)
$$E_{cell}=-\frac{RT}{6F}\ln\frac{[H^+]^6}{1}$$
By using the properties of logarithm, you can show this is equivalent to $-\frac{RT}{1F}\ln[H^+]^1$, and this is the equation if you took only 1 electron transfer in the reaction.
It does not matter what equation you write, since in the final analysis, the number of electrons taken are compensated by the power of the concentration term in the logarithm.


Synthesis of synthetic detergent with Friedel Crafts Acylation I propose the following synthesis.

Oxidation of aldehyde to a carboxylic acid. This sets me up for FC acylation once I ...

Attack the -OH with thionyl chloride ad form the acid chloride.

Reaction with vanilla benzene in the presence of Lewis acid catalyst. Benzene won't go after acid chlorides unless there is some catalyst in the system.

Now that the acid chloride is installed, I can go through the motions of getting rid of the pesky C=O group.

Given that alkyl groups are slightly activating and are ortho/para directing, I can sulfonate with fuming sulfuric acid and expect a mixture of products (including the one desired by the synthesis). Sterics might favor the para product over the ortho product given the sheer size of the SO3 group.

General comments:
I did NOT sulfonate first, as that is an electron-withdrawing group akin to the nitro group. It deactivates the ring and is also a meta-director. I want a para-substituted product. Not a meta-substituted product.
Therefore, how reasonable is my synthesis? Is it correct?
Also would sulfonation before getting rid of the C=O work? Something tells me no; the C=O is able to withdraw electrons from the ring via resonance, and that would make it a meta-director.

Your synthesis looks reasonable, and it is correct to perform sulfonation as the very last step, because the alkanoyl group would be meta-directing for the reason you have provided.

You can simplify your synthesis by reducing the carbonyl group away in one step with the Clemmensen reduction, which works well for aryl-alkyl ketones produced by Friedel-Crafts acylation.

Alternatively, the Wolff-Kishner or Mozingo reaction could be used. Reducing the number of steps in your synthesis would in practice have the benefit of increased overall yield for the final product, as the yield of each single step is usually < 100 %.


Why is the eclipsed conformer more stable than the bisected conformer in 1-butene? I have just started learning conformational analysis, and a major doubt came into my mind.

In simple alkanes such as ethane, the staggered conformer is much more stable than the eclipsed conformer. This fact may be explained as follows:

In the staggered conformer, the repulsion between the hydrogens is minimized as they are "anti" with respect to each other.
We have a case of favorable overlap between sigma bonding MO of the C-H bond, and sigma anti bonding orbital of the C-H bond of the second Carbon.

I have no issues with this part. Then, I started reading about alkenes. In the simple example of propene, two cases may arise, as shown in the figure.

Note that (2) can be obtained from (1) by rotating the C-C bond by 60 degrees. Unlike in the case of alkanes, the eclipsed form is more stable, not the bisected form. This can be easily accounted for:
In the eclipsed form, the Hydrogen is oriented in the right way to allow for effective Hyperconjugation. This involves delocalization of electrons from C-H sigma bond to the C=C pi antibonding orbital.

My main question is about when it becomes more complicated. Let me take the example of 1-butene. Here, we have twice the number of eclipsed and bisected conformers. I have labelled them as E1, E2, B1 and B2 (E= eclipsed;B=bisected)

The correct order of stability is E2 > E1 > B1 > B2.
I do not understand this order. When I attempted to find the order of stability, I got: E2 > B1 > B2 > E1. I based this mainly on steric effects.

E2 is clearly the most stable as we have Hyperconjugation due to the C-H bond which is oriented in just the right way.
I couldn't get the rest of the order. Why is E1 the second most stable? I understand that it is eclipsed, but shouldn't that lead to greater torsional strain? Is Hyperconjugation the reason for stability, again?
Can Hyperconjugation take place when we have a C-C bond(when there is no alpha Hydrogen with respect to the double bond), or is it restricted to C-H bonds(where we have an alpha Hydrogen)?

In alkanes staggered conformers (all anti) are local minima, while eclipsed conformers represent transition states. The first is "stable", while the latter is not. The rotational barrier has been measured and calculated often, it is around 12 kJ/mol.[1]

In alkenes the eclipsed conformers are local minima, while the staggered, or more accurately bisected, conformers are transition states. The rotational barrier for propene is only about 8 kJ/mol.[2]

I have performed a full, relaxed rotational scan of 1-butene around the central carbon-carbon bond axis (blue). The minima and maxima of this scan have been reoptimised and thermal corrections at 298.15 K and 1 atm have been added (orange).

The chosen level of theory, DF-BP86+D3(BJ)/def2-SVP, is not the most accurate, but it will give you a general idea. You can see the bisected conformations are maxima along the rotational coordinate and the eclipsed ones are minima.
There is dispersion correction included, which makes the E1 conformer slightly more stable than the E2 conformer.
Including the thermal corrections, the energy difference between these two structures is negligible.
$$\begin{array}{crrr}
\text{Conformer} &
\angle_\mathrm{D} &
\Delta E~/\mathrm{kJ\,mol^{-1}} &
\Delta G~/\mathrm{kJ\,mol^{-1}}\\\hline
\mathrm{E_1}     &   0.0             &     0.0  &   0.0   \\
\mathrm{B_2}     &  60.0             &     9.9  &  10.1   \\
\mathrm{B_1}     & 120.0             &     1.2  &  -0.2   \\
\mathrm{E_1}     & 180.0             &     9.7  &   9.4   \\\hline
\end{array}$$

The eclipsed forms are stabilised by two π-σ/π-σ hyperconjugation interactions. Here is the visualisation for E1 in the natural bond orbital picture. Occupied orbitals are depicted blue/orange, virtual orbitals are depicted red/yellow. The phases are assigned randomly.

In the B2 conformer the overlap between those orbitals is not ideal. The nodal planes in the anti-bonding orbitals are in the way.

The representations for E2 are analogue to E1, and B1 is similar to B2. Because of time reasons I have not been able to create the pictures yet.

Taking the maximum difference in the energy we can approximate the rate of the reaction with transition state theory (assuming $\kappa=1$):
$$k = \kappa\frac{k_\mathrm{B}T}{h}e^{\frac{- \Delta G^{\ddagger }}{RT}}
\approx 1\times10^{11}~\mathrm{\frac{1}{s}}$$
This is ultrafast and at room temperature we consider it a free rotation. With spectroscopic methods we will not see a difference. We would probably have to cool the system down to below 1 K to actually see two (E1, E2) conformations.

Steric effects are often electronic effects in disguise, they offer tempting explanations, but are often rooted in a more complex electronic explanation. In this example the possibility of hyperconjugation explains the slightly higher energy the bisected forms compared to the staggered forms.
Hyperconjugation can come in different forms, the most common explanation is the overlap of an empty anti-bonding orbital with a filled (non-)bonding orbital. In this scenario it does not matter which elements are involved in these interactions.
There is even negative hyperconjugation, but this is a little beyond the scope of this post - if there are more questions about the topic, you are always welcome to ask them here.

Footnotes

For a quite recent molecular orbital study see: Ramiro F. Quijano-Quiñones,  Mariana Quesadas-Rojas,  Gabriel Cuevas and  Gonzalo J. Mena-Rejón, Molecules 2012, 17(4), 4661-4671. (open access)
For an educational approach see: Henry Eyring, D. M. Grant and H. Hecht
J. Chem. Educ. 1962, 39 (9), 466.
I did not have the time to search thoroughly, but the following publication contains the value. Leland Cullen Allen, Eugene Scarzafava, J. Am. Chem. Soc. 1971, 93 (2), 311–314.


Best way to dissolve and consume cysteine HCl monohydrate powder Does cysteine break down in filtered water?  Should this made more acidic?

If I mix cysteine powder in juice (something acidic maybe), does it remain at cysteine?  It seems that it can oxidize into cystine and lose some of its qualities.

I wouldn't worry at all about the oxidative dimerization of cysteine to cystine when you consume freshly prepared solutions.

In the lab, you'd use oxidants, such as $\ce{FeCl3}$ (in the presence of air) or hydrogen peroxide to achieve this reaction; both of which you're not likely to add to your amino acid diet!

As far as the article which you mentioned is concerned: The stabilization methods in question will only be relevant to maintain the composition of prefabricated and sterile-packed solutions over a long period of time.


Should one pronounce “periodic” the same in “periodic acid” and “periodic table”? In “periodic table”, the adjective is related to the noun period, and comes from Ancient Greek περίοδος through French périodique. In “periodic acid”, it is formed from the prefix per- and iodic (like peroxide and permanganate).

Wiktionary lists their respective UK pronunciations of as /pɪə(ɹ).iˈɒdɪk/ and /ˌpɜːraɪˈɒdɪk/, markedly different: pɪə (as in piece) vs. pɜː (as in perfume); then i (as in it) vs. aɪ (as in eye).

However, is that distinction really made in practice? Would a native US/UK/Aussie speaker make the difference when talking in the lab?

I am an Australian English speaker and yes, this distinction is made in practice.

Possibly the only reasonable opportunity to use the pɪə pronunciation in the name of a compound is in the case of the entertaining molecule periodane, which is actually named after the periodic table. This molecule (and later a number of different plausible isomers) was identified computationally by a methodology called 'mindless chemistry' which optimises randomly generated molecular graphs. Periodane is a stable configuration of each atom on the second row of the periodic table, with the exception of neon (although some people are working on that).


Sodium metal reduction of chloroform In discussing how best to deuterate chloroform, it occurred to me that we could use sodium metal to reduce hydrogen into hydrogen gas. If this could actually happen, this would be a great thing - we wouldn't have much if any residual hydrogen in the liquid phase of the system after treatment.

Is this possible? I had a possible mechanism in mind.

The problems I foresee are:

1) Won't reduction of chlorine to chlorine gas be a competitive, side reaction? From this page (http://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)) I see that the reduction potential for chloride ion to to chlorine gas is -1.36 V (unfavorable) and reduction of hydride ion to hydrogen gas is a +2.23 V (favorable). Am I interpreting and applying the data correctly?

2) What happens to the $\ce{CCl_3}$ radicals after hydrogen is reduced? Is dimerization going to occur to a large extent?

3) How then do we introduce deuterium and effectively incorporate it with $\ce{CCl_3}$ radicals?

Sodium + Chloroform:
https://www.youtube.com/watch?v=du3ksrfnCcg

To prepare deuterochloroform I'd recommend the haloform reaction.


Electrophilic aromatic substitution of napthalene It is known that the $\alpha$-position (or 1-position) for napthalene is the preferable position for substitution to occur at. However, I am getting two different explanations as to why this is the case, one of which I am not able to make sense of.

The first explanation which I have come across and which is also the one I am able to make sense of is given by Hepworth, Waring and Waring (2002). They explain it by using the concept of Clar's rule. With an attack at the $\alpha$-position, there would be two resonance structures in the intermediate with preservation of the aromatic sextet. With an attack at the $\beta$-position, there would only be one resonance structure in the intermediate with preservation of the aromatic sextet. As such, the $\alpha$ attack would be more preferable since the resultant intermediate would be more stable.

The second explanation which I have come across just today comes from Carey and Sundberg (2007). An image of the section of the text discussing the issue is shown below.

I am not able to understand why they have mentioned that the intermediate resulting from $\beta$ attack is not under allylic stabilisation because it seems to me that it is under such stabilisation, after drawing out the resonance structures. I would like to clarify if this explanation is accurate.

References

Carey, F. A., & Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.

Hepworth, J. D., Waring, D. R., & Waring, M. J. (2002). Aromatic Chemistry. United Kingdom: The Royal Society of Chemistry.

Essentially, both descriptions are identical. However, they use different phrasings.

What Carey and Sundberg mean when they say no allylic stabilisation is essentially that the cation itself is not in an allylic position; that there is no single (non-phenylic) double bond in α-position. To reach the allylic double bond in the β-attack case, you would need to resonate across at least part of the phenyl ring (meaning you’re already making use of benzylic stabilisation) while in the case of α-attack you have access to allylic resonance without interrupting the phenyl resonance.


How do I interpret a flux quoted in cm⁻² s⁻¹ in atmospheric chemistry? I'm reading a paper about the atmospheric chemistry on Titan {1}. It quotes "column photochemical and loss rates" for various species in units of  $\mathrm{cm^{-2}s^{-1}}$. These figures are the overall rates at which these species are generated photochemically according to a model, and I would like to convert them into units of $\mathrm{mol/s}$ or $\mathrm{mol/s/m^2}$.

However, I don't know how to interpret the given units of $\mathrm{cm^{-2}s^{-1}}$. It feels like it should be something like CCs of gas produced per $\mathrm{cm}^2$ of the planet's surface area - but that would be $\mathrm{cm\,s^{-1}}$ rather than $\mathrm{cm^{-2}s^{-1}}$. No matter how I look at it I can't see where the $\mathrm{cm^{-2}}$ comes from - is anyone able to explain this?

{1} P. P. Lavvas et al. (2008) Planetary and Space Science 56, pp. 67-99.

The "strange" unit describes the photon flux, which is the number of photons (of unspecified wavelength) that hit an area (thus $\mathrm{cm^{-2}}$) in a time interval (thus $\mathrm{s^{-1}}$).

  It feels like it should be something like CCs of gas produced […]

Not really. Not every photon leads to an excited state, not every excited state triggers a chemical reaction.


White hot hydrogen gas, like that during a big bang Difficult to know for certain which field this is. According to what I can discover about it, the big bang explodes into being as white hot hydrogen gas, millions of degrees hot.

I'm okay with that my question is what colour would it fade to as it cools down? Is there a spectrum I can refer that would illustrate that. I've seen such diagrams for metals as they cool down.

Difficult times people, thank you for your time and input, stay safe. Cheers!

(This might get more/better answers on Physics Stack Exchange)

At the high temperatures of the early Big Bang, you have individual protons and electrons, not hydrogen atoms and molecules.  This is because the $kT$ thermal energy is much, much more than the binding energy of the hydrogen atom.

In that case, the moving individual charges create a continuous black body spectrum:

(from Sun.org)

The light will look like a 10,000K source: a bit on the blue side.  But most of the energy will be at wavelengths much shorter than visible light.  The Wien displacement law shows that:

$$\lambda_{\rm{peak}} = {{b}\over{T}} = {{3\times 10^{-3}m \cdot K}\over{1\times 10^6 K}} = 3\times 10^{-9}m$$

A 3nm wavelength corresponds to soft X-rays (or perhaps the very hardest ultraviolet light).

You'll note that the vertical axis below is logarithmic.  At 100X temperature to the 10000K like, the peak will be at the shorter (left) wavelengths of X rays, but also much higher.  The energy flux goes as $T^4$, so a 100X temperature is a factor of $10^8$ in intensity:  There will be a lot of X rays at these early temperatures.


Products of silicate synthesis I'm attempting to make a hard refractory cement to apply over kiln brick made of sodium silicate and Perlite. This creates a very good kiln brick, however they are susceptible to abrasion.  I've been making my own sodium silicate, and thought I might be able to make it harder with the addition of some oxides.

I added silica gel and water; then I added magnesium oxide and sodium hydroxide.   There was a extremely strong exothermic reaction.

What would this be called? I imagine it's called something like $\ce{Na2SiO4Mg}$.  Tomorrow I will try some combinations of sodium silicate with Aluminum Oxide, and magnesium oxide. What would these be called?

If you have a recipe for a hard refractory cement to protect the Perlite mix from abrasion I would appreciate the input.

I imagine it's called something like $\ce{Na2SiO4Mg}$.  Tomorrow I will try some combinations of sodium silicate with Aluminum Oxide, and magnesium oxide. What would these be called?

Much simplier than that. Sodium hydroxide dissolves incredibly exothermally:

$$\ce{NaOH (s) -> NaOH (aq)}\qquad{\mathrm{\Delta} H = \pu{-44.5 kJ mol-1}}$$

This is what heated your mixture.


Finding the amount of nitrous oxide produced in the reaction between lead sulfide and nitric acid Lead(II) sulfide dissolves in excess nitric acid according to the equation below. Calculate the volume of $\ce{NO{(g)}}$ at $\pu{27 ^\circ C}$ and $\pu{1.10 atm}$ produced from $\pu{4.7 g}$ of $\ce{PbS(s)}$.
$$\ce{3PbS{(s)} + 2NO3–{(aq)} + 8H+{(aq)} -> 3Pb^2+{(aq)} + 3S{(s)} + 2NO{(g)} + 4H2O{(l)}}$$
a) $\pu{0.29 L}$
b) $\pu{0.44 L}$
c) $\pu{0.66 L}$
d) $\pu{30 L}$
e) $\pu{45 L}$

I used the ideal gas formula and converted the mass of $\ce{PbS}$ into number of moles of $\ce{PbS}$ however got the answer b) $\pu{0.44 L}$ which is incorrect. The correct answer is a) however I am unsure about. Why this is?

The first step is to calculate the number of moles of $\ce{PbS}$:
$$n(\ce{PbS}) = \frac{\pu{4.7 g}}{\pu{239 g mol-1}} = \pu{0.0197 mol}$$
The next step is to calculate how many moles of $\ce{NO}$ will be produced based on the stoichiometry of the reaction. As stated in a comment, it looks like this is the step that was left out of your calculations. Because only $\pu{2 mol}$ of $\ce{NO}$ are produced for every $\pu{3 mol}$ of $\ce{PbS}$, only $2/3$ of $\pu{0.0197 mol}$, or $\pu{0.0131 mol}$ $\ce{NO}$ will be formed.
At this point we know all of the variables required to invoke the ideal gas law:
$$
\begin{align}
V(\ce{NO}) &= \frac{nRT}{p}\\
 &= \frac{\pu{0.0131 mol} \times \pu{0.0821 L atm mol-1 K-1} \times \pu{300K}}{\pu{1.10 atm}}\\
 &= \pu{0.29 L}
\end{align}
$$
And this gives us the correct answer of a. Again, as your answer was off by a factor of $2/3$, it is pretty clear that the second step above regarding the stoichiometry of the reaction is the only step that you left out of your calculations.


Mass of iron from the reduction of iron(III) oxide Given the equation
  $$\ce{Fe2O3 + 3H2 -> 2Fe + 3H2O},$$
  determine how many grams of $\ce{Fe}$ can be produced from $\pu{16.5 g}$ of $\ce{Fe2O3}$.

I've already determined the masses of each component. I believe the next step is convert the masses into moles, but after that, I'm not sure what formula to use.

Any help is greatly appreciated.

After finding the moles of $\ce{Fe2O3}$ we see that after balancing the equation, $2$ moles of $\ce{Fe}$ are formed for every mole of $\ce{Fe2O3}$.

But that would be considering that the equation moves completely to the RHS, which is assumed if nothing is explicitly mentioned but not true.


Why is 2-methylpropene less in energy than its alkene counterparts? Other isomers include 1-butene, cis 2-butene, and trans 2-butene. Why would 2-methylpropene be less in energy if there is more steric hindrance?

There are two things we need to understand before we can answer the question.

1) More highly substituted double bonds are generally more stable than less substituted double bonds.

This is because the $\ce{sp^3}$ hybridized carbon in the alkyl group is electron donating towards the $\ce{sp^2}$  hybridized carbon in the double bond. Electron density likes to flow from low to high s-character orbitals since the more s-character in an orbital, the lower the energy of the electrons in that orbital.  As the electrons flow from the $\ce{sp^3}$ carbon to the $\ce{sp^2}$ carbon the energy of the electrons (and therefore the energy of the molecule) is lowered.

A "shorthand" way to illustrate this electron flow is by drawing hyperconjugated resonance structures like the following.

Consequently 2-butene (where we can draw 6 hyperconjugated resonance structures) is more stable than 1-butene (where we can only draw 2 hyperconjugated resonance structures).

2) Carbanion stability decreases in the order primary > secondary > tertiary

As we've just shown above, alkyl groups are electron releasing.  If we have a negative charge on a carbon (a carbanion), the last thing we want is to have more alkyl groups attached to that carbon releasing more electron density onto an already negative carbon atom.

Now we can return to your question and try to understand why 2-methylpropene is more stable than trans-2-butene.   Carbon-hydrogen hyperconjugation is at the center of the answer.  Let's compare some of the hyperconjugated resonance structures that can be drawn for these two molecules.

Remembering that a) more highly substituted double bonds are more stable and b) that primary carbanions are more stable than secondary carbanions, we can compare hyperconjugated resonance structures A and B. Resonance structure A contains a disubstituted double (more stable) and a primary carbanion (more stable), while resonance structure B contains a monosubstituted double bond (less stable) and a secondary carbanion (less stable). Therefore, resonance structure A should be more stable than resonance structure B.  Since resonance structure A is more stable than resonance structure B, 2-methylpropene will be more stable than trans-2-butene.

Finally, we note that cis-2-butene is less stable that trans-2-butene for steric reasons.  This allows us to place these $\ce{C_4}$ isomers in the following order of stability:

2-methylpropene is more stable than trans-2-butene which is more stable than cis-2-butene, which is more stable than 1-butene. If we examine the heats of combustion for these isomers, we see that our order of stability is confirmed.

image source


How to recycle 60 tons/hour of CaCO3? I hope this is the right place to ask. I'm not a chemist so my question may seem obvious but it's not for me.

I work in the egg industry and we produce every hour 60 tons of dried egg shells as waste.
Those shells come in small grains (few millimeters) but can be further reduced to powder (like talco or marble powder).

This powder is $97~\%\ \ce{CaCO3}$ and $3~\%$ proteins.

What I'm interested in is an idea to recycle this product in a cost-effective way. At the moment a small part is used in cultivations but the rest is just stored in abandoned mines and I don't like the environmental impact that this solution has.

60 tons on $\ce{CaCO3}$ per hour means half a million tons per year. That's quite some calcium carbonate.

Hens need $\ce{CaCO3}$ to produce eggs. So the most natural idea to recycle it is to produce feed lime.

Or consider producing agricultural lime. The annual consumption of agricultural lime is about 2.5 million tons in Germany, and about 25 million tons in Brazil. Your production is in this scale. The presence of 3% proteins is probably of little concern if not beneficial.


What is the chemistry behind this "petrol from air" technology? A recent news report in the UK claimed a breakthrough in making a petrol equivalent from carbon dioxide and water:

  A small British company has produced the first "petrol from air" using a revolutionary technology that promises to solve the energy crisis as well as helping to curb global warming by removing carbon dioxide from the atmosphere.

It sounds like the news reports are over-hyping it a little (and I'm not sure it is that new), so what is the real chemistry behind it? What are the key reactions and how much input energy is required?

Step one: trap $\ce{CO2}$. This can be done using specially selected amines, like triethanolamine. Aminosilicones (a common chemical used in hair conditioner), can also do this.

Step two: release trapped $\ce{CO2}$. In case of triethanolamine you'll have to boil mixture.

Step three: mix $\ce{CO2}$ with hydrogen (hydrogen is produced by electrolysis of water). Volume proportion should be somewhere 1:3.

Step four: threat the mixture with proper catalyst. This step may be divided into two stages: reverse water-shift reaction  and some variant of Fischer–Tropsch process with some type of alcohol synthesis as very interesting candidate. There are some multi-functional catalysts that can perform both stages at once.

Another variation on step 4 is the Sabatier process, which involves generating methane from $\ce{CO2}$ and hydrogen.  Typically the catalyst is Rainy nickel (powder).

The required energy in form of electrical current is probably from 1.5 to 3 times of heat of combustion of resulting fuel, which is in turn at least twice more that can be converted into actual work again.


Does milk drinking prevent long-term chemical poisoning? I have heard some rumors that drinking milk prevents chemical poisoning. I have done a bit research and some sources confirm that.

Corrosive Poisons
The best first aid is to dilute the poison as quickly as possible. For acids or alkalis (bases), give the patient water or preferably milk or ice cream — one (1) cup for victims under five (5) years; or one (1) to two (2) glasses for patients over five (5) years. Milk or ice cream is better than water because it dilutes and helps neutralize the poison. Water only dilutes the poison. Cornell University

In Case of Poison Ingestion: Drink Milk?
As @Jan pointed out in the case of chemical poisoning the paramedics are immediately called.
So my question is, does milk helps chemical poisoning, e.g., for people that daily work with chemicals or are working in chemical facilities (working with or be in the presence of $\ce{HCl}$, $\ce{HF}$, $\ce{NaOH}$, $\ce{HNO3}$, $\ce{H3PO4}$, etc.)? Or what would be beneficial? And I mean long-term exposure and not immediately threat to life or health.

This post only concerns chronic poisoning, i.e. exposure to a harmful chemical in low amounts for an extended period of time. The answer is not applicable to acute poisoning; in that case a doctor should immediately be called.

Basically there are three types of chemicals that have a harmfulness threshold to which one can be exposed over extended periods of time.

Type one includes substances such as mercury that are primarily aggregated in the body rather than excreted. In rare cases, drinking milk may enhance excretion thereby decontaminating the body. I haven’t heard of any such examples, though, and I would assume it to be a very rare case so neglegible.

Type two includes those substances that the body can actually metabolise to harmless products. This would include most protein poisons. However, the metabolism shouldn’t be enhanced by drinking milk so it will likely have a non-effect.

Type three includes those that, while they aren’t metabolised, just don’t do enough damage at low concentrations. That would basically include your list of acidic and basic substances but also non-aggregating metals at low concentrations. For these, excretion is not significantly enhanced or reduced by drinking milk so again no effect is expected.

Comparing the acute exposure to acids to the chronic one, in the former giving a person milk to drink is actively encouraged. This is because milk contains a lot of proteins that can act as buffers, increasing the body’s buffering potential and reducing the effects of the acid. The body does have more than enough buffering capabilities to cope with chronic cases though, so even there regularly drinking milk would not help.

Tl;dr: Milk is a great drink but cannot act as a type of preventive antidote.


Measurement of Water Clarity Turbidity is essentially the clarity of water. Turbidity can be measured in many ways such as a turbidimeter, secci disk or a turbidity tube (which incorporates a secci disk at the bottom). If you had a small turbid water sample - 20mL, how could you measure the turbidity without an electronic device??

Turbidity is measured by the examining the light passing through a sample.  A sample's turbidity is formally measured using standardized methods (e.g. ISO 7027 or "Standard Methods," No. 2130B. source).  Standardized methods require the instrumentation called for by the standard.

An informal measure of visible differences in turbidity can be obtained by comparing the 20mL sample to a set of calibrated standards.  An example of a set of calibration standards for turbidity are shown in the following link http://www.water.ncsu.edu/watershedss/info/turbid.html.


Why there is a huge difference between the basicity of urea and guanidine? The two molecules have very similar structures, but urea has a $\mathrm{p}K_\mathrm{b}$ of $13.9,$ while guanidine have a $\mathrm{p}K_\mathrm{b}$ of $0.4.$ If guanidine cation can be stabilized by resonance, then why can't urea cation do so?

Urea and guanidine are structurally similar except that the carbonyl oxygen of urea has been replaced by an imine $\ce{NH}$ in guanidine. When it come to protonation, carbonyl oxygen of urea is always protonated over its 2 $\ce{NH2}$ groups (Ref.1 and 2). In guanidine, it is the imine $\ce{NH}$ protonated first. This preference in guanidine is not because a lone pair of $\mathrm{sp^2}$ nitrogen is more basic than that of $\mathrm{sp^3}$ nitrogen (actually it is other way around), but the lone pair of $\mathrm{sp^2}$ nitrogen is the most available pair for protonation over those of two $\mathrm{sp^2}$ nitrogens (lone pairs of two $\mathrm{sp^2}$ nitrogen atoms are contributing to resonance stabilization at the time). However, when protonated, guanidinium ion was highly stabilized by resonance as depicted in the diagram:

Thus, basically, it  comes down to protonation ability of oxygen versus nitrogen. Oxygen is more electronegative than that of nitrogen, thus like to retain its lone pare better than nitrogen (cf. basicity of water and ammonia). Therefore, in that sense alone, guanidine is a better base than urea.

References:

Nanping Wen, Murray H. Brooker, “Urea protonation: Raman and theoretical study,” J. Phys. Chem. 1993, 97(33), 8608-8616 (https://doi.org/10.1021/j100135a013).
B. Valentine, T. E. St. Amour, D. Fiat, “A $\ce{^{17}O}$ NMR study of the protonation of urea,” Organic Magnetic Resonance 1984, 22(11), 697-700 (https://doi.org/10.1002/mrc.1270221105).


How does H's ionization energy relate to its transition energy (Bohr's Model)? I am currently reviewing some material about orbital energy levels. In my review book there is a short snipet that reads:

  The IE of $\ce{H}$ from its ground state ($n=1$) is $1312\ \mathrm{kJ/mol}$. Because of the squaring of the principle energy level, the ionization of an $\ce{e-}$ in $\ce{H}$ from the $n=2$ level is $1/4$ of that value ($328\ \mathrm{kJ/mol}$). The transition energy from the $n=1$ level to $n=2$ level is the difference between the two values, $984\ \mathrm{kJ/mol}$.

I am confused about how exactly an atom's ionization energy relate to its transition energy. If IE is the energy required to pluck off the first e- from an atom and TE is the energy required to change between different orbitals, does this mean that an atom with greater IE will need to absorb a greater amount of energy to jump orbitals?

Lastly, is there a equation that directly ties in both ionization energy with transition energy?

Ionization energy of some arbitrary atom is not related whatsoever to its transition energy. Also, there are many different transitions (even in a hydrogen atom), each with its own specific energy. It is just that this particular transition of this particular atom happens to have the energy of 3/4 its ionization energy, one of the few things the Bohr model managed to get right. You can't generalize that to other atoms and other transitions.


Why do double bonds break So today we performed an experiment, where we mixed cyclohexene with Bromine water. The equation was

which seems to make sense. The Bromine water we were using was decoloured as evidence that this equation took place, whereas when we mixed cyclohexane with Bromine water nothing happened (to show how double bonds are more 'reactive'). But what I don't understand is why the double bond was broken. I mean, wasn't it happy just being the way it was? Doesn't it still take energy to break the double bond, even if it is less than the energy required to break a single bond? It isn't like it WANTS to take the bromine? (or is it?) The only place where this energy could have come from in my mind is either the vibrational energy due to temperature, or else the kinetic energy from pouring it in? (may be taking too much of a physics approach). If so, doesn't that mean that if I got the cyclohexane, and poured it in from a great height, will it react with the Bromine water? Or is the energy difference between double bonds and single bonds so great that it can only perform under special circumstances? (i.e. under UV light)

Keep in mind, that the reason reactions occur spontaneously is because there's an overall driving force: $\Delta G = \Delta H - T\Delta S$. Since there are fewer molecules on the right hand, we can guess that $\Delta S$ isn't driving this reaction. ($G$ is Gibbs free energy, $S$ is entropy, $H$ is enthalpy, and $T$ is temperature.)

So then your question boils down to "why is this reaction favorable for $\Delta H$?"

Even before I look up the bond enthalpies for each of the bonds involved, I have an intuition. Yes, we're breaking the double bond in cyclohexene. We're also breaking the $\ce{Br-O}$ bond in $\ce{Br-O-H}$. So that's two bonds that are broken.

You're right -- that costs energy, so there's clearly something more energetically favorable on the other side. Well, in the product, I'm forming a $\ce{C-Br}$ and a $\ce{C-OH}$ bond.

So the net reaction breaks two bonds, and forms two bonds. As you say, the general intuition is that the "double" part of a double bond isn't quite as strong as the standard $\sigma$ single bond.

Let's look up the bond enthalpies also through Wikipedia:

$\ce{C=C}$ (vs. $\ce{C-C}$ which remains in the product): ~146 kcal/mol vs. ~84 kcal/mol = ~62 kcal/mol
$\ce{Br-O}$ = ~56 kcal/mol (I had to look this one up, so it's a bit iffy)
$\ce{C-Br}$ = ~69 kcal/mol
$\ce{C-O}$ = ~85 kcal/mol

Net $\Delta H \approx$ -36 kcal/mol.

Even if some of the bond enthalpies are off a bit, it's clearly energetically favorable. It has nothing to do with how the species are mixed, or really the temperature. (Well, my guess is that $\Delta S$ is not favorable here, since two molecules form one, so high temperature might make the reaction less favorable from a $\Delta G$ standpoint.) Nor do you need UV light or anything like that -- the difference in bond strengths will push this reaction.


Explain on the basis of valence bond theory that tetracyanonickelate(II) ion with square planar structure is diamagnetic Answers on the internet suggest that tetracyanonickelate ($\ce{([Ni(CN)4]^{2-}}$) with square planar structure is diamagnetic because $\ce{CN-}$ is a strong field ligand and causes pairing, but my book says that VBT does not distinguish between strong and weak field ligands. So can anyone explain the logic using VBT?

@M.L has already provided the solution as per VBT, but that answer seems to have some flaws when it comes to CFT (looks like he has corrected the mistakes now). So, I would like to take some time to answer this question using CFT.

According to crystal field theory, the degeneracy of the orbitals of a same orbital is lost in presence of a ligand and the pattern in which they are distorted depends upon the nature of the ligand. The spectrochemical series helps us understand the nature of a ligand.
In general there are two types of ligands, strong field and weak field. Strong field ligands have a greater power to distort the degeneracy of the sub-orbitals whereas weak field have a weaker power of distortion.
As per the spectrochemical series, $\ce{CN-}$ is a strong field ligand and thus creates a greater distortion in the degeneracy of $\mathrm d$ orbitals.
The central metal ion here is the $\ce{Ni^2+}$ with its $\mathrm d^8$ electronic configuration.
A $\mathrm d^8$ system prefers a square planar approach of ligands if the ligands are strong field. In such a condition the the degeneracy is disrupted and the energy diagram of the $\mathrm d$ orbitals is as below:

Now when trying to fill this diagram with electrons we have to take into consideration that an electron in the $\mathrm d_{x^2-y^2}$ orbital can pair up with the electron in $\mathrm d_{xy}$ in case of a strong field ligand. This is because the extent of splitting done by a strong field ligand is so high that it can overcome the energy gained due to pairing of electrons.
Thus the energy diagram filled with electrons looks as following:

Now the ligands donate their lone pairs in the $3\mathrm d_{x^2-y^2}$, $4\mathrm s$, $4\mathrm p_x$ and $4\mathrm p_y$ sub-orbitals.
In reality, no hybridization takes place but if forced to say the hybridization of this complex, we can say $\mathrm{dsp}^2$ because of the orbitals involved in the $\sigma$ bonds between the metal and ligands.
As you can see the energy diagram does not contain any unpaired electron and thus the complex is diamagnetic.


How to prepare a Gaussian input file for a binary solvent? The UV absorption energy spectrum of a molecule has been recorded experimentally using binary solvent (DMSO:Water=9:1 v/v). I am trying to calculate the absorption energy of the same molecule using computational methods. For that I want to give the same solvent mixture (DMSO:Water=9:1 v/v) in my DFT calculation.

How can I prepare the input file for Gaussian? What are the keywords that are used in Gaussian for binary solvents?

I have calculated the dielectric constant of the binary solvent (DMSO:Water=9:1 v/v) using the following expression
$$\epsilon_\mathrm{m}= x_1 \epsilon_1 + x_2 \epsilon_2,$$
where $x_1$, $x_2$ are the mole fractions of each solvent in the binary mixture, subscripts $1$ and $2$ denote the different solvents, and $\epsilon_1$/$\epsilon_2$  their dielectric constants.

I am unaware that this feature exists in Gaussian. It is certainly something that stretches the possibilities of a polarisable continuum model. I am quoting (liberally) from an email to the computational chemistry list (CCL, http://www.ccl.net/), because I found no evidence, that this has changed for Gaussian 16:

Actually, there is no standard approach for solvent mixtures to you use it directly in Gaussian03/09. A good suggestion is to define the solvent parameters as a linear combination of the solvents in the proportions of the mixture.  So for toluene:water in a 4:1 mix use 4/5 of the toluene values plus 1/5 of the water values.
It is recommended to use the SCRF=SMD approach here, but that is not recommended for properties other than solvation free energy. It may be give satisfied results in the case of geometry optimization or energetic properties.  It is especially not recommended for derivative properties, because the cavity for the SCRF=SMD model has a number of continuity issues.  So for the default SCRF=IEFPCM model with the smooth cavity you mainly need the static (or zero-frequency) dielectric constant and the dynamic (or optical) dielectric constant.  You would specify EPS and EPSInf in the additional SCRF input section after the blank line which terminates the structure input.

For example:

{%chk=}
#P {Method, Basis, additional keywords} SCRF=(Solvent=Generic,Read)

{title card}

{charge, multiplicity}
{molecule specification}

EPS=(Add the value of linear combination of both solvents)
EpsInf=(Add the value of linear combination of both solvents)

See the Gaussian Manual for more details.

Another approach is to first do an optimisation and related properties calculation in both pure solvents, and then compare these results. For example the difference in the structures, or spectra.
Since you are using nine parts DMSO and one part water, it might well be that the DMSO calculation is a well enough approximation to the mixture.
Solvent calculations are tricky and often enough unreliable. One should never ever expect a quantitative agreement. It is a qualitative description you should be after.
If you need more accuracy, you have to use an explicit solvent model with MD calculations, or you can look into COSMO-RS.


Explaining the Relative Energies of Various Vibrational Modes Specifically, I'm wondering why it is that the asymmetric stretch in most (although there are exceptions to this I believe) molecules is higher in energy than the symmetric stretch.

Similarly, I would like a nice, physical (ideally quantitative/mathematical) explanation of why bending modes are usually lower in energy than stretching modes.

I can rationalize this second question on the basis of electrostatic forces. That is, any distortion from the minimum of the potential energy well will either cause more coulombic-type attractive forces or more coulombic-type repulsive forces. These forces should not be as large for a lateral displacement (bending) than for a stretch of some kind. This makes sense because the change in radius will not be as large when simply moving to the side from an equilibrium position. Of course, the change in radius is significant because electrostatic potential energy goes as $\frac{1}{r}$.

Anyways, that's a pretty classical answer, so I wonder how both of those questions can be explained more quantitatively as well as quantum mechanically.

I don't know how to explain the bending mode, but there is a nice mathematical explanation for the difference in stretching modes.

Take $\ce{CO2}$, carbon dioxide, as an example. I'll use B3LYP/6-31G(d,p) values. $\nu_{1}$, the symmetric stretch, is 1372 cm$^{-1}$, and $\nu_{3}$, the asymmetric stretch, is at 2436 cm$^{-1}$. Now consider $\ce{CO}$, a single carbonyl group (the molecule won't work for this example). Looking at an IR table, the average stretch will be about 1750 cm$^{-1}$. $\ce{CO2}$ looks a lot like two $\ce{CO}$ bonds that are linear and share a carbon. But, clearly $\ce{CO2}$ doesn't have a doubly-degenerate stretch mode of any kind at that frequency, so something is causing the local modes to not be identical to the normal modes. Because both local modes have the same symmetry, they are allowed to mix with some coupling between them, just like molecular orbitals. A local mode is a vibration that is completely uncoupled from all other modes; its motion does not influence the motion of other atoms and vice versa.

Set up a Hamiltonian for a degenerate two-level system with some coupling between them. Keeping with a Huckel-like picture, I'll use $\alpha$ for the energy level of each degenerate local mode and $\beta$ for the coupling between them.

$$
\hat{H} = \begin{pmatrix}
\alpha & \beta \\
\beta & \alpha
\end{pmatrix}
$$

Obtaining the normal modes and their energies is equivalent to diagonalizing this Hamiltonian, where the eigenvalues are the normal mode energies and the eigenvectors are the normal modes themselves as linear combinations of the local modes. Doing so gives the normal mode energies:

$$
\alpha - \beta, \, \alpha + \beta
$$

and the normal modes themselves:

$$
\begin{pmatrix}
-1 \\
1
\end{pmatrix},\,
\begin{pmatrix}
1 \\
1
\end{pmatrix}
$$

By convention in Huckel theory, $\beta$ is usually negative, so the 2nd eigenvalue is the one lower in energy.

If you take the dot product between each normal mode eigenvector and the vector of local modes:

$$
\begin{pmatrix}
\phi_{1} \\
\phi_{2}
\end{pmatrix}
$$

out falls the symmetric and antisymmetric combinations leading to the symmetric and asymmetric stretches. Leaving out the normalization factor of $1/\sqrt{2}$:

$$
(-\phi_{1} + \phi_{2}), \, (\phi_{1} + \phi_{2})
$$

Setting $\alpha = 1750$ cm$^{-1}$ and $\beta = -500$ cm$^{-1}$ gives a symmetric stretching frequency of 1250 cm$^{-1}$ and an asymmetric stretching frequency of 2250 cm$^{-1}$. Not bad for a model with a single parameter.

This is how the energy levels (eigenvalues) split with increasing the coupling between the initially degenerate local modes into distinct normal modes:

Here is a link to the Jupyter Notebook containing the math I did. This is a simplified (math-wise) version of what we did in a paper, since the conceptual point is identical.

This sort of description is how any two-level time-independent system works. See how the two spin energy levels of a free electron change in energy when an external magnetic field is applied (Zeeman effect):


The meaning of the sulfite anion attached to the benzene ring I needed to work with PEDOT:PSS solution, so I also looked up it's chemical composition and the wikipedia page threw up this pic.

I need to understand how the sulfite anion is connected to the benzene ring in the structure on the right. I have studied it in chemistry in my school days but I have not really used it afterwards, hence I have forgotten. Your explanations will be appreciated.

If your wondering about the bond, the line going to the center means that the sulfonyl group is attached at any available position (2, 3, or 4) on the ring, and no particular position is given.


How can we tell if an ion is amphiprotic If I had the following ions:
$$\ce{HCO3^-}$$
$$\ce{H2PO4^-}$$
$$\ce{CH3COO^-}$$

Which ones would be amphiprotic?

Looking at my $ K_A$ chart, it looks like each ion is formed by the disassociation of weak acids, so It looks like all of them are amphiprotic because they could all technically act as a bronstead-lowry base.

The answer is the first two ions, but why?

Thanks for answering :)

To be amphiprotic means that the chemical species can donate or accept H+ ions.

$$\ce{H2CO3<=>[H+] HCO3^- <=>[OH-] CO3^{2-}}$$
$$\ce{H3PO4 <=>[H+] H2PO4^- <=>[OH-] HPO4^{2-}}$$
$$\ce{CH3COOH <=>[H+] CH3COO^- <=>[OH-] \text{No Change}}$$

Thus since the acetate anion can't donate a proton, it is not amphiprotic.


What does ³D mean in energy level diagram for helium? I'm coming from a Physics background and am trying to make sense of the energy level diagram for para-helium and ortho-helium.

From what I've gathered each column shows the different energy levels for different orbitals so that for example the leftmost column with $^1S$ written on top shows the energy levels of 1s, 2s, 3s etc.

What I don't get is what the different denotations on top mean, what does for example $^1P$ and $^3D$ mean?

The index 1 or 3 is used for saying that the spins of the two electrons are parallel or anti-parallel. One spin is up or +1/2, and the other one is down, or -1/2. This is the case in the ground state : the 2 electrons of Helium have two antiparallel spins (one +1/2 and one -1/2). The atom has no total spin, as the two electrons cancel the magnetic effect of their spins. The total spin of the atom is then 1/2 - 1/2 = 0. In such a configuration, the He atom is called para-helium. This remain true if one electron is excited and goes to an upper orbital like 2s, 3s, etc. provided that the two electrons maintain their spins (one up and one down). The total spin is zero, and the atom is an excited state of para-helium.

It is difficult, but not impossible, to change the spin of one electron, so that both electrons have the same spin. In this configuration, the helium is called ortho-helium. The total spin for the whole atom is 1/2 + 1/2 = 1. In that case, there are three possibilities for the projection of the total spin in an external magnetic field. +1 in the sense of the field, -1 in the opposite sens, and zero, if the total spin is perpendicular to the magnetic field. This number 3 is reported in upper index. The differences of the three projections must be an integer.


Why do bases taste bitter? My chemistry teacher told us that acids taste sour and bases taste bitter. After reading Do acids without hydrogen taste sour? and https://en.wikipedia.org/wiki/Taste#Sourness I learned that sourness in fact is the taste of acidity. But why is it that bases generally taste bitter?

Some bases taste bitter, some do not. Between bitterness and basicity is correlation, not causation.
The strongest bases like hydroxides of alkali metals and alkali earth metals, or alkali metal carbonates, are not bitter.
Near all if not all bitter  bases are organic compounds with basic groups containing nitrogen, like aminogroups or heterocyclic nitrogen. They are often potentially poisonous alkaloids, so evolution came with unpleasant sensation for such compounds. The typical example is quite bitter caffeine.
OTOH, not all bitterness is caused by basic compounds. Grapefruits are bitter because of the glycoside naringin, which does not contain nitrogen and is not a base. Glycosides are frequently bitter, regardless of basicity.
Sensory impact is rather a topic for biology, not chemistry.


How many resonating structures of Carbon Monoxide are there? Anywhere I look, I get 3 resonating structures for CO molecule, like in this answer.
However, according to the rules stated for drawing resonating structures in this site, I wonder why there can't be this resonating structure as well?

Although it will be very unstable (due to lesser number of covalent bonds, positive charge on more electronegative atom), yet a minute contribution of this structure must in present in the resonance hybrid of the CO molecule. Then why is this structure not shown anywhere?

You have said it yourself. There are many possible resonance structures but most of them have a negligible contribution to the overall structure. Your proposed structure is one of them.

The major resonance contributors of carbon monoxide are: $$\ce{\!\overset{\ominus}{C}#\overset{\oplus}{O}\! ~<->~ \!C=O\! ~<->~\!\overset{\oplus}{C}-\overset{\ominus}{O}\!}$$


Copper acetate solution turning clear I made a solution of copper acetate using 5% Acetic acid 3%hydrogen peroxide and copper wire and copper pipe and it had produced a blue solution of copper acetate. I left it over night to continue reacting, but when I came back In the morning the solution turned clear. I didn’t do any Stoichiometry so I don’t know if they’re was a excess of H2O2 or a excess of acetic acid. I’m wondering if you know what happened

Wikipedia states that upon heating, copper(II) acetate reacts with metallic copper to make copper(I) acetate, which would be colorless.  While this is no guarantee of what happened with your solution, you can mix a sample of the solution with saltwater (chloride ions).  White precipitate = copper(I).

In light of a comment, I am compelled to expand on the possibility of copper(I) in solution.  According to the Wiley Online Library, this salt is soluble in water with hydrolysis to cuprous oxide.  At the same time we have the copper Pourbaix diagram reported from Ref. 1, indicating that the oxide is stable only above a pH of 4.

This means any significant leftover acetic acid could prevent the hydrolysis and keep the copper(I) in solution, apparently stabilized by complexing with the acetate ion given that disproportionation is not reported.  (Such a stabilization is not included in the above diagram but would contract the Cu2O region further.)

Since this is only indirect evidence, again testing for a copper(I) chloride precipitate by mixing a sample with saltwater is recommended.

Reference

1.
Oh, Youn-Jin & Park, Gyung-Soon & Chung, Chan-Hwa. (2006). Planarization of Copper Layer for Damascene Interconnection by Electrochemical Polishing in Alkali-Based Solution. Journal of The Electrochemical Society - J ELECTROCHEM SOC. 153. 10.1149/1.2200288.


When can a Secondary standard solution be used for standardising something instead of a Primary? I am writing an assignment on 'Preparing and Using Standard solutions for Use in Analysis in Titrations and Spectroscopy' and one part is asking about primary and secondary standard solutions.

I know the different between them and have stated what they are and some examples, but one question I have to answer is 'Explain why Primary standard substances are not always used in standardization, and why Secondary standard substances are used instead'.

I have looked at many sources online and they all just tell me the differences between primary and secondary, was wondering if anyone could shine some light on why a secondary standard solution would be used to standardize a solution instead of a primary?

If you are doing pH titrations, and your primary standard is an acid - say - potassium hydrogen phthalate, you can't use that to standardize an acid directly. Instead, you would make a secondary standard of - say - sodium hydroxide, titrating against your primary standard. Then, you are ready to standardize the acid you want to use in your titration.

Secondary standards are also used if the primary standard is expensive. Hydrochloric acid solutions are cheap and stable, but you can't make them by weighing in a solid. So instead, you make an expensive primary standard and a lot of cheap hydrochloric acid of unknown concentrations, and determine the titer of the hydrochloric acid. Then, you can use it as a secondary standard.


α versus β D-ribopyranose I understand that sugars are named α/β according to the arrangement of the OH on the anomeric C with respect to the hydroxymethylene group (Ex. glucose), but I'm a little confused how you determine α/β when you don't have that group (Ex. ribopyranose, below).

α-D-ribopyranose

 β-D-ribopyranose

Thanks!

It's actually not the hydroxymethelene group that determines the α/β status of a sugar, instead it's the relative orientation of the anomeric hydroxyl to the stereocenter which determines the absolute stereochemistry (D/L) of the sugar. This just so happens to match up with the same side/opposite side heuristic for the common six carbon pyranoses, but that's a quick rule of thumb, not the official definition:

  α (alpha), β (beta): Relative stereodescriptors used in carbohydrate nomenclature to
  describe the configuration at the anomeric carbon by relating it to
  the anomeric reference atom. For simple cases the anomeric reference
  atom is the same as the configurational reference atom. Thus in
  α-d-glucopyranose the reference atom is C-5 and the OH at C-1 is on
  the same side as the OH at C-5 in the Fischer projection.

IUPAC Gold Book

  The configuration at the anomeric centre (that derived from the
  carbonyl carbon) is denoted alpha- (α-) or beta- (β-) by reference to
  the stereocentre that determines the absolute configuration. In a
  Fischer projection, if the substituent off the anomeric centre is on
  the same side as the oxygen of the configurational (D- or L-) carbon,
  then it is the α--anomer. If it is directed in the opposite direction
  it is the β-anomer.

CAZypedia - includes diagrams and examples, including for five carbon pyranoses.


Nomenclature and bonding of the complexes ions Why the nitrogen and not the sulphur in the SCN is bonded to the chromium?

The bonding of the $N$ end of the thiocyanato group to the Metal centre $\ce{Cr^3+}$ can be explained by both HSAB Theory and Symbiotic effect.According to HSAB Theory, Hard acids prefer to bind with  hard bases, while soft acids prefer to bind with soft bases. Here the metal centre is $\ce{Cr^3+}$, which has $3d^3$ electronic configuration. So,  $\ce{Cr^3+}$ satisfies the criterions for a Hard acid centre, as it has very less no. of $d $ electrons, high positive charge density on it, smaller radius and greater polarisation. So, the metal centre in this complex acts as Hard acid centre. And, now if you consider the properties of a hard base, you will find that $N$ is the harder basic centre, whereas, $S$ is the softer base centre. $N$ has higher electron charge density on it and also smaller in size than $S$ which makes it harder base. Thus, due to the presence of the harder metal acid centre, the ligand tries to bind with the harder basic centre, which in this case is $N$. The attachment of the $N$ site can also be explained by the Symbiotic effect. Observe that, in the complex, you already have two $\ce{NH_3}$ ligands donating through the $N$ site, which actually compells the thiocyanate ligands also to coordinate through the $N$ site, which is the main idea of this effect. Actually, donation through $N$ sites of ammonia makes the metal centre even more hard, and therefore it prefers also the $N$ site of other ligands.  But a remark which is necessary is that the HSAB theory doesn't always predict the correct coordinating site for an ambidentate ligand. If even stronger effects like Competitive $\pi$-Bonding effect start to interfere due to presence of $\pi$-accepting and $\pi$-donating ligands, HSAB theory may not give a correct answer at all. So, in the absence of those effects, HSAB theory  works well.


Why can't we calculate the atomic radii of noble gases? My book says that only the Van der Waal's radius of noble gases can be measured as these gases generally do not form compounds (except Xenon.) Then why can't we just calculate the atomic radii by computation, say solving the Schrödinger equation? Can't we make computer simulation models of isolated atoms of noble gases?

Covalent radius is measured as the distance ($r_\text{cov} = d/2$) between the nuclei of two bonded atoms (covalent). But, if you try to do the same for noble gases/inert gases (good luck!), as they have fully filled $np$ orbitals, they will repel each other, hence the closest distance between the two atoms is taken (high pressure, low temperature) as the van der Waals radius.


What is the exact meaning of GAMESS (US) SCF density convergence criteria? In Gaussian the SCF convergence criteria is set to $10^{-N}$ by

SCF=(Conver=N)

Note that this criteria applies to the root mean square change in density matrix between two SCF cycles, while the maximum change convergence criteria is set to $10^{-N+2}$. For instance, for the default case of $N=8$ the output says

 Requested convergence on RMS density matrix=1.00D-08 within 128 cycles.
 Requested convergence on MAX density matrix=1.00D-06.
 Requested convergence on             energy=1.00D-06.

Now in GAMESS (US) the SCF convergence criteria is set to $10^{-N}$ by

$SCF CONV=1.0d-N $END

However, neither manual, nor the output do not explicitly specify does this criteria refers to the root mean square change in density matrix or to the maximum change in it? Judging from the default value (1.0d-05), it is the last one, the maximum change in density matrix, otherwise it would be too small for a reliable convergence. Thus, SCF=(Conver=N) in Gaussian corresponds to $SCF CONV=1.0d-(N+2) $END in GAMESS (US). But I'm not sure, so can anyone confirm that?

P.S. I tried to look at the source code, but it is an old FORTRAN mostly undocumented spaghetti mess with good-old six-lettered totally undescriptive variable names. :D

The difference is calculated in the subroutine DDIFF (scflib.src) and corresponds to the largest absolute change in the density matrix


Fehling's Test meaning for partial formation of a precipitate? Whilst doing tests on two unknown solutions, I used Fehling's test. The solutions could be a mixture of compounds or just one compound dissolved in water/ another solvent - I don't know. Upon adding Fehling's solution both solutions (which were originally colourless) turned a blue colour. I then heated the solutions very gently for about 5 minutes and after this time one had completely formed a red precipitate, whereas the other still had the blue solution left in it (while forming some red precipitate). After waiting for about 20 minutes both solutions had completely formed a brick red precipitate and were the same. I assume this is to do with rate of the oxidation reaction occurring. Is there any explanation as to the rate of reaction in each solution or an explanation which is not to do with rate at all perhaps?

The reason is thus:

For aldoses (like glucose), the copper sulfate in the reagent easily oxidizes it into the carboxylic acid forming copper oxide (a precipitate).

For ketoses (like fructose), the ketone cannot be oxidized, so there is no color change. Over time at elevated temperature, the α-hydroxy ketone isomerizes to a α-hydroxy aldehyde. This aldehyde is then oxidized to the carboxylic acid, producing a red precipitate.


Fries Rearrangement Products During Fries Rearrangement, oxygen donates its lone pair to $\ce{AlCl_3}$ forming $\ce{-O^+-AlCl3}$. Since this is a strong electron withdrawing group, shouldn't the final product be meta-substituted?

The above is the mechanism for Fries Rearrangement for a brominated ester( doesn't matter really, for any ester the mechanism essentially remains to be the same).As per what I infer from your question, I think you mean to ask why even in the presence of $\ce{O^+-AlCl3}$ which is a strongly deactivating group, the major product is not meta-substituted.
As you can clearly see, in the next step itself a stable acylium is formed and the formal charge on $\ce{O}$ is no longer positive.So, as phenoxide ion is an activating group and o,p-director, two possible products are expected which are ortho and para substituted.However, at higher temperatures, the major product is ortho substituted which is driven by chelation.
NOTE:
As far as I know, both intramolecular and intermolecular reaction mechanisms are able to explain the major products and it is still not confirmed which one is the actual mechanism.
Image source:
https://www.synarchive.com/named-reactions/fries-rearrangement


Stereo selective reaction products Pyruvic acid on reduction with yeast gives (-) lactic acid while with $\ce{NaBH4}$ it reduces to (+) lactic acid. How?

I know this has something to do with stereoselectivity. But I don't know which one would be a correct answer.

It would be more helpful if you could explain a bit about stereo-selective reactions. I know what it is, what I want to know is, how to find which of the products will be stereochemically preferred one, just like in this question.

Enyzmatic reduction (as would occur in yeast) could easily be a stereoselective process - probably via selective delivery of a hydride from NADH.

As the other respondent notes, there is simply no way that combining achiral reagents (pyruvic acid and NaBH4) can yield anything other than a racemic product.


How can I find out a substance name based on its structural formula? When visiting a botanical garden in Stockholm, I stumbled across the following mosaic tile:

I have tried to identify it using MolView, but it does not seem to identify it. Also, I have tried their advanced search (similarity), but that provided no results.

I have also tried searching for it based on SMILES formula on Chemspider.

I am wondering if there any other tools that can help me in finding what substance it is.

Extra context information: it was next to a DNA structure artwork, so I guess it is somewhat related to the evolution, DNA etc.

Question: How can I find out a substance name based on its structural formula?

That's betanin, from which beets derive much of their deep purple color, and which may have neuro-protective (Ref. 1) features. The following is the structure retrieved from the wikipedia (on 2019-10-5), which closely matches that in the mosaic:

I used the editor available on the Sigma-Aldritch website to generate a structure and submit a search:

The program identified betanin as the top hit despite my sloppy structure.

The SMILES code is: C(OC(C1O)OC(C(=CC2[N+]3=CC=C(CC(N4)C(=O)O)C=C4C(=O)O)O)=CC=2CC3C([O-])=O)(C(C1O)O)CO

The IUPAC name (retrieved from the wikipedia) is (2S)-1-{2-[(2S)-2,6-dicarboxy-2,3-dihydropyridin-4(1H)-ylidene]ethylidene}-5-(β-d-glucopyranosyloxy)-6-hydroxy-2,3-dihydro-1H-indol-1-ium-2-carboxylate

As noted in a comment, the stereochemistry in the wikipedia image above and that in the mosaic does not match that of the structure in the Sigma-Aldrich catalogue. I used ACD/ChemSketch freeware to generate the following image that does match the one in the catalogue:

American Chemical Society. "Vegetable compound could have a key role in 'beeting' Alzheimer's disease." ScienceDaily, 20 March 2018.

As an aside, it's interesting to note that work on L-dopa, a biosynthetic precursor of betanin, led to a Nobel prize award for Swedish scientist Arvid Carlsson. Also, quoting from the wikipedia:

  The 2001 Nobel Prize in Chemistry was also related to l-DOPA: the Nobel Committee awarded one-quarter of the prize to William S. Knowles for his work on chirally catalysed hydrogenation reactions, the most noted example of which was used for the synthesis of l-DOPA.

The structure-drawing engine (and presumably also search engine) provided through the Sigma-Aldrich structure search web service is JSDraw2 (JSDraw V3.2.3) by Scilligence.


Reaction intermediates of MnO2 catalyzed H2O2 decomposition reaction Manganese dioxide catalyzes the decomposition of hydrogen peroxide to water and oxygen gas.

But what are the intermediates in this catalyzed reaction?

Reaction conditions

As Watts et al. have shown, the decomposition products of this Fenton-like reaction strongly depend on the $\mathrm{pH}$ of the solution.1

If performed in acidic conditions, the reaction generates mostly hydroxy radicals, but no reductants (which would be the hydroperoxide and superoxide anions). If, conversely, the reaction is held in neutral conditions, Watts et al. have shown that the reaction produces significantly more of the aforementioned reductants.

Do et al. have conducted further research into the reaction mechanism at $\mathrm{pH}=7$ and concluded that the observed reaction order may be approximated as a pseudo 1st order reaction, where the ratio $\ce{[H2O2]}/\ce{[#MnO2]}$ is vital to the description of the decomposition rate.2 They have further shown that reactive intermediates, such as superoxide and hydroperoxide anions are generated by the reaction. After slightly modifying the pH towards alkaline conditions, the production rates for the reactive anions increased drastically.

Reaction mechanism

Do et al. have meticulously assembled a table with a proposed reaction pathway:

And finally, I quote one of their closing statements:2

  The existence of hydroperoxide/superoxide anion implies that the suggested reaction mechanism could be explained by hydrogen peroxide being decomposed, not only directly on the surface of manganese oxide, but also through a propagation reaction involving intermediates such as hydroperoxide/superoxide anion in solution.

References

Watts, R., Sarasa, J., Loge, F., and Teel, A. Oxidative and Reductive Pathways in Manganese-Catalyzed Fenton’s Reactions. J. Environ. Eng., 131(1), 2005, 158–164, DOI link.
Si-Hyun Do, Bill Batchelor, Hong-Kyun Lee, Sung-Ho Kong. Hydrogen peroxide decomposition on manganese oxide (pyrolusite): Kinetics, intermediates, and mechanism. Chemosphere, Volume 75, Issue 1, March 2009, Pages 8-12, DOI link.


Why is HF considered a covalently bonded compound? Why is the bond between H and F in HF considered polar covalent whereas HCl, HI, and so on are all ionic? The electronegativity difference between them suggests that it too should be ionic, yet all textbooks say that HF is covalent. Is there a good reason why?

I don't think I've ever seen a reputable source claiming the bonds in HCl, HBr and HI are mainly ionic. The fact that they are all gasses at room temperature clearly suggests otherwise. HF has the most ionic character of all the hydrogen halides, but it too has a rather low boiling point (just below room temperature), which is uncharacteristic of ionic compounds.

It is best to think of all hydrogen halides as covalent polar molecules, with the polarity increasing in the order HI < HBr < HCl < HF, as suggested by the electronegativity differences.


What is the correct reaction of sulfur dioxide with chlorine? I'm confused in reaction of $\ce{SO2}$ with $\ce{Cl2}$. Some source states the given reaction (1), while others state the reaction (2):

$$\begin{align}
\ce{SO2 + Cl2 &-> SOCl2 + Cl2O \tag{1}}\\
\ce{SO2 + Cl2 &-> SO2Cl2 \tag{2}}
\end{align}$$

While I read in NCERT Chemistry 12 part-II page-198 that chlorine oxidises $\ce{SO2}$ to $\ce{SO3}$. I wish to know the specific condition in which the particular reactions will occur.

Sulfuryl chloride ($\ce{SO2Cl2}$) is prepared by the reaction of $\ce{SO2}$ and $\ce{Cl2}$ over activated charcoal, details of the preparation are here 1.

Thionyl chloride may be prepared from $\ce{SO2}$ and $\ce{Cl2}$ but the reaction requires an equivalent of $\ce{SCl2}$ to be present.


What is the effect of the temperature on the no of moles in an equilibrium system? Here's something that makes me confuse. What will happen to the total no of moles in a system if the temp is raised. Will it increase? If so how is that possible.?
Say we have an equlibrium,
$$\ce{2P <=> Q + R}$$
The temp is $300\:\mathrm{K}$ and total no of moles is 6 moles.
If we increase the temp upto $500\:\mathrm{K}$ will the total no of moles change to 10 moles when the volume and pressure kept constant?

Well, your equation contains the same number of moles either side, so temperature does not matter.

But if the equation had different moles on either side, it should be given if the reaction is endothermic or exothermic.

For eg- if a reaction is exothermic and the temperature is increased, then according to Le Chatelier's principle, the reaction should go in the backward direction (To compensate for the increase in temperature).
So, if the reactant side has more moles than product side, the number of moles will be increased. You can apply similar logic if the reaction is endothermic.


Why can cloud's bottoms be flat? I've always wondered why clouds can have flat bottoms, and have debated to myself whether it was a difference in atmospheric pressures, or some water molecule tendency? Could someone shed light?

That many types of clouds have flat bottoms is a function of decreasing air temperature with increasing height, and the water vapor pressure, commonly in this context called the dew point, of the air. When some force causes a non-cloudy parcel of air to rise, moisture will condense out as cloud particles just as soon as the air temperature reaches the dew point of the air mass.  The colder air (higher up) then contains a cloud, where the warmer air (even just slightly above the dew point) below the cloud remains clear.


What is the difference between a positive and negative electron orbital? Why are do some electron orbitals have a negative number in the subscript (i.e. f-1), some a positive (f1), and some have a 0 (f0)? What do these numbers tell you about the atom? I know that these orbitals have different shapes, but I was wondering if they had any other effect on the atom/ion and if there was a reason that they were put in the order that they are in now.

There are no such things as positive or negative electron orbitals. The number in the subscript is $m$, known as magnetic quantum number and running from $−\ell$ to $\ell$. It doesn't tell you much about the atom. It is a property of the orbital, not of the atom. Every atom has all these orbitals at once (some of them occupied and some empty); also, there are versions of each for different $n$.

In a free atom, all orbitals with the same $n,\;\ell$ and different $m$ have the same energy, so this number does not mean much for an orbital either (at least unless you put the atom in a magnetic field).

Moreover, the pictures are fake anyway. Orbitals which are the true eigenfunctions of $m$ (that is, correspond to specific $m$) are complex-valued and thus can't be properly represented in a plot. As for the orbitals featured on this diagram (and all over the textbooks), these are the linear combinations of complex orbitals for different $m$, hence each of them does not correspond to any particular $m$. It's not that they are wrong per se (absolutely not!), it's just that you can't point at them and tell which is $p_1$ and which is $p_{-1}$.


Are there any "catalysts" that decrease the rate of reaction Catalysts are things you can add to a reaction that provide an alternate reaction pathway for the reaction that decreases the overall activation energy of the reaction meaning the reaction is faster but the catalyst isn't consumed in the reactions. Are there any materials, in any reactions where a similar material can be added to a reaction that will decrease the rate of reaction by providing a more favorable reaction pathway but one that takes longer.
Maybe it isn't possible for a reaction to be more favorable but require a higher activation energy and this is why this kind of "pseudo catalyst" doesn't exist.

Well, such a "negative catalyst" cannot increase the activation energy of already existing path.
An alternative with the higher activation energy would not be significantly taken. You can detour a river making it flow easier by a side channel, but not if you lead it across a hill.
It could act differently, as an  inhibitor, blocking the low activation energy path by reaction with the key intermediate product or existing catalyst, or by competitive inhibition.
Many redox reactions, including food deterioration, are catalyzed by traces of transient metals acting as redox catalysts, typically copper. An additive, forming metal complexes like EDTA, in large extend inhibits these catalyzed reactions.


Gas Law problems with stoichiometry Consider the following reaction : $\ce{H2 + Cl2 -> 2HCl}$

A) If you have 15.00 liters of hydrogen reacting with 10.00 liters of chlorine at STP, how many liters of HCl would you produce?

B) What would be the limiting reagent?

C) How many moles of excess reagent would remain?

Please explain.

Molarity = moles/Liters
to check for $\ce{H2}$: M = 2/15.00 = .13334
to check for $\ce{Cl2}$: M = (35.45 x 2)/10.00 = 7.09

But I do not think that is correct. can somebody show me the correct way to do it.

Avogadro's law says that $V/n = \text{constant}$ at constant temperature and pressure.   That means your volumes can be used "as if" they were moles in the chemical reaction stoichiometry.   So, if the mole ratio is 1:1 for H$_2$ and Cl$_2$ reacting, then the limiting reagent is obviously the smaller volume of 10.0 L for Cl$_2$.

For (b) you can easily calculate the volume of HCl since 1 mole of Cl$_2$ produces 2 moles of HCl.  So, you get 20.0 L of HCl.

For (c) the excess reagent is the H$_2$.  Since it reacts with 10.0 L of Cl$_2$ in a 1:1 mole ratio, then you have 15.0L - 10.0 L = 5.0 L of H$_2$ left over.   Since this all ends up at STP, you can use the ideal gas law to calculate the moles of H$_2$ left over:

$n_{H_2} = \displaystyle \frac{p V}{R T} = \frac{(1 \text{atm})(5.0 \text{L})}{R \times 273.15 \text{K}} = 0.2230749373505515$ mol

Don't forget sig. figs., which in this case is determined by the 2 sig. figs. in the volume of 5.0 L, giving you

$n_{H_2} = 0.22$ mol

For Review check out:
gas laws: http://www.grandinetti.org/Teaching/Chem122/Lectures/Gases

If you have a mac, check out PhySyCalc. It will automatically do all the unit conversions in your calculations.

https://itunes.apple.com/us/app/physycalc/id644105995


Is 3,4-diethyl-3-hexene cis or trans? I'd really appreciate it if someone could explain this one, please! According to my answers sheet it is not cis nor trans. But it looks like it is cis? Why, according to my answers sheet, is this neither?

Thank you in advance!

As each of the groups surrounding the double bond are all identical there is no optical isomers of this molecule, as no matter which way you rotate or swap groups around you will always have the same molecule. therefore as it has no optical isomers it is neither cis nor trans


Is there a difference between hydrated oxides and hydroxides? My guess is that they are different.

I think hydrated oxides would have a lattice with the cation and oxide anions as lattice points and molecules of water trapped in interstitial spaces, while hydroxides would have the cation and hydroxide anions as lattice points with no water molecules in interstitial spaces.

Hydrated oxide is a compound of an oxide with water. And Hydroxide is a diatomic anion with chemical formula OH− ( a hydroxyl functional group). you can refer to this link for better understanding : http://www.memidex.com/hydrated-oxides. Also looking at the definition of hydration might be helpful.


Concerted mechanism of saponification The energetics of the mechanism of saponification were previously under debate, but I will admit, I misrepresented a point.

Here is the revised mechanism of saponification proposed by someone other than myself. It is the one on top.

The previously suggested mechanism is in the middle.

The textbook mechanism is the bottom one.

Now, it does seem that if the mechanism were concerted then proton abstraction would be much more plausible from a thermodynamic standpoint. If the second mechanism were akin to scaling Mt. Everest, then the first one looks to me to be more like climbing the rock wall at the gym.

We start with the same reactant - an ester - but in the second mechanism we are going through a dianion intermediate. In the first mechanism however we are simply going through a concerted transition state en route to the stable carboxylate.

So, again, I ask - which is (kinetically) correct?

The question borders on a semantic argument. In the third mechanism (textbook), we would assume that the deprotonation event is very fast. The first mechanism (concerted) is basically saying that the deprotonation is so fast that it cannot be distinguished from the alkoxide leaving.

We know that acid-base reactions are fast. We also know that an anionic oxygen 'kicking out' an alkoxide is fast, at least relative to nucleophilic attack on a carbonyl (the rate determining step for any of these reactions AND the related trans-esterification is nucleophilic attack on a carbonyl to give the tetrahedral intermediate). What we're trying to answer is how fast is fast?

Here is my argument for the third mechanism. While acid-base reaction is fast, the top mechanism is essentially an E2 reaction, which does have a stereoelectronic preference for the hydrogen to be anti-periplanar to the leaving group. I have no doubt that conformation is readily accessible (the C-O bond will be rotating freely). The step is also bimolecular. In the third mechanism, an alkoxide kicking out a leaving group has no stereoelectronic preference. There are no alternate conformations to even consider. The C-O bond may be rotating, but it doesn't matter: there is always an electron pair arranged properly to kick out the alkoxide. Furthermore, the step is unimolecular.

The above is speculation, because I have never seen an alternative to the standard textbook saponification mechanism before. To be honest I never considered the concerted mechanism, and although I think it is a reasonable mechanism, I still prefer the textbook mechanism.


Wavelength of X-rays What will be the wavelength of an X-ray, if we apply a potential difference of, say $\pu{20 kV}$ across the ends of the X-ray tube?
Alright, I started with the modified de-Broglie equation-
$$\Lambda=\frac{h}{\sqrt{meV}}$$
I pinned the values of Planck constant, electronic charge and potential difference but can't guess what would be the mass $m$ here.
The answer comes $\pu{0.62*10^-10m}$ if solved perfectly. Can anybody confirm?

Firstly as @MaxW pointed out, using the given information, it is possible to find the shortest wavelength (or maximum frequency) X-ray only.

In an X-ray tube, electrons are accelerated in a vacuum by an electric field and shot into a piece of heavy metal (e.g., $\ce{W, Rh, Mo, Cu, Ag}$) plate. X-rays are emitted as the electrons decelerate in the metal. The output spectrum consists of a continuous spectrum of X-rays, with sharp peaks at certain energies as in the graph. The continuous spectrum is due to bremsstrahlung (German for "deceleration radiation"), while the sharp peaks are characteristic X-rays associated with the atoms in the target.
The spectrum has a sharp cutoff at low wavelength (high frequency), which is due to the limited energy of the incoming electrons (which is equal to the voltage on the tube times the electron charge). This cutoff applies to both the continuous (bremsstrahlung) spectrum and the characteristic sharp peaks, i.e. there is no X-ray of any kind beyond the cutoff. However, the cutoff is most obvious for the continuous spectrum.
Hence, the Duane-Hunt law comes handy in here, which gives the maximum frequency of X-rays that can be emitted by Bremsstrahlung in an X-ray tube by accelerating electrons through an excitation voltage $V$ into a metal target.
The maximum frequency $\nu_\text{max}$ is given by-
${\displaystyle \nu _{\text {max}}={\frac {eV}{h}},}$
which corresponds to a minimum wavelength
${\displaystyle \lambda _{\text {min}}={\frac {hc}{eV}},}$
where $h$ is Planck's constant, $e$ is the charge of the electron, and $c$ is the speed of light.
Thus putting the values given-
$$\lambda _{\text {min}}={\frac {hc}{eV}}={\frac {\pu{1240 eV nm}}{\pu{20*1000 eV}}}= \pu{0.62*10^-10 m}$$
For further references check the Wikipedia page.


Salt analysis of Ag salts So I was doing analysis of Ag salts and I found Agbr is curry white partially soluble in ammonia , AgI is yellowish and insoluble . BUT silver phosphate is also yellow . So can anyone tell whether or is soluble in ammonia or not ?

Silver phosphate is soluble in aqueous ammonia. AgBr is only slightly soluble in ammonia. AgI is not soluble in ammonia.


Pourbaix diagram – writing reactions How would you go about writing the reaction between water and chromium metal under standard acid and base conditions?

I know you would have to look at the vertical lines at pH 0 and 14 but I don’t know how you would know which species to look at and whether or not you write the reduction or oxidation of water.

Also, at pH 14 how would you calculate the ${\Delta G}$ value for the reduction between water and chromium metal?

The red line in your diagram corresponds to the redox potential of hydrogen, which depends on $\mathrm{pH}$:

$$\begin{alignat}{3}
\ce{2H+ + 2e- \;&<=> H2}\quad\quad &&E^\circ = +0.000\ \mathrm{V}\quad\quad &&&(\mathrm{pH}=0)\\
\ce{2H2O + 2e- \;&<=> H2 + 2OH-}\quad\quad &&E^\circ = -0.828\ \mathrm{V}\quad\quad &&&(\mathrm{pH}=14)
\end{alignat}$$

The blue line in your diagram corresponds to the redox potential of oxygen, which depends on $\mathrm{pH}$, too:

$$\begin{alignat}{3}
\ce{O2 + 4H+ + 4e- \;&<=> 2H2O}\quad\quad &&E^\circ = +1.229\ \mathrm{V}\quad\quad &&&(\mathrm{pH}=0)\\
\ce{O2 + 2H2O + 4e- \;&<=> 4OH-}\quad\quad &&E^\circ = +0.401\ \mathrm{V}\quad\quad &&&(\mathrm{pH}=14)
\end{alignat}$$

Therefore, only areas between these lines are stable in aqueous solutions.

Your initial species is elemental chromium ($\ce{Cr}$). It is shown below the red line; therefore, it is unstable with respect to oxidation to $\ce{Cr^3+}$ at $\mathrm{pH}=0$

$$\begin{alignat}{2}
\ce{2H+ + 2e- \;&<=> H2}\quad\quad &&E^\circ = +0.000\ \mathrm{V}\\
\ce{Cr^3+ + e- \;&<=> Cr^2+}\quad\quad &&E^\circ = -0.408\ \mathrm{V}\\
\ce{Cr^2+ + 2e- \;&<=> Cr}\quad\quad &&E^\circ = -0.913\ \mathrm{V}\\\\
\ce{2Cr + 6H+ &-> 2Cr^3+ + 3H2}
\end{alignat}$$

and to $\ce{Cr(OH)3}$ at $\mathrm{pH}=14$

$$\begin{alignat}{2}
\ce{2H2O + 2e- \;&<=> H2 + 2OH-}\quad\quad &&E^\circ = -0.828\ \mathrm{V}\\
\ce{Cr(OH)3 + 3e- \;&<=> Cr + 3OH-}\quad\quad &&E^\circ = -1.33\ \mathrm{V}\\\\
\ce{2Cr + 6H2O &-> 2Cr(OH)3 + 3H2}
\end{alignat}$$


Why does adding salt to boiling water cause it to flare up suddenly? I was expecting that the water would become less "boily" since the ions in $\ce{NaCl}$ would require energy to disassociate. Instead it turned whitish and the water level suddenly rose and then subsided. I think the white color can be attributed to the fact that table salt is white but why did the water suddenly flare up ?

The salt does not instantly dissolve so the surface of the crystals suddenly provides a lot of nucleation sites for the water to form vapour - hence the surge as it boils from these surfaces. The white is due to turbulence from the boiling just like waterfalls look white.


Reaction of cyclohexane-1,2,4-triol with acetone What is the product formed when cyclohexane-1,2,4-triol reacts with 1 equivalent of acetone in an acidic medium?

There are 3 possible hemiacetal products. The major one is formed via nucleophilic attack by a less hindered hydroxyl group:

There is one possible cyclic acetal formed by the two adjacent $\ce{- OH}$ groups:

But which of the two products will be formed in a larger quantity? I have heard that cyclic acetals are more stable than acyclic ones but don't know the reason (because of entropy change or something?).

As long as you have a proton source in your solution (hint: all the time) these acetals will equilibrate back and forth until they find the thermodynamic hole. So you will never observe a hemiketal like the one you drew; rather, it will always move forward with the elimination of water (there is only little water around if acetone is the solvent, so liberation of water is entropically favourable) to the formation of the $\ce{O{,}O}$-ketal.

Note that you can form two different ketals if the stereochemistry of the hydroxy groups is correct: If they are 1,3-syn, the six-membered cyclic ketal can also be formed.

Which one you end up with will likely depend solely on the reaction conditions. Kinetically, the five-membered ring is favoured. Thermodynamically, I assume the six-membered ring to be more stable, if it is able to find a conformation with little steric stress.


Use of axis of symmetry Does an axis of symmetry determine chiralty?

Is axis of symmetry considered a true symmetry?

From the above two links, I read that a compound having axis /alternating axis of symmetry is NOT necessarily achiral. For example, trans-1,3-dimethyl cyclopentane has an axis of symmetry but it is still chiral and hence optically active.

My doubt is, then why would we look for alternating axis of symmetry in the first place? Where/How is it useful?

Alternating axes are not axes, much like a Guinea pig is not a pig.

Having axes of symmetry is irrelevant to chirality.

Having alternating axes is a necessary and sufficient condition of non-chirality, that's why we look for them.

So it goes.