Why is the first ionisation energy of a sodium cation bigger than that of a neon atom? I understand that the 1st ionization energy gets bigger for elements along a period from left to right and along a group from down to up.

But why is the 1st ionization energy of $\ce{Na+}$ bigger than $\ce{Ne}$?

I would have guessed the opposite.

A sodium cation $\ce{Na+}$ and a neutral neon atom $\ce{Ne}$ are isoelectronic species; meaning they have the same number of electrons and also the same electronic configurations.

Yet, the two have different nuclei: most prominently, sodium always has an additional proton when compared to neon. Therefore, only one of the isoelectronic species (neon) is neutral, the other is positively charged.

It may now become obvious why the ionisation enthalpy of $\ce{Na+}$ is much larger than that of neon. Different ways to express this all boil down to the fact that it is harder to remove a negatively charged electron from a positively charged species than it is to remove one from a neutral species.


How can I figure out which coupling constants match which hydrogen? I am looking for some guidance assigning the $J$ coupling constants to the corresponding hydrogens. Here is my data:

I can obviously see that shift 3.56 is for the methyl group. I can also see that the 4.63 is for the hydrogen adjacent to the methoxy group as it can only couple ($J=\pu{0.8Hz}$) to one hydrogen next to it. Is the approach then to see that the hydrogen at shift 4.04 also has $J=\pu{0.8Hz}$ so this must be adjacent (going clockwise from C of methoxy, next C along)?

Coupling constants are mutual. If nuclei $\ce{A}$ and $\ce{B}$ couple and this coupling results in a coupling constant $J_\ce{AB} = \pu{7.4Hz}$ on $\ce{A}$ then $\ce{B}$ will also be split into two signals with a coupling constant $J_\ce{BA} = \pu{7.4Hz}$.

Therefore, you can indeed start at the anomeric hydrogen (which you can identify not only by the coupling pattern but also by its chemical shift) and work your way along the ring, noting whether two hydrogens are in gauche or anti configuration along the way.


Is the density of a two-component mixture always between the densities of its individual components? There exist eutectic mixtures, in which the freezing point of the mixture is below those of either component of the mixture, i.e., the freezing point of the mixture is not intermediate between that of each component, but is instead lower than both.
Can the same occur for density? Are there mixtures of substances $\ce{A}$ and $\ce{B}$ such that either $\rho_\mathrm{mix}$ is greater than both $\rho_\ce{A}$ and $\rho_\ce{B},$ or $\rho_\mathrm{mix}$ is less than than both $\rho_\ce{A}$ and $\rho_\ce{B}?$
Or is the density of a two-component mixture always bracketed by (i.e., some intermediate value between) the densities of its individual components?

No, there is no law that requires the density of a mixture to fall between the densities of pure components. It does so most of the time, but the exceptions are not unheard of. Here's one.
Water: density $1.00\ \rm{g/cm^3}$
Hydrazine: $1.02\ \rm{g/cm^3}$
Hydrazine hydrate: $1.03\ \rm{g/cm^3}$
So it goes.


Is there a relationship between viscosity and enthalpy of vaporization? It would be useful to be able to predict relative evaporation rates of liquids.  For small reduced temperatures the heat of vaporization is relatively constant, but it is not one of the more common properties cited for substances.

Is there any sort of functional relationship between enthalpy of vaporization and common properties like viscosity, melting point, and boiling point?

Yes. The phase envelope (or equilibrium curve) of most liquids have a similar shape and can be approximated, to a reasonable degree, by using the Clausius-Clapeyron equation:

$$
\Delta h_v=T(v^V-v^L)\frac{dP^s}{dT}
$$

which requires finding ${dP^s}/{dT}$ from e.g. a vapor-pressure-temperature correlation like the Antoine equation, and also a separate estimate for $\Delta v$ before $\Delta h_v$ can be obtained. This is an exact thermodynamic relationship, but is often simplified by using the ideal gas law for $v^V$ and neglecting the liquid volume $v^L$ i.e. $v^V \gg v^L$, giving

$$
\Delta h_v=\frac{RT^2}{P^s}\frac{dP^s}{dT}
$$

However, it is very important to note that the term $(v^V-v^L)$ becomes increasingly decisive as you go above 1 bar, where the assumption $v^V \gg v^L$ is no longer valid. In such a case, separate correlations for $v^L$ and $v^V$ are required before the correct curvature is obtained. $v^L$, for instance, can be calculated from e.g. a density correlation like the Rackett equation, and $v^v$ from an equation of state like Peng-Robinson.


The thermal decomposition of FeCO3 The thermal decomposition of siderite equation: $\ce{FeCO3 -> FeO + CO2}$ looks correctly balanced to me, you can notice $\ce{1 Fe , 1 C , 3 O}$ on the Left side and the same number of molecules on the other side.

However my text book writes it as $\ce{2FeCO3 -> 2FeO + 2CO2}$
and I wanted to know why mine is not correct.

Well, just like any other reactions, it depends upon the the heating temperature and the reaction environment. Contrary to what the reaction tells, the decomposition products of the reaction is actually iron(II,III) oxide, $\ce{Fe2O3}$ if it is heated in oxygen environment$\ce{^{[1]}}$ and $\ce{Fe3O4}$ if it is heated in $\ce{CO2}$ environment$\ce{^{[2]}}$.  $\ce{FeO}$ also formed during the reaction but for a small period of time. Hence it is an intermediate product.

  Figure 1 shows the thermal decomposition of iron carbonate at
  temperature at 600°C. It is shown from one DTA curve in Fig. 1 one
  exothermic peak at 495°C accompanied with a weight loss of about 32%
  (theoretical value 31.1%). This step is due to the decomposition of
  iron carbonate to $\ce{Fe2O3}$.

  X-ray diffraction patterns of the decomposition products of iron
  carbonate were determined for the solid calcined at 550 and 800°C
  Figure 3 (a) shows X-ray diffractogram for the sample dried at room
  temperature corresponding to iron carbonate. The diffractogram of the
  sample calcined at 550°C consists of very poorly crystallized α-Fe2O3
  phase. .Heating of ferrous carbonateat 800°C led to the formation of
  well crystalline $\ce{α-Fe2O3}$

  In air, the X-ray pattern of $\ce{FeCO3}$ was replaced by that of
  $\ce{Fe3O4}$ as the sample was heated. The change was essentially
  complete at 500°C. No further changes were noted. If, however, the
  furnace atmosphere was changed from air to $\ce{CO2}$ or helium above
  1000°C. the sample transformed slowly to $\ce{Fe3O4}$. Restoring the
  air atmosphere again resulted in the pattern of $\ce{Fe2O3}$. When
  $\ce{FeCO3}$ was decomposed in a $\ce{CO2}$ atmosphere, the first
  decomposition product observed was $\ce{Fe3O4}$.[...]the brief
  appearance of $\ce{FeO}$ was suspected in the decomposition. By
  oscillating the Geiger counter through a very small arc encompassing
  the strongest line of FeO, the presence of this oxide was indicated.
  The line appeared for only a very short time and was at no time very
  strong. In the inert atmosphere, the products of decomposition were
  $\ce{FeO}$ and $\ce{Fe3O4}$. The $\ce{Fe3O4}$ in the sample could not
  be converted to $\ce{FeO}$ even by heating to the highest temperature
  available with the furnace (about 1150°C. in helium).

So, the reaction indeed forms $\ce{FeO}$ but it gets oxidized depending on the environment the reaction is proceeded.

$$\ce{FeCO3 -> FeO + CO2 }$$

$$\ce{4FeO + O2 -> 2Fe2O3}$$

$$\ce{3FeO + CO2 -> Fe3O4 + CO}$$

References

A.A. El-Bellihi, "Kinetics of Thermal Decomposition of Iron
Carbonate", Egypt. J. Chem. 53, No. 6, pp.871- 884 (2010) DOI: https://ejchem.journals.ekb.eg/article_1268_82097e00ce1c4e475f17cfec61edb6d2.pdf
H. E. KISSINGER, H. F. McMURDIE, and B. E. SIMPSON,Thermal Decomposition of Manganous and Ferrous Carbonates,DOI: https://ceramics.onlinelibrary.wiley.com/doi/pdf/10.1111/j.1151-2916.1956.tb15639.x


In biochemistry, when do rate constant units matter in calculating free energy changes? In my experience, it is always been the ratio of products to reactants that mattered, so as long as your units are consistent; what they are doesn't matter. However, an in-class problem today required that the units be changed to molar (M) from millimolar (mM). Can you summarize when/why this matters in some cases, but not in others?

$K$ should be a unitless value. Otherwise, you'd get a nonsensical unit when you try to take the log of the unit. Each component of the product $K$ is usually an activity, which is frequently similar in value to the molar concentration, or a fugacity, which is frequently similar in value to the partial pressure of a gas.


Determine the amount of substance of calcium chloride used Determine the amount of substance (in moles) of calcium chloride used in this lab. (Hint: $0.20\rm~M$ means $0.20\rm~moles:1~L$)

I made $96$ milliliters of $\ce{CaCl2}$ which would be $0.096\rm~L$.
$$\rm0.096~L~of~CaCl_2\times\frac{0.20~mol}{1~L} = 0.0192~mol~of~CaCl_2$$

I was wondering if I solved this right.

Yes, you did it correctly, though there is a small issue with significant figures if that is a concern.

$$\mathrm{Molarity}=\frac{\mathrm{moles}}{\mathrm{liters}}$$

So for your values:

\begin{align}
0.20 &= \frac{n}{0.096}\\
n &= \pu{0.20 M} \times \pu{0.096 L} \\&= \pu{0.0192 mol}\, \ce{CaCl2}
\end{align}

However, both your molarity and volume have only two significant figures and you're multiplying those values, so the result should only be reported to two significant figures as well: $n = \pu{0.019 mol}\,\ce{CaCl2}$.


"HBr is donating a proton to LiOH": Why doesn't it produce LiOH2? In the section Brønsted-Lowry acid and base theory from Khan Academy:

In the reaction $\ce{LiOH(aq) + HBr(aq) -> H2O(l) + LiBr(aq)}$, $\ce{HBr}$ is donating a proton to $\ce{LiOH}$, thus acting as a Brønsted-Lowry acid. Since $\ce{LiOH}$ accepts the proton, it is acting as a Brønsted-Lowry base.
Image source: https://i.stack.imgur.com/fjV8R.png

I don't understand how $\ce{HBr}$ is donating a proton to $\ce{LiOH}$, because I think it would produce something like "$\ce{LiOH2}$", but the reaction gives $\ce{LiBr}$.
So where is my misunderstanding?

I think it would produce something like "LiOH2,"

You are missing the charge. It should be something like $\ce{LiOH2+}$, which is an unconventional way of writing $\ce{Li+(aq) + H2O(l)}$.

So where is my misunderstanding?

You have to get used to interpreting the (aq) designation after ionic compounds. To help you out, here is the equation written with the ions shown as individual species (more typing, but closer to the actual situation):
$$\ce{Li+(aq) + OH-(aq) + H+(aq) + Br-(aq) -> Li+(aq) + H2O(l) + Br-(aq)}$$


Magnesium Sulfate (MgSO4) back to Magnesium Oxide (MgO) + Sulfuric Acid (H2SO4)? Is it possible to go back from magnesium sulfate ($\ce{MgSO4}$) to magnesium oxide ($\ce{MgO}$) and sulfuric acid ($\ce{H2SO4}$) ?

How difficult or easy would it be?

The reason I ask is because epsom salts ($\ce{MgSO4}$) are really cheap (i.e. less than 50 euro cents / kg). Whereas magnesia ($\ce{MgO}$) of $98\%$ or better grade is much more expensive (i.e 3-4 euro / kg).

I was wondering how easy and cheap it would be to revert the reaction and separate out the sulfur from the magnesium oxide.

Yes, magnesium sulfate can be converted back to magnesium oxide by thermal decomposition or use of reducing agents.

Thermal decomposition

$$\ce{2MgSO4 ->[\Delta] 2MgO + 2SO2 ^  + O2 ^}$$

  The thermal decomposition of magnesium sulfate to produce magnesium
  oxide, sulfur dioxide and oxygen. This reaction takes place by heating
  to the temperature of 1137°C.(Chemiday 1)

Use of reducing agent

$$\ce{2MgSO4 + C ->[\Delta] 2MgO + 2SO2 ^  + CO2 ^}$$

  Magnesium sulfate react with carbon (as reducing agent) to produce
  magnesium oxide, sulfur dioxide and carbon dioxide. This reaction
  takes place at a temperature of 700-900°C.   (Chemiday 2)

The products other than magnesium oxide are gaseous and escapes leaving behind pure magnesium oxide.


Why does an anion not only bind to a single cation? When $\ce{Na}$ binds with $\ce{Cl}$ to form $\ce{NaCl}$, why does the Na bind with multiple $\ce{Cl}$'s and not only one? A Na-ion is positively charged due to donating an electron and a $\ce{Cl^-}$ion is negatively charged due to accepting an electron. That's why they bind. But salts form a grid where each Na-ion binds with 6 $\ce{Cl^-}$ions and vice versa.

Shouldn't be the Na-ion already be "used" after binding with a $\ce{Cl^-}$ion? Or does it stay charged and attracts all ions that are negatively charged?

I hope someone can help me to understand.

In the vapour phase where NaCl can exist (obviously at high temperature) then the molecule is a diatomic, with a covalent bond, albeit with a lot of ionic character.

In the crystal, to which I assume you refer, there is exactly one $\ce{Na^+}$ to each $\ce{Cl^-}$. You can see this if you look at the structure of the crystal, however to achieve this in the cubic crystal each ion has to be surrounded by six of the opposite charge as you correctly state. However, the charges are not 'used up' with one neighbour but shared equally between them, so a whole perfect crystal is electrically neutral. (If you want you can think of the charges as electric fields spreading out radially from each ion, they cancel each other out at some point between ions making a complicated shaped surface of zero potential). The bigger chloride ions form a simple cubic array with the sodium in the gaps. The crystal has this structure, as opposed to another (say face centred cubic), because this is the one than minimises the total energy, by keeping (large) chloride ions far enough away form one another so that their repulsion does not destroy the crystal.

In aqueous solution the water, which is very polar, means that it is more energetically favourable to dissociate into ions and solvate them rather than keep NaCl as a diatomic molecule, i.e. form $\ce{Na^+(aq) + Cl^-(aq)}$ where each ion is surrounded by several water molecules.


Why do we use Δ to mean heating? To express heating, we usually use $\Delta$ over reaction arrows like
\begin{align}
\ce{CaCO3 ->[\Delta] CaO +O2}.
\end{align}
Why do we use $\Delta$ to mean heating during chemical reactions? I thought up with one idea that $\Delta$ symbolizes fire. But actually I have no information about the origin of this usage. Does anybody know the origin or reason of this usage of $\Delta$?

Δ symbolizes difference. For chemical reactions, it's usually difference between energy that reactants already have and energy needed for reaction. Then Δ = $E_a$ - activation energy of the reaction. Activation energy is like a "potential barrier" for reaction to happen. And Δ means that we've added this amount of energy so energy of reactants changed by Δ (difference).


Reason for inconsistency in specific gravity of calcium choride added to water I am seeing a significant change in specific gravity of products upon addition of $\ce{CaCl2}$ to water. The mass and volume of what goes in doesn't match my total specific gravity of what comes out. I am curious as to why that is the case?

When you say, "of products", do you really mean "of the solution"? If so:

The solvent mass and the solute mass add up to the mass of the solution.

But solvent volume and solute volume often don't add up to the volume of the solution. This is because the forces between molecules in the solution are different from the forces between molecules in the pure solute and the pure solvent.

In your solution, the calcium ions and chloride ions attract water molecules around them. The water molecules are packed more tightly than they would be if the ions weren't there. That can increase the density of the solution relative to the total volume of the unmixed solvent and solute. Another effect is that the ions in solution are packed less tightly than they are in solid $\rm CaCl_2$.

You can see which effect wins by dumping some calcium chloride into a volumetric flask. Fill the flask up to the mark with water, and then mix until all the calcium chloride dissolves. The solution probably doesn't quite come up to the mark anymore, because the calcium and chloride ions have drawn the water molecules in around themselves, making the solution denser than it would be otherwise.

More detail than you may want below:

The volume of the solution $V$ is given by the following equation:

$$V = n_1 \overline{V}_1 + n_1 \overline{V}_2$$

where $n_1$ and $n_2$ are moles of calcium chloride and moles of water in the solution, and $\overline{V}_1$ and $\overline{V}_2$ are the partial molar volumes of the calcium chloride and the water.

These partial molar volumes are actually derivatives of the volume of the solution with respect to moles of calcium chloride ($\overline{V}_1 = \partial V/\partial n_1$) and with respect to water ($\overline{V}_2 = \partial V/\partial n_2$). They are related to each other (when one rises, the other falls) and they vary with the composition and temperature of the solution in a rather complicated way.


Do hydrogen sulfide and oxygen produce pure sulfur or sulfur dioxide? If both, under which circumstances does each scenario occur? If both of the following hydrogen sulfide and oxygen chemical equations can occur, under what circumstances is the harmless pure sulfur and the toxic sulfur dioxide produced?
$\ce{2H2S + O2 -> 2H2O + 2S}$
$\ce{2H2S + 3O2 -> 2H2O + 2SO2}$
Specifically I have a fresh water aquarium with sand, plants, and small creatures I collected from a creek. I want to prevent hydrogen sulfide building up by oxygenation with an air pump, but I want to make sure the environment does not result in the harmful sulfur dioxide in the tank.

Sulfur dioxide is not a problem ; in an aquarium it will make very dilute sulfurous acid which will likely neutralize with some carbonate mineral/gravel. H2S is lethal , extended exposure at 50 ppm for a day , or short exposure at 500 ppm. On the other hand , what you smell in aquarium sand is less then 1 ppm, the sense of smell is very sensitive to H2S. There are several answers on the internet but I expect most people can smell H2S at 0.1 ppm. An aerator will have no significant affect on H2S in an aquarium. When I clean my pond , which is well aerated ,the smell of H2s is strong from decomposing leaves, etc.
Regarding the first part of your question ; Hundreds of tons a day of elemental sulfur are made ,world wide, every day by burning H2S with limited air at high temperature. By recycling and control of gas compositions the only products are elemental sulfur and water,


Work done by system of ideal gas in isobaric expansion Consider a system of ideal gas in a container with piston and  the isobaric expansion of gas takes place.
As the process is isobaric, so initially the pressure of gas is equal to atmospheric pressure as the piston remains in equilibrium over the container. Then, first we increase the temperature and as a result pressure of gas increases and the piston rises till the pressure reduces to atmospheric pressure. So at initial and final state pressure of gas is constant thus the process is isobaric.
So, work done by gas should be
$dW=F_{applied\;by\;gas}(x).dx$
$dW=P_{applied\;by\;gas}(x)Adx$
$s.t.\;P_{applied\;by\;gas}(x_{initial\;state})=P_{applied\;by\;gas}(x_{final\;state})=P_{atm}$
So, total work done, $W=\int_{x_{initial}}^{x_{final}}P(x)Adx$
So, we should know pressure as a function of $x$ to calculate the work done by gas.
My question is that in books, they directly use the argument that work done by gas is dependent on $P_{external}$ and as it is constant, they give the expression of work as $dW=-P_{external}Adx$.
I don't understand how they do that and use external pressure in the expression of work done.
I think we should take into account pressure of gas as a function of $x$ to find the work done by gas.
Please help me in clarifying the doubt.

At the interface between the gas and its surroundings, by Newton's 3rd law, the pressure exerted by the gas on its surroundings is equal to the pressure exerted by the surroundings on the gas.  So we can use either.  For a reversible expansion, the gas pressure is given by the ideal gas law.  But the ideal gas law only applies at thermodynamic equilibrium and during a reversible process (which consists of a continuous sequence of thermodynamics equilibrium states).  For an irreversible expansion, the ideal gas law doesn't work (i.e., gives the wrong answer), and we can't use it to get the gas pressure at its interface with the surroundings (where the work is being done).  But, if we know the surroundings pressure during the irreversible expansion, we can still calculate the work (because, at the interface between the gas and surroundings), the two will match.  So using the external pressure is always a valid approach to getting the work.


Are pesticides used on fruit easily washed off? I was hoping there'd be someone who studies pesticides here.

Are they water soluble?
Do they get absorbed in to the fruit?
How well would you have to wash the fruit to fully remove the residue?

Pesticides never were my main field of research, so you might want to take the following with a grain of salt:

Most pesticides i'm aware of are insoluble in water, but you might want to cross-check that for particular substances in the huge Pesticide database of the EC.

This insolubility seems reasonable, otherwise the substances would get washed off after the next rain.

Concerning your second question, the answer unfortunately is yes. To my knowledge, different pesticides are typically applied at the various state of plant growth. In order to suppress weeds, one might spray fields in an early, pre-emergent state. Later, fungicides or other substances might be applied during blossom. In order to protect the bees, it is however not recommended to apply insecticides during that time. Reasonable and minimized use of pesticides will keep the load in the fruit close to the limits of detection or beyond. Depending on the origin of the fruit, this might however not be the norm. In these cases, high concentrations of pesticides, even those banned in your home country, are detected both in the fruit and its products.

This leaves us with the question what we can do as consumers. If the fruit is highly contaminated, we can't do anything. Therefore, it might be reasonable to avoid products from countries infamous for breaking environment protection rules and go "organic" instead. Honestly, i never peel apples, but i wipe the surface with a piece of cloth. But i avoid industrial sorts, buy from local farmers, or pick them from the ground of meadow orchards.

I hope that this contribution, although more personal and less scientific, was helpful.


Which is more basic, piperidine or pyrrolidine and why? Pyrrolidine seems to be less basic as it is a 5-member ring and +charge on N atom won't be stable due to steric factors as compared to piperidine but the results are opposite.

I think I may have a possible insight into the small difference. Wikipedia gets the pKa from J. Am. Chem. Soc. 1957, 79 (20), 5441–5444 and lists pyrrolidine at 11.27 and piperidine at 11.22.

Calculating the equilibrium structures (B3LYP 6-31g*) for both species in their neutral and protonated states, I found something interesting.

  For piperidine

Neutral and protonated species both achieve the optimal staggered conformation about all bonds.

  For pyrrolidine

For the neutral species, the C-C bonds have their hydrogens in an eclipsed conformation. In the protonated species, the ring gets a slight twist, which allows the C-C bonds to have their hydrogens in a staggered conformation.

  So...

The hydrogens being moved into a staggered conformation upon protonation of pyrrolidine could be the source of the extra stabilisation of the protonated form.

I'm also thinking if the cation form is favoured slightly more in pyrrolidine because its aliphatic portion is less, thus leading to a bigger release of solvation energy upon protonation (calculated to be 1 kJ/mol more favourable for pyrrolidine, but this could be within the error of the method).

If I had to pose arguments for this small difference in basicity, these would be the ones I'd choose.


Is protein folding thermodynamically favourable? If yes.. why? The second law of thermodynamics dictates that the sum of entropy of the universe is always increasing. Is the process of protein folding a spontaneous process which is increasing the entropy? Is this why it can be called thermodynamically favourable? Or is there something else I am forgetting..

What determines whether a process is spontaneous or not, is the change in free energy, not the increase of entropy. Entropy is one important factor, as is enthalpy. Entropy in this case is not increasing since the system is becoming more organized, however,enthalpy is greatly favoured since protein folding creates many Hydrogen bonds between its polar aminoacid fragments and surrounding water and also many hydrophobic interactions between their non-polar aminoacids constituents. Take a look at this page to have a better understanding about protein folding thermodynamics (introductory text, though it´s a good starting point).
http://www.chembio.uoguelph.ca/educmat/phy456/456lec02.htm


What is 'coprecipitated (Ni,Fe)S'? I am writing an essay on the origin of life and I was reading an article on hydrothermal vents. (Peptides by Activation of Amino Acids with CO on (Ni,Fe)S Surfaces: Implications for the Origin of Life).

It appears that (Ni,Fe)S has played an important role in protein synthesis, but I wanted to know more about the substance its self. It is a compound, a group of chemicals? Or is it simply nickel, iron, and sulfur individually? I'm not very clued up when it comes to chemistry and I haven't come across that sort of notation before.

When two materials precipitate at the same time, you'll often get an extremely intimate mix of the two. I'm assuming that the notation you've got here means a precipitate consisting of a mixture of both nickel (II) ions, iron (II) ions, and sulfide ions, or a mix of nickel and iron sulfides (NiS & FeS).

Iron and nickel (II) have similar ionic radii and charge densities, so it should be possible to create crystal domains containing both metal ions interchangeably, in any ratio.


How does the equilibrium shift when concentration of reactant and product are increased simulatenously Before answering my question, take a look at this example
$$\ce{2NOCl <=> 2NO + Cl2}$$
Let's suppose I have $\pu{2 mol/L}~\ce{NOCl},~\pu{2 mol/L}~\ce{NO}$, and $\pu{1 mol/L}~\ce{Cl2}$ in a closed system. If we were to calculate the equilibrium constant, it would be
$$K_c = \frac{(2)^2(1)}{(2)^2}$$
Now suppose I increase the concentration of both $\ce{NOCl}$ and $\ce{NO}$ by $\pu{1 mol/L}$. The equilibrium will stay the same,
$$K_c = \frac{(2+1)^2(1)}{(2+1)^2};$$
however, if I were to instead increase the concentration of both $\ce{NOCl}$ and $\ce{Cl2}$ by $\pu{1 mol/L}$, the equilibrium will shift to the product side
$$K_c = \frac{(2)^2(1+1)}{(2+1)^2}$$
Can someone explain this phenomenon in terms of rate of reaction or in terms of molecules?

Can someone explain this phenomenon in terms of rate of reaction or in terms of molecules?

In terms of rate of reaction, at equilibrium the forward rate is proportional to the concentrations of the reactant (as if this were an elementary reaction). Likewise, at equilibrium the reverse rate is proportional to the concentrations of the products (as this were an elementary reaction). If you change the concentration of a single reactant and a single product by the same factor (i.e. multiply by the same number, not add the same number) and if the stoichiometric factors are the same (i.e. the order of reaction in this special case), and the reactants are sufficiently dilute, the system will still be at equilibrium. In the second change described by the OP, neither is the case (one concentration increases by 50%, the other doubles, and the exponents are different).

In terms of thermodynamics, if the Gibbs free energy of the reactants is changed by the same amount (i.e. add or subtract a number) the system will remain at equilibrium. This will happen if one species on either side is changed by the same factor because the Gibbs free energy changes with the logarithm of the activity of a species. Again, this only works if the stoichiometric coefficient of the two species are the same.

Nomenclature K vs Q

In the equations shown by the OP, different concentrations are substituted into the equilibrium constant expression even when the reaction is not at equilibrium. The correct name for these expresssions is reaction quotient Q. If Q matches K, the system is at equilibrium. In the examples provided, Q stays the same in one case (still in equilibrium) and changes in the other case (no longer in equilibrium). What the equilibrium would be needs not to be calculated. Knowing that Q is unequal to K is sufficient for showing that the system is in fact not in equilibrium.


Calclulation of MP2 first order wave function My question is, how can I calculate the MP2 first order wavefunction, and thus construct the MP2 density matrix? Now here is what I understand so far:

The density matrix is given as:

$$P_{ij}=\sum_{a}C_{ia}(C_{ja})^*$$

Here $C$ is the wave function coefficient matrix. To get the density matrix for the MP2 first order wavefunction, their coefficients need to be found.

The first-order wave function is given as [1]:

$$\left|\left.\varPsi_{i}^{(1)}\right> \right.=\sum_{n}c_{n}^{(1)}\left|\left.\varPsi_{n}^{(0)}\right> \right.$$

Now this gives:

$$\left< \left.\varPsi_{n}^{(0)}\right|\varPsi_{i}^{(1)}\right> =c_{n}^{(1)}$$

Which by some more steps, that can be seen in Szabo and Ostlund leads to:

$$c_{n}^{(1)}=\left< \left.\varPsi_{n}^{(0)}\right|\varPsi_{i}^{(1)}\right> =\frac{\left< \varPsi_{n}^{(0)}\left|\mathscr{V}\right|\varPsi_{i}^{(0)}\right> }{E_{i}^{(0)}-E_{n}^{(0)}}$$

In Szabo and Ostlund the energy is derived to have the expression:

$$E_{i}^{(2)}=\sum_{n}'\frac{\left< \varPsi_{i}^{(0)}\left|\mathscr{V}\right|\varPsi_{n}^{(0)}\right> \left< \varPsi_{n}^{(0)}\left|\mathscr{V}\right|\varPsi_{i}^{(0)}\right> }{E_{i}^{(0)}-E_{n}^{(0)}}$$

Though it can also be written as:

$$E_{i}^{(2)}=\sum_{n}'c_{n}^{(1)}\left< \varPsi_{i}^{(0)}\left|\mathscr{V}\right|\varPsi_{n}^{(0)}\right>$$

Now forwarding to the energy expression for the MP2 energy:

$$E_{0}^{(2)}=\sum_{i<j,a<b}\frac{\left< \varPsi_{0}^{(0)}\left|\mathscr{V}\right|\varPsi_{ij}^{ab}\right>\left< \varPsi_{ij}^{ab}\left|\mathscr{V}\right|\varPsi_{0}^{0}\right>}{E_{0}^{(0)}-E_{ij}^{ab}}=\sum_{i<j,a<b}\frac{\left< ij\left|\right|ab\right>\left< ab\left|\right|ij\right>}{\varepsilon_{i}+\varepsilon_{j}-\varepsilon_{a}-\varepsilon_{b}}$$

Now I would like to think that I could identify the wave function coefficients by comparing the two above equations to get:

$$c_{ijab}^{(1)}=\frac{\left< ab\left|\right|ij\right>}{\varepsilon_{i}+\varepsilon_{j}-\varepsilon_{a}-\varepsilon_{b}}$$

I would have expected that I would have found a vector of $c$ such that I would be able to construct the coefficent matrix as:

$$C^{MP2}=c\cdot c^T \Rightarrow P^{MP2}_{ij}=\sum_{a}C^{MP2}_{ia}(C^{MP2}_{ja})$$

But instead I got a four index coefficent out. Where did my line of thinking go wrong? And how can I proceed to construct the first order wave function, and thus get the density matrix?

References to old papers where this have been done explicit is also appreciated, because I have not been able to find any thing but recent papers about MP2 and density matrix fitting, which is not what I am looking for.

[1] Szabo, A.; Ostlund, N. S. Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory; Dover: Mineola, NY, 1989.

The main problem with the last statement,

$$
\begin{align}
C^{MP2} &= c\cdot c^T \\
\Rightarrow P^{MP2}_{ij} &= \sum_{a}C^{MP2}_{ia}(C^{MP2}_{ja}),
\end{align}
$$

is that one never has "MP2 MO coefficients" in this way; you would have MP2 natural orbital (NO) coefficients that come from diagonalizing this density, so it's the other way around:

$$
P^{\text{MP2}}C^{\text{MP2 NO}} = n^{\text{MP2 NO}}C^{\text{MP2 NO}},
$$

where $n^{\text{MP2 NO}}$ is the diagonal matrix of natural orbital occupation numbers. This still leaves the question of how to calculate $P^{\text{MP2}}$. Conceptually, the way I view it is that the set of $\{c_{ijab}^{(1)}\}$ are the amplitudes describing how to correct the uncorrelated ground-state wavefunction; they don't specify the wavefunction on their own, so $C^{\text{SCF}}$ will probably have to appear in the total density.

All of the following assumed a restricted wavefunction. $i,j,k,\dots$ are occupied MO indices, $a,b,c,\dots$ are virtual/unoccupied MO indices, $p,q,r,\dots$ are general (any) MO indices, and $\mu,\nu,\dots$ are general AO indices. The 2nd-order correction to the density, specifically the occupied-occupied and virtual-virtual blocks are given by Eqn. (11a) in Trucks [1]:

$$
\begin{align}
P_{ij}^{\text{MP2}} &= -\frac{1}{2} \sum_{kab} c_{jk}^{ab} c_{ik}^{ab} \\
P_{ab}^{\text{MP2}} &= +\frac{1}{2} \sum_{ijc} c_{ij}^{ac} c_{ij}^{bc},
\end{align}
$$

where I switched to using your notation. This is indeed similar to your idea from above; accumulate over both indices of the opposite space, then do a matrix multiplication.

The total density is then [Eqn. (7)]

$$
\mathscr{P} = P^{\text{SCF}} + P^{\text{MP2}},
$$

where $P_{pq}^{\text{SCF}} = \delta_{pq}$ because so far we're working in the MO basis. The back-transformation to the AO basis will be the same as for any other 2-index MO-basis quantity:

$$
\begin{align}
P_{\mu\nu} &= \sum_{pq}^{\text{all MOs}} C_{\mu p} P_{pq} (C_{\nu q})^{T} \\
\Rightarrow \mathbf{P}^{\text{AO}} &= \mathbf{C} \cdot \mathbf{P}^{\text{MO}} \cdot \mathbf{C}^{T}.
\end{align}
$$

Then, we can evaluate molecular properties that are represented as energy derivatives [Eqn. (6)]:

$$
\left.\frac{\partial \mathscr{E}}{\partial \lambda}\right|_{\lambda=0} = \sum_{\mu\nu} \mathscr{P}_{\mu\nu} Q_{\mu\nu}^{\lambda}
$$

where $\mathscr{E} = E_0^{(0)} + E_0^{(2)}$ and $Q_{\mu\nu}^{\lambda}$ is the matrix representation of the external perturbation $\lambda$ in the AO basis. This is analogous to the SCF result:

$$
\left.\frac{\partial E^{\text{SCF}}}{\partial \lambda}\right|_{\lambda=0} = \sum_{\mu\nu} P_{\mu\nu}^{\text{SCF}} Q_{\mu\nu}^{\lambda}
$$

Unrelaxed vs. relaxed density

The above definition of the density is incomplete, since it describes the unrelaxed correlated density. Our usual formulation of correlated methods is to describe correlation as some to-be-determined mixture of the virtual orbitals with the occupied orbitals. There is no such mixing in the above equations. In reality, $P_{ia}^{\text{MP2}} \overset{!}{\neq} 0$, so there is something that looks like a "transition density" with non-zero occupied-virtual blocks, in order to correct for the fact that our $C^{\text{SCF}}$ has no correlation. Allowing this mixing (which corresponds to orbital response) leads to the relaxed density. However, one has to solve the coupled-perturbed Hartree-Fock (CPHF or CPSCF) equations.

See between Eqns. (8) and (9) in [1] for the CPHF equations and how to solve them for the occupied-virtual mixing coefficients and finally the $P_{ia}^{\text{MP2}}$ block. Generally, note the similarities to the CIS density equations.

CPHF equations and first-order properties

As stated, when differentiating the energy with respect to some perturbation $\lambda$ to calculate first-order properties, one needs to contract the density with the operator property matrix. If you transform Eqn. (6) from the AO into the MO basis,

$$
\begin{align}
\left.\frac{\partial \mathscr{E}}{\partial \lambda}\right|_{\lambda=0} &= \sum_{pq} \mathscr{P}_{pq} Q_{pq}^{\lambda} \\
&= \sum_{pq} P_{pq}^{\text{HF}} Q_{pq}^{\lambda} + \sum_{pq} P_{pq}^{\text{MP2}} Q_{pq}^{\lambda},
\end{align}
$$

the SCF contribution simplifies due to the nature of the SCF density matrix in the MO basis,

$$
P_{pq}^{\text{HF}} = \left\{
\begin{array}{lr}
2 & \text{if }p = q\text{, p is an occupied MO} \\
0 & \text{otherwise}
\end{array}
\right.,
$$

leading to the expression

$$
\begin{align}
\left<\hat{\lambda}\right> &= \sum_{pq} P_{pq}^{\text{HF}} Q_{pq}^{\lambda} \\
&= 2 \sum_{i} Q_{ii}^{\lambda} \\
&= \mathrm{tr}\left(\mathbf{P}^{\text{AO}}\mathbf{Q}^{\text{AO}}\right) \\
&= \mathrm{tr}\left(\mathbf{P}^{\text{MO}}\mathbf{Q}^{\text{MO}}\right)
\end{align}
$$

for one-electron operators and wavefunctions that obey the Hellmann-Feynman theorem. I absorbed the 2 into $\mathbf{Q}^{\text{MO}}$, but this depends on the program implementations and their definition of the density. The point is that this contribution is relatively simple, and illustrates how for one-electron operators (dipole, quadrupole, angular momentum, ...), evaluating them for a converged Hellmann-Feynman wavefunction is very fast: one the density and property integrals are in the same basis, either do a matrix multiply and then trace, or do an element-wise (Schur) product and accumulate all the elements, so it scales as $O(N^3)$. No solution of CPHF equations is required. The same goes for the gradient with respect to nuclear coordinates $\frac{\partial E^{\text{SCF}}}{\partial x}$; it isn't a one-electron operator, and requires a mathematical trick to make $\frac{\partial P^{\text{SCF}}}{\partial x}$ disappear, but the SCF gradient doesn't require solving the CPHF equations either. This isn't true for $\frac{\partial E^{\text{MP2}}}{\partial x}$, which requires $\frac{\partial P^{\text{MP2}}}{\partial x}$, where you need to solve the CPHF equations to even form the full $P^{\text{MP2}}$.

Consider the remaining MP2 part of the property evaluation; the total density contains contributions from the occupied-occupied (oo) block, the virtual-virtual (vv) block, and the occupied-virtual (ov) block. I'm going to switch from $P^{\text{MP2}}$ and $c$ to $D$ and $t$ to make typing easier.

$$
\sum_{pq} D_{pq} Q_{pq}^{\lambda} = \sum_{ij} D_{ij} Q_{ij}^{\lambda} + \sum_{ab} D_{ab} Q_{ab}^{\lambda} + 2\sum_{ia} D_{ia} Q_{ia}^{\lambda}
$$

The first two terms can already be calculated from the unrelaxed density; $D_{ia}$ is the term that must come from solving the CPHF equations.

In general, the form of solving CPHF equations looks like $\mathbf{AU} = -\mathbf{V}$, where $\mathbf{A}$ is the orbital Hessian (2nd derivative of the energy with respect to orbital rotations), $\mathbf{V}$ is the gradient of some perturbation $\hat{V}$. In [1], it's $\lambda$, but Neese [3,4] and the Scandinavians usually use $V$ or $Q$ when the perturbation is associated with an observable. This perturbation gradient, termed the right-hand side (or RHS), describes how the perturbation might cause occupation of virtual orbitals, and the orbital Hessian describes how the response of one MO to some perturbation is coupled to the other MOs [5,6].

Start by rewriting $\sum_{ia} D_{ia} Q_{ia}^{\lambda}$ as the combination of a perturbation-dependent response $\mathbf{U}$ and the perturbation gradient $\mathbf{X}$:

$$
2\sum_{ia} D_{ia} Q_{ia}^{\lambda} = 2\sum_{ia} X_{ai} U_{ai}^{\lambda}
$$

$\mathbf{X}$ serves the same purpose as the $\mathbf{V}$ mentioned above, but rather than coming from an external perturbation, it is an intermediate quantity that describes the MP2 wavefunction.

First, the perturbation-dependent response. This part is not specific to MP2, and is also important for higher-order SCF-based molecular response [3].

$$
\begin{align}
U_{ai}^{\lambda} &= \sum_{bj} (\mathbf{A}^{-1})_{ai,bj} Q_{bj}^{\lambda} (\varepsilon_{j} - \varepsilon_{b})^{-1} \\
A_{ai,bj} &= 1 + \left( \left<ab||ij\right> - \left<aj||ib\right> \right) (\varepsilon_{a} - \varepsilon_{b})^{-1}
\end{align}
$$

The $\varepsilon$ are MO energies. If we don't assume the use of canonical MOs, this energy denominator is actually $F_{ab}S_{ij} - F_{ij}S_{ab}$.

Jumping ahead, the relationship between the ov density, the orbital Hessian, and the MP2 RHS is [Eqn. (9)]

$$
\sum_{bj} D_{bj} (\varepsilon_{j} - \varepsilon_{b}) A_{bj,ai} = X_{ai}
$$

See how this is structurally similar to solving for $U$, but the RHS is different. To show the equivalence between all quantities,

$$
\begin{align}
2\sum_{ai} X_{ai} U_{ai}^{\lambda} &= 2\sum_{ai} X_{ai} \sum_{bj} (\mathbf{A}^{-1})_{ai,bj} Q_{bj}^{\lambda} (\varepsilon_{j} - \varepsilon_{b})^{-1} \\
&= 2\sum_{bj} D_{bj} Q_{bj}^{\lambda}.
\end{align}
$$

The explicit equation for $X_{ai}$ is [Eqn. (11b)]

$$
X_{ai} = \sum_{jk} D_{kj} \left<ki||ja\right> + \sum_{bc} D_{bc} \left<bi||ca\right> - \frac{1}{2} \left[ t_{kj}^{ab} \left<ib||kj\right> + t_{ij}^{cb} \left<cb||aj\right> \right],
$$

which is in fact the MP2 Lagrangian, $L_{ai}$ [2, Eqn. (11)]. Again, notice the similarity to the CIS Lagrangian. The Lagrangian is required because the wavefunction doesn't need to be relaxed in order to calculate the correlation energy, so you pay the price here. These equations must be evaluated for MP2 geometry optimizations (during the gradient part) or for fully-relaxed MP2 molecular properties. To be clear, if one is only interested in the density and not molecular properties, only $\mathbf{DA} = \mathbf{X}$ needs to be solved for, and not the equations involving $\mathbf{U}$. In terms of scaling, SCF-based CPHF is about $O(N^4)$ (after back-transforming the $\mathbf{A}^{-1}\mathbf{Q}$ terms to be AO-direct) but might require fewer iterations than the SCF itself. According to [2], the MP2 gradient is something like $O(ON^4) + O(O^{2}V^{3})$, but they can do similar tricks during iterations to cut down on the number of fifth-order steps. In my experience, MP2 gradient calculations are about as expensive as an SCF polarizability calculation.

References

Trucks, G. W.; Salter, E. A.; Sosa, C.; Bartlett, R. J. Theory and implementation of the MBPT density matrix. An application to one-electron properties. Chem. Phys. Lett. 1988, 147 (4), 359-366. DOI: 10.1016/0009-2614(88)80249-5
Frisch, M. J.; Head-Gordon, M.; Pople, J. A. A Direct MP2 gradient method. Chem. Phys. Lett. 1990, 166 (3), 275-280. DOI: 10.1016/0009-2614(90)80029-D
Neese, F. Prediction of molecular properties and molecular spectroscopy with density functional theory: From fundamental theory to exchange-coupling. Coordination Chemistry Reviews 2009, 253 (5-6), 526-563. DOI: 10.1016/j.ccr.2008.05.014
I cannot recommend the above paper enough; it contains excellent descriptions of how electric and magnetic response are calculated with explicit equations that are actually usable. There, you can see how CPHF is used for molecular response properties and the connection with excitation energies, found by solving $\mathbf{AU} = \mathbf{0}$ (diagonalization rather than solving a set of linear equations).
Toulouse, J. Introduction to the calculation of molecular properties by response theory.
Parrish, R. M. Psi4 Development: Theory Supplement, CPHF for Dipole Polarizabilities.


Converting Chlorine Dioxide concentration to Chlorine Looking for a little guidance. I have been given some water quality data as part of a project. Specifically I'm interested in the concentration of Chlorine in river water. The analysis results indicate concentrations of  0.6 mg/l of chlorine dioxide but not chlorine. However, there are no water quality standards for chlorine dioxide, only chlorine. Can I calculate the concentration of chlorine in the water from the concentration of chlorine dioxide?

Thanks

You can, in fact, calculate the chlorine concentration from the concentration of chlorine dioxide. Chlorine will normally be in some compound, so when concentration of chlorine is given, it is usually concentration of chlorine compounds.

The concentration of chlorine dioxide is 0.6 mg/L. First convert the mg to g. this results in .006 g. Divide that value by the molar mass of chlorine dioxide (67.45 grams/mole) to get .00008895 moles of chlorine dioxide, or .00008895 moles of chlorine. Multiply this by the molar mass of chlorine (35.45 grams/mole) to get .003153 grams of chlorine. Convert this to milligrams to get 3 mg (one significant figure), so the concentration of chlorine from chlorine dioxide is 3 mg/L.


Ionization isomerism in coordination compounds Ionization isomerism is defined as complexes which have same empirical formula but ionize to different ions in solution.

Consider the complex $\ce{[Pt(NH3)4Cl2]I2}$. The other possible ionization isomers are:
$\ce{[Pt(NH3)4I2]Cl2}$ and $\ce{[Pt(NH3)4ICl]ICl}$.

But I have not come across the complexes similar to the last one with two different anions outside the coordination sphere.
So, do these sort of complexes exist?

Moreover, what can be the IUPAC name for this compound?

tetraamminechloridoiodoplatinum(IV) chlorineiodine seems wrong.

I think it should be -

tetraamminechloridoiodoplatinum(IV) chloride iodide.

The rules of the naming of counter ions are pretty much the same as in general Inorganic Chemistry. I do know of a compound AgClI - named as Silver chloride iodide. So, I think the counter ions here - chloride and iodide should be named as by the nomenclature rules of anions i.e with an -ide suffix and alphabetically if there is the presence of more than one anion

Proof of Reference - Silver chloride iodide (NCBI PubChem)

EDIT: My teacher did give an example of it being a valid complex. I think it is not generally stated in the preliminary books of this topic to avoid confusion among students.I imagine it would be present in some more in-depth books of this topic.


Dibromomethane bond angles With regard to the above professor's claim, I disagree. Bent's rule, which states that atomic p-character is concentrated in orbitals directed toward more electronegative substituents, implies that the $\ce{Br-C-Br}$ bond angle is going to be smaller than 109.5 degrees since the "$\ce{sp^3}$" hybrid orbitals involved in C-Br bonding have more p-character than 75%.

Therefore, the $\ce{H-C-H}$ bond angle should be bigger than 109.5 degrees.

Questions

1) Does Bent's rule work here (i.e. am I correct?)

2) Does Bent's rule always work?

Revision

HCH bond angle is ~110 degrees.
HCBr bond angle is ~108 degrees.
Br-C-Br bond angle is 112 degrees.

Rationales

vdW repulsions play a role in Br-C-Br bond angle expansion, although the amount of p-character in the C-Br bonds might be >75%.
HCH angle is expanded because of Bent's rule; s-character in the H-C bond exceeds 25%.
The expansion of the Br-C-Br bond angle and the expansion of the H-C-H bond angle squeezes together the remaining H-C-Br bond angle.

Here's the B3LYP/6-31G* geometry. On something like this, I'd suspect this is fairly reliable, although I'll try a few other methods in a second. It gives the Br-C-Br angle as a healthy 113.4 degrees, and H-C-H as 112.5, both in agreement with experiment.

I don't see how the Br-C-Br angle is ever going to be smaller than 109.5. It's the Br-C-H angles that shrink (107.7 degrees), since breaking the symmetry forces the molecule into a "squashed" tetrahedron (point group $D_{2d}$ for the squashed tetrahedron, $C_{2v}$ for $CH_2Br_2$ in particular)

So your questions:

  1) Does Bent's rule work here (i.e. am I correct?)

Certainly not in the way you're thinking. As suggested in the other answer, the Br-Br repulsion opens that bond angle, and there's likely a Br-H electrostatic interaction shrinking that bond angle. (That's what I'd think.)

  2) Does Bent's rule always work?

I typically consider VSEPR and Bent's rule to be very useful heuristics. Indeed, it's used in one of my favorite force field methods, UFF.

I'm holding back on "always" because chemistry is incredibly varied. A quick Google search of "Bent's rule fail" turns up this interesting nugget from Weinhold's book:


The mechanism of demethylation of a quaternary ammonium The problem is such:

I know that the t-Bu locks the ring in equatorial position as I have drawn out:

Could someone please point me in the right direction as how I can get rid of a Me in the NMe3+ group?

Is it possible that tBuO- is acting as a nucleophile in this case?

As you point out, the tert-butyl group wants to remain in an equatorial position, else both the tert-butyl group and the trimethylammonium group end up being axial which is disfavoured for two bulky groups.

The obvious role of tert-butoxide:

The tert-butoxide would obviously rather act as a base than a nucleophile (in-fact under ordinary circumstances we would say that tert-butoxide is a non-nucleophilic base in the same way we refer to LDA).

In order for the tert-butoxide to act as a base, and do the elimination, the conformation of the ring would first have to put both groups axial, which we have already said isn't favourable.

There is an additional reason why this is unable to happen however, since the protons that you need to abstract to get elimination are 1,3-diaxial to the tert-butyl group. This would be highly sterically disfavoured and so is unlikely to happen.

The less obvious role of tert-butoxide:

Since the tert-butoxide can't act as a base, there is the possibility of it attacking one of the methyl groups on the trimethylammonium group to give the product as drawn in an SN2 type reaction. Whilst this seems a little odd, it does have the advantage of neutralising the charges on both the tert-butoxide and the trimethylammonium group to give two neutral molecules.


Why is isobutane more thermodynamically stable than n-butane? I would like to understand the term stability in organic chemistry. We know empirically that isobutane (2-methylpropane) is more stable than n-butane. For example, their heats of formation (taken from the NIST WebBook) are $\pu{-134.2 kJ/mol}$ and $\pu{-125.6 kJ/mol}$ respectively, putting isobutane at ca. $\pu{10 kJ/mol}$ lower in energy.

But why is this so? Is there a physical or chemical explanation for this?

I would like to understand the term stability in organic chemistry.

We can speak of kinetic stability or thermodynamic stability when we discuss chemical compounds.  If a compound is kinetically stable, that indicates that it has a high barrier (large activation energy) to cross and is therefore less reactive than a compound that has a low barrier (small activation energy) to reaction.  In the following chart we can say that the reactants are kinetically less stable and the products are kinetically more stable since the barrier to the forward reaction is smaller than the barrier to the back reaction.

(image source)

If one compound has a lower heat of formation than another compound, we can say that the former compound is thermodynamically more stable than the latter compound.  Again using the above chart, we can say that the product is thermodynamically more stable than the reactant.

  We know empirically that i-butane is more stable than n-butane. But
  why is it more stable?

Steric interactions do not provide the answer (reference).

We use arguments based on carbon-hydrogen hyperconjugation to explain many things.

For example, why more highly substituted double bonds are more stable than less substituted double bonds.  Carbon-carbon hyperconjugation can also be used to explain various chemical phenomena.  In the case at hand, let's compare carbon-carbon hyperconjugated resonance structures for n-butane and 2-methylpropane.

The hyperconjugated resonance structure for butane involves a primary carbocation, while the hyperconjugated resonance structure for 2-methylpropane involves a more stable secondary carbocation.  This more stable carbon-carbon hyperconjugated resonance structure explains the increased thermodynamic stability of 2-methylpropane compared to the isomeric n-butane.


Why isn't HCl called hydrogen monochloride? Since HCl is a covalent compound, then it would be expected to be called hydrogen ***mono***chloride following the rules for naming. Instead it is named as if it were an ionic compound, without any prefixes. Why?

Why not call it monohydrogen monochloride? If it's unambiguous you leave away anything that could describe "mono-".


How can I make a graphene supercapacitor? I have heard that a super capacitor can be made with a dvd burner. What materials do I need to to get and what software for my DVD burner do I need to make this amazing device?

Chemistry World UK have made this YouTube Clip that has a professor from UCLA explaining what is needed and how to go about it - even though the video is about 2.5 minutes, but it is a good demonstration.  The associated article is DVD player burns graphene to disc

Additionally, according to this JensLabs page DIY Graphene, it states that the following readily available materials are needed:

  Graphite oxide
  Light scribe capable DVD burner
  Light scribe DVD
  Substrate
  Sonicator - the article suggests that a ultrasonic cleaner would suffice
  Pipette
  Ion-porous separator
  Electrolyte


Why can higher voltages lead to less efficient electroplating? I was reading up on electroplating and I came across some things which really confused me:

Faraday's First Law says that the amount of metal plated onto the cathode is proportionate to the current, so it should be that a higher current= more metal plated onto the cathode (I assume).
And from Ohm's Law, a higher current= higher voltage so (by syllogistic reasoning) a higher voltage should give you more metal plated onto the cathode.

However this isn't the case, because I did an experiment recently where the highest voltage didn't produce the highest amount of metal plated onto the cathode. I did some reading on why, and I read (on the internet) that it's because when you increase the current by a lot, the ions formed at the anode can't keep up with the rate of electron flow and electrons just end up combining with other things in solution and no metal forms.

I have some issues with this though; I used a solution with ions already in it, so shouldn't there be no problem for me? Even if the ions cannot be formed at the anode fast enough, I have ions already in solution, so I thought it would be fine. Plus, the experiment was run for only 25 minutes, so I doubt they would've all been used up or anything.

Also, even if it were true that the current was just too high for the ions to keep up; wouldn't the amount of metal formed be at least equal to or greater than the amount formed for lower voltages? It's just that you're forming more metal in less time, ie it gets faster (I think); so you should get the same amount at say 5 V that you do that 3 V, even if towards the end of the experiment in 5 V it gets inefficient and you stop plating. Because the amount of ions or anything doesn't change.

Sorry for the huge amount of writing, I just thought it'd be easier if I explained my thought process. I'd appreciate it a lot if anyone could help me! Thanks so much. :)

When you apply the required voltage for the desired cell, then your desired product is being produced, but if you provide an overpotential then you forced the other reaction to also occur(the undesired one).

For eg. If in a reaction two species can be oxidized then, of the required potential is applied then you will get the required product but if an over potential is applied then the other species will also get oxidized and you will have two oxidation products.


Why does adding a charged group to an organic molecule decrease its stability? As I understand it, during glycolysis, glucose is converted into fructose 1,6-bisphosphate which greatly destabilizes the molecule. This is so it can be divided into pyruvates more easily.

Now I don't understand why adding charged groups such as phosphates destabilize molecules, what actually happens to the molecule when a charged group associates with it? Why does it become destabilized?

The quick and general answer is electrostatic repulsion. Fructose-6-phosphate is already a dianion. Phosphorylating it a second time will mean adding another two negative charges to the dianion. Like charges repel each other, so the two dianionic phosphate groups are pushed apart electrostatically.

However, you cannot generalise this. Sometimes, the additional negative charge may serve to better balance charges overall. It could also be so far away from an existing negative charge that electrostatic interactions are minimal. Finally, there are many ions and positively charged organic molecules dissolved in the cell which altogether stabilise negative charges relatively with respect to a simple solution of fructose-1,6-bisphosphate. (The same can be said for the stabilisation of positive charges.)

Finally, please note that the additional negative charge does not really assist the fragmentation of fructose-1,6-bisphosphate. The reaction — a retro-aldol — is enzymatic and catalysed principally by a lysine and aspartate residue. The former is used to convert the keto functionality into an imine which has a greater tendency to form an enamine while the latter serves as a proton shuttle. The phosphate groups are merely there so the enzyme recognises the correct molecule; modifying the enzyme’s structure to allow for unphosphorylated fructose to be consumed would not alter the reaction rate significantly.


Why are fluorides more reactive in nucleophilic aromatic substitutions than bromides? In nucleophilic aromatic substitution reactions, why do fluorides react faster than bromides?

Ordinarily bromide is a better leaving group than fluoride, e.g. in $\mathrm{S_N2}$ reactions, so why isn't this the case here? The only thing I can think of is that fluorine is more electron-withdrawing (via the inductive effect), which could stabilise the Meisenheimer complex formed as an intermediate.

The key point to understanding why fluorides are so reactive in the nucleophilic aromatic substitution (I will call it SNAr in the following) is knowing the rate determining step of the reaction mechanism. The mechanism is as shown in the following picture (Nu = Nucleophile, X = leaving group):

Now, the first step (= addition) is very slow as aromaticity is lost and thus the energy barrier is very high. The second step (= elimination of the leaving group) is quite fast as aromaticity is restored. So, since the elimination step is fast compared to the addition step, the actual quality of the leaving group is not very important, because even if you use a very good leaving group (e.g. iodine), which speeds up the elimination step, the overall reaction rate will not increase as the addition step is the bottleneck of the reaction.

Now, what about fluorine? Fluorine is not a good leaving group, but that doesn't matter as I said before. It is not the leaving group ability of $\ce{F-}$ which makes the reaction go faster than with, say, bromine or chlorine, but its very high negative inductive effect (due to its large electronegativity). This negative inductive effect helps to stabilize the negative charge in the Meisenheimer complex and thus lowers the activation barrier of the (slow) addition step.
Since this step is the bottleneck of the overall reaction, a speedup here, speeds up the whole reaction.

Leaving groups with a negative mesomeric effect (but little negative inductive effect) are not good at stabilizing the Meisenheimer complex, because the negative charge can't be delocalized to them.


Gravimetric analysis of barium or sulfate Which is a better way to obtain barium sulfate precipitate for gravimetric analysis: by adding $\ce{Ba^2+}$ ions to sulfate ion solution, or the other way round? I think it is better to precipitate sulfate ions by $\ce{BaCl2}$ solution, as this causes chloride ions to coprecipitate.

I think $\ce{BaCl2}$ should be added later. This is because $\ce{BaCl2}$ is soluble in water, but the metal sulphate you will be given may not be soluble. So if you add $\ce{M2SO4}$ later, you may get a wrong result as you have to make sure that the whole of $\ce{BaCl2}$ is used up. On the other hand all chloride salts are soluble. So while filtering the $\ce{MCl}$ will be removed easily along with the solvent and you will get $\ce{BaSO4}$.


How far-reaching is the effect of conjugation? Due to conjugation, it is possible for 1,4-addition, or conjugate addition, to occur. I was thinking, since conjugation and resonance effects do no diminish with distance, unlike inductive effects, does that mean that we can indefinitely conjugate and even have 1,6-addition? And even 1,8-addition and so on? My chemistry teacher mentions that 1,6-additions are possible but is there really no limit as to the effects of conjugations? Perhaps, there is increasing "resistance" as the carbon chain lengthens, like how a longer wire would have a larger resistance? I wonder if that would be an appropriate analogy.

1,6-additions are indeed possible (1) but since you have competing 1,4-addition and 1,2-addition it becomes more difficult. To push it to the 1,6-addition bulky organometallic catalysts are used. Reaction pathways are also determined by the electrostatic interactions, and these diminish the further away you get from the carbonyl.

(1) https://onlinelibrary.wiley.com/doi/epdf/10.1002/adsc.201601342, https://www.organic-chemistry.org/abstracts/lit1/353.shtm


Positive charge on a zwitter ion at pH=7? I faced this question. But I am having some problems understanding the solution to this:

How can the negative charge on the peptide vanish on $\pu{pH = 7}$? Suppose, we have some $\ce{H+}$ donor on either $\ce{R1}$ or $\ce{R2}$. Then the $\ce{H+}$ donor can make the carboxyl radical neutral. But by donating a $\ce{H+}$ ion, the donor becomes negatively charged. Maybe, the donor can take the $\ce{H+}$ from the $\ce{NH3+}$ radical. But the result, a completely neutral tetrapeptide, doesn't satisfy the condition of the question.

Need a hint.

To sum up: of the side chains listed, only the aminoalkyl ones will be positively charged at pH 7, so four peptides -- 4, 6, 8 and 9 -- will have a net positive charge at pH 7.


Pubchem: list all compounds for which Kovats retention indices are available Pubchem recently started listing Kovats retention indices, e.g. for
alpha ionone:
https://pubchem.ncbi.nlm.nih.gov/compound/5282108

I was wondering though if there is a way to get a list of all Pubchem compound ids which have this info available? Ideally via a PUG REST query, and a JSON or XML export option?

Note:
doing a search

https://www.ncbi.nlm.nih.gov/sites/myncbi/searches/save?db=pccompound&qk=14

https://www.ncbi.nlm.nih.gov/sites/myncbi/searches/save?db=pccompound&qk=12

will get me all records uploaded by NIST and NIST Chemistry Webbook - this will already get me close. But how can I get the list of all these pubchem IDs, ideally using some PUG REST query? (it has to be used programmatically from R)

Ha NCBI helpdesk just replied with this:

The easiest way to do this is to use the PubChem Classification Browser to access the list of Compound records with Kovats data.

You can find the PubChem Classification Browser here:  https://pubchem.ncbi.nlm.nih.gov/classification

1) Click on the "Select classification" pull-down and select "PubChem Compound TOC".
2) Under "Data type counts to display" click on "Compound".
3) Scroll down to the TOC listing and expand the "Chemical and Physical Properties" section and then "Experimental Properties".
4) Look for the entry "Kovats Retention Index" with the number of counts next to it.
5) Click on the number of counts (currently 79,656) to take you to PubChem Compound results list populated with those records shown. (https://www.ncbi.nlm.nih.gov/pccompound?DbFrom=pchierarchy&Cmd=Link&Db=pccompound&LinkName=pchierarchy_pccompound&IdsFromResult=1856976)

To save the list of CIDs for these records there are two options....

1) Click "Send to" (near the top).
2) Select "File".
3) Select "UI List"
4) Click "Create file"

OR

Click "Structure Download" (on the right)& follow the directions....to download the SDF or XML formats of these records.

If you don't need the 3D versions of the records/images this will work fine, HOWEVER if you want the "3D Records/Images" versions of these - there is a limit of 50,000.  So you'd need to download the CID list and split it in two to upload in this tool for downloading the file.

I hope this helps.

NCBI User Services


If we start heating a substance, does it melt when we break its intermolecular bonds? I was reading the 2 following posts:

Do substances melt when we break their intermolecular bonds?
Is melting/boiling point of ionically bonded substance higher than of covalently bound?

Altough they both contain wonderful answers, I did not understand if melting can be caused just by the break of the intermolecular weak bonds, while we're heating the substance.

I'm somewhat befuddled as to the confusion. I think it is perhaps the problem is in the phrase "intermolecular weak bonds."

If you look at a diamond crystal there aren't any diamond "molecules." Rather the whole crystal is one big "molecule."

Likewise we write the chemical formula for sodium chloride as $\ce{NaCl}$, but in solid and liquid sodium chloride there are no sodium chloride molecules. True sodium chloride molecules can only exist in the gaseous state.

If you look at $\ce{H2O}$ the hydrogen atoms are extremely labile. So the two hydrogen atoms and the one oxygen atom that were a molecule when the ice froze might not be part of the same molecule when the ice melts.

So to fit the overall criteria of "intermolecular weak bonds," we need a molecule which has stronger intramolecular bonds than intermolecular bonds. Say methane for example. So when methane freezes the same molecule that freezes in the solid will be the same molecule that is freed when the solid methane melts.

So in the case of methane adding energy in the form of of heat to solid methane will eventually reach the melting point of methane. At that point the weaker intermolecular bonds of the crystal break and the bulk solid turns into liquid methane.

So the overall notion is that the solid form exists as colder  temperatures than the liquid form and that the liquid form exists at cold temperatures that the gaseous form. So generally we take solid and heat it. When (IF!) it turns to a liquid then the temperature of the transition is the melting point. Likewise if the liquid can be turned into a vapor then the temperature of the transition is the boiling point.  Thus solid methane and sodium chloride will both "melt," but for different reasons.


Thermal decomposition of magnesium bicarbonate I was trying to solve a problem from a regional contest where it was a mixture of bicarbonates of $\ce{Na}$ and $\ce{Mg}$ which was put at high temperature. What are the reactions of decomposition?

I've thought that the reactions are:
$$\ce{2NaHCO3 -> Na2CO3 + H2O + CO2}$$
and $$\ce{Mg(HCO3)2 -> MgCO3 + H2O + CO2}$$

But, I found out that the second reaction is wrong. Actually, the reaction is

$$\ce{Mg(HCO3)2 -> MgO + H2O + 2CO2}$$

I searched on the web and I found the decomposition of other bicarbonates, like $\ce{Ca(HCO3)2}$ is similar to the first reaction. Is there any rule? When is carbonate and when is oxide? Or is $\ce{Mg}$ an exception?

Thanks!

I can quite easily remember from my class XI studies that carbonates of alkali metals decompose on heating to give carbon dioxide and the corresponding metal oxide.

Moreover the thermal stability of the alkali metal carbonates increases with increasing cationic size because carbonate ion is big in size and increased cationic size leads to better bonding and hence greater stability.

In this regard the beryllium carbonate is the least stable alkali carbonate or in other words, an unstable alkali carbonate and readily decomposes into $\ce{BeO}$ and $\ce{CO2}$. $\ce{BeCO3}$ is so unstable that it can only be kept in an atmosphere of $\ce{CO2}$. $\ce{MgCO3}$ is also a bit unstable taking into consideration the above fact.

So the reaction you mentioned basically gives $\ce{MgCO3}$ as the product but $\ce{MgCO3}$ decomposes readily to give $\ce{MgO}$ and $\ce{CO2}$.
$$\ce{Mg(HCO3)2 -> MgCO3 + H2O + CO2}$$
and
$$\ce{MgCO3 -> MgO + CO2}$$
giving
$$\ce{Mg(HCO3)2 -> MgO + H2O + 2CO2}$$


Error in formula for the solution to a chemical problem The problem in my textbook goes like this:

  Burning $0.68\ \mathrm{g}$ of an unknown substance, we obtained $1.28\ \mathrm{g}$ of sulfur oxide (IV) and $0.36\ \mathrm{g}$ of water. Find the chemical formula of the burned substance, providing that it was a complete combustion.

It's quite easy to see that $0.2\ \mathrm{mol}$ of both $\ce{SO2}$ and $\ce{H2O}$ have been produced, there are $0.04\ \mathrm{mol}$ of hydrogen in the water and $0.02\ \mathrm{mol}$ of sulfur in the oxide. Calculating the masses and summing them up, we get $0.68\ \mathrm{g}$, meaning the original substance contained only sulfur and hydrogen, and it's $\ce{H2S}$.

However, the solution volume offers this equation as part of the solution process:

$$\ce{S_{$x$}H_{$y$}O_{$z$} + \frac{x + y}{4} O2 -> $x$ SO2 + \frac{$y$}{$z$} H2O}$$

The volume is full of typos (apparently a plus sign missing here), but still there must be some explanation for the coefficients which I don't get. Why do the coefficients for $\ce{O2}$ and $\ce{H2O}$ equal $(x+y)/4$ and $y/z$? If there's zero oxygen in the first term, won't we get division by zero?

How do we come up with coefficients of this sort? Maybe these particular coefficients are erroneous? (the solutions volume is quite low in quality)

The equation is a little dubious, as I mentioned in the comments.

  How do we come up with coefficients of this sort?

Let's go from scratch. We don't know the coefficients at this point so we'll just call them $a$, $b$, and $c$. Note that the coefficient of $\ce{S$_x$H$_y$O$_z$}$ is taken to be 1. It doesn't have to be 1; however, for the sake of simplicity you might as well take it to be 1. If you take it to be 2, then you just have to double all the other coefficients.

$$\ce{S$_x$H$_y$O$_z$ + $a\,$O2 -> $b\,$SO2 + $c\,$H2O}$$

In the reactants, we have $x$ sulfur, $y$ hydrogen, and $z + 2a$ oxygen. In the products, we have $b$ sulfur, $2c$ hydrogen, and $2b + c$ oxygen.

Our aim is to express $a$, $b$, and $c$ in terms of $x$, $y$, and $z$. Equating all three, we get

$$\begin{align}
x &= b \\
y &= 2c \\
z + 2a &= 2b + c
\end{align}$$

which immediately tells you that $b$ and $c$ are equal to $x$ and $y/2$ respectively. Substituting these into the last equation,

$$\begin{align}
z + 2a &= 2x + \frac{y}{2} \\
a &= \frac{4x + y - 2z}{4}
\end{align}$$

So the full equation is:

$$\ce{S$_x$H$_y$O$_z$ + $\frac{4x + y - 2z}{4}$O2 -> $x\,$SO2 + $\frac{y}{2}\,$H2O}$$


Does this NMR show the carboxy group? This is the NMR for the molecule shown. I was told that carboxy group will be in the above 10ppm range. Does that tiny signal at about 12.9 ppm indicate the carboxy group though? I dont understand why the methoxy would show large integration while carboxy barely shows.
Also my general question is, how much absorption is necessary for it to indicate some group present?  Is there "noise" on NMR spectra or does every signal indicate a group?

Indeed, the very small signal at 12.9 ppm is due to the carboxyl group. The reason why it's so low is because, as explained here, at room temperature the proton exchange makes it nearly undetectable. At higher temperatures, you would probably not even see that spike.

The methoxy group, on the other hand, is not affected by such effects.

There will always be a tiny amount of "noise" on experimental NMR spectra, because a solution always contains unwanted elements; but usually anything that looks like a well defined spike is a significant signal.


How much ascorbate to neutralize one liter of chlorinated tap water? In Las Palmas de Gran Canaria, tap water is chlorinated. In an Internet forum, someone   claims that in 2008 the water had a chlorine concentration of 0,5 - 1mg/l. On the official site of Emalsa, the water provider, I couldn't find any information about chlorine concentration. I tried contacting them online, but the contact form is broken.

In another forum, I read that ascorbic acid (vitamin C) can be used to neutralize the chlorine: My taste buds say it works!

How much ascorbate/l do I need to neutralize the above amount of chlorine?

This is just a very short explanatory note on the article suggested by Greg E..

The reaction between ascorbic acid and hypochlorous acid is as follows:
$$ \ce{C5H5O5CH2OH + HOCl → C5H3O5CH2OH + HCl + H2O}$$
so for 1 mole of hypochlorous acid you need 1 mole of ascorbic acid. This can be recalculated into a mass ratio by the use of their molecular masses:

$$M(\ce{C5H5O5CH2OH})=176\;\text{g/mole}$$
$$M(\ce{HOCl})=52.5\;\text{g/mole}$$
From this we can see that if the molar ratio is $1:1$, the mass ratio is $176:52.5$ which is $3.3:1$ somewhat higher than the article claims.

I am not exactly sure of the reason, it could be that there is some side reaction of ascorbic acid that also removes some additional hypochlorous acid although I wouldn't know which. Perhaps the article is talking about the amount of ascorbic acid you need to get a concentration which you don't taste anymore?


Is didehydrobenzene valid IUPAC nomenclature? I don't know if this compound exists in real life or not. If it did, what would be the correct IUPAC nomenclature for it? According to me, it could be:

1,2-didehydrobenzene
cyclohexa-1,3-dien-5-yne

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC name (PIN) for the compound that is given in the question is 1,2-didehydrobenzene. The name cyclohexa-1,3-dien-5-yne is acceptable for general nomenclature. The formerly used name benzyne is no longer recommended.


Acidity of primary and secondary amines I was reading a book on organic chemistry by David Klein and I came across this question, but I wasn't able to understand why one amine is more acidic than the other.

Have you taken into account that there is a $\ce{-CH3}$ group attached to the top N which has +I effect i.e., electron donating effect?
$\ce{-CH3}$ group will destabilize the negative charge on N formed after the release of proton.
Electron releasing group decreases acidic strength of a compound. So in this question proton of primary amine is more acidic and that of secondary amine is less acidic.


4-fold chemical compounds I have been reading some papers and one mentioned "4-fold triethylamine based on PEG-diol end groups", "4-fold acryloyl chloride." What is the meaning of "4-fold" in those compounds? Also, can I find these chemicals for sale or must I synthesize them myself?

Assuming the "4-fold" is not referring to "use four equivalents of X per molecular function Y to obtain the intended chemical reaction", it refers more often to a specical class of star shaped polymers, called dendrimers.  Commonly, around a nucleous, you have several times the same chemical group, as, for example, drawn here:

In this example, instead of a linear chain of polyethylene glycol (PEG), four branches are stitched together; the fourth example may relate the closest to your question.

Source of the picture: Zhu, Biomaterials 31 (2010) 4639-4656, doi: 10.1016/j.biomaterials.2010.02.044


Nernst Equation for Lithium ion battery For electrochemical cells we know
$$
\Delta G = \Delta G^{\ominus} + RT\ln Q=- RT\ln Keq + RT\ln Q=-nFE
 $$
So for lithium ion batteries the reactions are
Cathode
$$ \ce{CoO2 + Li+ + e- <=> LiCoO2} $$
Anode
$$ \ce{LiC6 <=> C6 + Li+ + e^-}$$
Overall
$$\ce{ LiC6 + CoO2 <=> C6 + LiCoO2} $$
Is it safe to say the cobalt oxide is a solid as is the Lithium and can be ignored in the reaction quotient? As in a Zinc copper battery i do not think the mass of the zinc and copper affect the cell potential only the aqueous concentrations. For a lithium ion battery the cell potential is a function of the state of charge and temperature. but what are the concentrations in the reaction quotient for a lithium ion battery as most of the products and reactants are solids, is it not accurate to ignore them due to intercalation and are not exactly solids? What is nernst equation for lithium ion batteries

The Nernst equation for the anode is :$$E_a = E°_a + \frac{RT}{F}ln{\frac{[Li^+]_e [C_6]_s}{[LiC_6]_s}}$$ where the concentrations $[C_6]_s$ and $[LiC_6]_s$ are  defined in the solid phase of the anode, and not in the electrolyte. The concentration $[Li^+]_e$ is defined in the electrolyte. When the anode is working, the concentration $[LiC_6]_s$ decreases in the solid phase and the concentration $[C_6]_s$ increases.

The same reasoning may be done on the Nernst equation at the cathode :  $$E_c = E°_c + \frac{RT}{F}· ln{\frac{[Li^+]_e[CoO_2]_{s'}}{[LiCoO_2]_{s'}}}$$ where $[CoO_2]_{s'}$ is the concentration of $CoO_2$ in the solid phase of the cathode, and $[LiCoO_2]_{s'}$ is the concentration of $LiCoO_2$ in the solid phase of the cathode.

The overall voltage of the battery $\Delta E$ can be written :  $$\Delta E = E°_c - E°_a + \frac{RT}{F} · ln\frac{[LiC_6]_s [CoO_2]_{s'}}{[C_6]_s [LiCoO_2]_{s'}}$$ As a consequence, both $[LiC_6]_s$ and $ [CoO_2]_{s'}$ decrease when the battery is working. So the overall voltage decreases.


NO2 gas was detected in FTIR We have a laser chamber full of $\ce{F2}$ gas (no air should be in the chamber) and several parts including steel, wires, ceramics, zinc.... This is for laser generation so the temperature is pretty high in the chamber. We have detected $\ce{NO2}$ gas in our FTIR specturm which caused our laser to fail and there’s a good chance that the $\ce{NO2}$ may be associated with the failure. The questions are

How might $\ce{NO2}$ be formed in a chamber? If it’s from air, why would $\ce{NO2}$ be formed and not $\ce{NO}$ or $\ce{N2O}$? Why would $\ce{NO2}$ persist after $\ce{N2}$ and $\ce{O2}$ are removed by chamber refills with fresh $\ce{F2}$?
How does $\ce{NO2}$ interact with fluorine? What other species might be formed when they react?
Can $\ce{NO2}$ be dissociated by UV light (i.e. 193 nm)? What is its UV absorption spectrum?

Fluorine reacts with practically everything, especially at high temperatures, including zinc, wires, some ceramics, polymers, etc., so the system you describe may be prone to contamination. I don't see any obvious source of NO2, but it could easily form from nitrates, if present, in an acidic environment, which could result from fluorine attacking metals present.


How to purify water with activated carbon? If I am not mistaken, distilling water does not remove all possible impurities. Chemicals with boiling points less than $100\ \mathrm{^\circ C}$ will be transferred as well. Does the usual tap water contain such (I think there is at least fluoride.)? I read that activated carbon can be used to purify water as well. How is it done in practice? Suppose I first use activated carbon and then distill the water (or in the reverse order). How much impurities there would be still left?

The main reason activated carbon is used in water treatment is that its large surface area is useful for adsorbing volatile organic contaminants and chlorine. It is not effective at removing inorganic ions (dissolved minerals, including fluoride), but these are removed quite well by distillation. I think distillation first would make more sense—distillation will remove less volatile contaminants so that less has to be adsorbed by the carbon.

Tap water has to meet certain standards (that depend on your area) for dissolved minerals and volatile organic compounds, so depending on the end use of the water, removing the small amount of remaining contaminants may not be necessary. Formerly, double-distilled water was the standard for making highly purified water, but certain applications are more or less tolerant of certain types of contaminants, and double-distillation requires a lot of energy, so it has largely been replaced for many purposes.

Consumer water filters usually contain filters to remove particulates, ion exchange resin to remove inorganic ions, and activated carbon to improve the taste of the water, and as far as purification for laboratory use, combinations of filtration, deionization, reverse osmosis, ultrafiltration, etc. are common these days. Different processes are better at removing some types of contaminants than others, so combinations are used. Distillation of tap water followed by treatment by activated carbon will produce quite pure water by consumer standards, but might not be pure enough for say semiconductor microfabrication or biomedical applications.


What's the relationship between chemical polarity and vapor pressure？ A question which confuses me now.
Chemical polarity is one of microscopic properties for certain molecules.
Vapor pressure is a macro thermodynamic properties for certain materials.

But I often read about some violate organics are polar material. So, it has a  relatively small vapor pressure and  hard to evaporate.

How to connect these two concepts together? If I can access the volatile ability based on the polarity, it will help me to recognize some kind of VOCs which are eager to condense into particle phase.

The vapor pressure of a liquid is the equilibrium pressure of vapor above the liquid at a given temperature. Liquids with higher vapor pressures must have less strong forces holding the molecules together in the liquid phase, thus more of the molecules are able to escape the liquid and be in the vapor phase. This is the connection between inter-molecular forces and vapor pressure.

Polar molecules can attract each other through dipole-dipole and dipole-induced dipole interactions in addition to the induced dipole-induced dipole intermolecular interactions that all molecules experience. Thus, all else (molar mass, for example) equal, a polar molecule will generally have a lower vapor pressure than a non-polar molecule.


Concentration of mercury in body Symptoms of mercury poisoning become apparent after a person has accumulated $\pu{20 mg}$ of mercury. If a person ingested $\pu{30 mg}$ of mercury, what concentration of mercury in parts per million, are in his body? Assume the person has a mass of $\pu{63.5 kg}$."

My thought process, using dimensional analysis, was to convert the amount of $\pu{mg}$ the person ingested into $\pu{g}$. Then I took it ($\pu{0.03 g}\ \ce{Hg}$) and divided by the molar mass of mercury to find the number of moles of mercury there are.

I then divided the amount of mercury by $\pu{63500 mL}$ because I assumed $\pu{1 g} = \pu{1 mL}$. I did not use the $\pu{20 mg}$ information that it got me at all. Thus what I got was:

$$\pu{0.03 g} \cdot \frac{\pu{1 mol}}{\pu{200.59 g}} \cdot \frac{1}{\pu{63500 mL}} = \pu{2.3e-9 mol L-1}$$

However, I completely messed up, because first, my answer was in $\pu{mol L-1}$, and I'm not sure if you're supposed to use the molar mass of mercury here. The answer was $\pu{0.47 ppm}$, but I don't know how to get to here. Could someone explain the thought process for this?

There are 1,000,000 mg in a kg (this is, mg/kg = ppm), so all you need to do is divide the mass of $\ce{Hg}$ by the mass of the person as-is (the units work out to ppm):

$$\left({30\,{\rm mg}\over 63.5\,{\rm kg}}\right)\cdot\left({{\rm ppm}\over {\rm mg\cdot kg^{-1}}}\right) = 0.47\,{\rm ppm}\,\ce{Hg}$$


Relative humidity at constant temperatures It’s known that Relative humidity = (water pressure in a certain volume/ saturated water pressure in the same volume)* 100% “at the same temperature”

So my question is why does the amount of water isn’t max all the time instead it differs so we calculate humidity?

The unusual thing for me is that we didn’t change the temperature so what makes it differs?

Thanks in advance

To have 100% relative humidity at constant temperature, 3 conditions have to be met:

There must be water source to saturate space with vapour.
There must not be present materials absorbing humidity below 100%.
There must be allowed enough time to reach saturation.

E.g., if there is solid table salt exposed to air, the equilibrium relative humidity would not be higher than about 75%, even if water is there. As salt absorbs water and starts dissolving in it. Either all water evaporates and  humidity is still <= 75%, either all salt dissolves and continues absorbing vapour and diluting itself.

But even if all salt dissolves, 100% could be reached only if salt solution absorbs so much vapour we can neglect salt content in the salt solution.

For real life air conditions, varying temperature of free air leads to cycling of evaporation and condensation, resulting in air usually unsaturated. If kept closed at constant temperature, without evaporation or condensation, it stays unsaturated.


How to determine the reaction of a strong electrolyte? I'm told the $\ce{KMnO4}$ is a strong electrolyte.

How can I figure out how the reaction looks like?

$\ce{HMnO4}$ is s a strong acid (compare oxidation state with perchloric acid).  Its anion is then a weak base.  The salt is completely dissociated in water solution.


How can I measure the scale in water? I have hard water in my house. All the taps have tartar around. It generates some problems in some of my home appliances so I have bought polyphosphate salt but when I boil water I don´t get the expected results.
I´m looking for some product or filter that would make the water in my house better for drink and use. That´s is why I´m evaluating to install a filter with polyphosphate salt or a floating dispenser (like a buoy).
I´m now trying to make a test before installing the floating dispenser I have recently bought in my water tank so, this is what I have done:
First I have filled a pan with water and boil it until it got empty and this was the result:

Then, I have use the floating dispenser in a pan with polyphosphate salt for 30 mins. After that I took out the dispenser and boil it until it got empty. This was the result:

It seems like there is no difference.
How can I make a better measurement or treatment? Is this the right product for my problem? 30 minutes is enough to the polyphosphate salt? Do you think that will be better a floating dispenser or the filter with that product? Maybe I have to buy another kind of filter?

Most drinking water has residue like what you are showing.  Boiling water shows there are ions in the water, but it does not show what kind of ions are present.  Boiling water is practically useless when determining scaling potential and it is not safe to do using cooking pots and pans.  Part of the reason laboratory equipment is more expensive than cooking pots and pans is because lab equipment has much better fire prevention measures.  Boiling down water in this manner has the potential to start a fire.
pH is a better measure for scaling potential.  If you want to get technical, water pH is used to determine the Langelier Saturation Index or the Ryzner Index which are indicators of scaling potential.
Your water is probably too alkaline. Alkalinity and acidity need to be balanced because if water is too alkaline then too much scale forms; however, if the water is too acid pipes corrode.
For a household, filtering water with an activated carbon filter first, then an ion exchange filter might work.  Ion exchange filters contain a resin which exchange scale forming ions like calcium with ions like sodium. These filters can be easily fouled by organic material, so the organic material will need to be removed by something like an activated carbon filter.


What is the molecular structure of the coating on cast iron cookware known as seasoning? When a piece of cast iron is repeatedly covered in fat and exposed to heat it develops a hard, black, non-stick coating referred to as "seasoning".

What exactly is this coating made of molecularly? How is it bonded to the cast iron substrate? What is the chemical process that transforms a liquid fat into this coating? And what if any characteristics of the fatty acids or triglycerides used have an effect on the final coating or how well the it bonds to the iron?

Some hints:

if you mean that "what exactly" in the original sense of the words,
then the answer is: nobody knows and will ever know.
If you are content with some more vague answer, then
look at the pyrolysis chemistry occuring from 150 °C
up with fats, proteins and starch/sugars, and all that
at once. This is a vast field of chemistry, and all
is gives to a trained chemist is a idea of the structures
of that coating.
Bonds to iron will not occur, first because on cast iron and
on steel pans there is always a layer of iron oxide,
on cast iron  you find a glaze formed by fusion of the
molds finish layer (often containing graphite and clay)


Modeling activation energy function I'm quite new to the field, and actually am a mathematician. I was wondering if there is a way to model (find the equation of) the reaction progress (more specifically, the activation energy functions) using some equation?

Optimally I'd like to use some graphing calculator like Desmos to write the equation and get a similar function.

Yes, you definitely can, but it's complicated. Because you are a mathematician it might be easier to understand.
Introduction
On your picture is one-dimensional representation of the energy function of multiple variables. Variables are coordinates of all atoms in the molecular system. This function define the potential energy surface or PES. Sometimes you can simplified it and represent as a three-dimensional surface using a linear combination of coordinates. Your picture is the energy profile of the PES. Where minimums are equilibrium states of the system and saddle points (minimax) are transitional states.
The reaction progress on the picture is the intrinsic reaction coordinate (IRC) which is a one-dimensional representation of the a linear combination of vectors of your molecular system vibration.
Real energy profile
Consider, for example, the reaction $N_2O_2$ with $H_2$. Atoms in both molecules can oscillate in a specific well-defined way which gives you the molecular vibration. You need to solve quantum differential equations for the molecular system to get vibrational frequencies and vibrational vectors. To solve it analytically is incredibly difficult, but you can solve it numerically using quantum chemistry software. It will compute energy over intrinsic reaction coordinate (IRC) as a set of points with a certain displacement or step. Using simple spline function you can get a real energy profile.

Approximation
It is possible just to use a square function. According to the harmonic approximation it is precise at the vicinity of equilibrium states, but useless with transitional states. Sometimes to avoid this problem people use energy profile diagrams. Usually people are not interested in the form of function but the energy differences (activation energies, Gibbs energies and etc.)

If precision is not the case you just need to know energies of initial separated molecules, the transitional state energy (at the saddle point) and the energy of the product (which you can compute with quantum chemistry software) and join square functions at the found energies. You also need to remember that the energy profile is made for the elementary reaction.
Calculation
There are a lot of tutorials how to compute energies of initial systems, transitional states and products with quantum chemistry software. It should be mentioned that you will need to sum equilibrium energy of computed separately molecules to get your initial energy. The most difficult part is to compute the transitional state energy. You need to find a molecular geometry which would be very close to your potential transitional state. You need to calculate Hessian (second derivatives) and start minimax optimization (saddle point optimization).


What is 'desorption hysteresis'? I am currently reading a paper titled "Polystyrene Nanoplastics-Enhanced Contaminant Transport: Role of Irreversible Adsorption in Glassy Polymeric Domain" (link here).

In the paper, the authors mention the term 'desorption hysteresis' as the mechanism of the interaction between nanoplastics and contaminants. Here's the relevant excerpt (from the introduction section):

    ... at low nanoparticle (i.e., “carrier”) concentrations significant facilitated
    transport of organic contaminants requires not only strong adsorption of contaminants to the carriers, but also significant desorption hysteresis (a collective term referring to both slow desorption kinetics and thermodynamically irreversible adsorption) of contaminants from the carriers. Desorption hysteresis can either be due to the physical entrapment of contaminants in the complex matrices of the carriers, or strong specific adsorptive interactions that lead to irreversible binding of contaminants to the carriers.

I am having some difficulty in understanding what do they mean by adsorption hysteresis. From what I've read, hysteresis is how the system in its current state is related to some past state, but I don't understand this concept fully.

Could someone please explain what is desorption hysteresis? I couldn't even understand the explanation in parentheses (right after the term in bold above). Is it about the rate of desorption of contaminants from nanoparticles?

Hysteresis occurs when the state of the system depends on its past states---that is, the state of the system is memory-dependent.

The canonical example is a ferromagnet: if you apply and slowly ramp up an external magnetic field to a ferromagnet for some time, and then slowly ramp the magnetic field back down for the same amount of time, you might expect the ferromagnet to return to its original state, but it doesn't. Instead, if you track the magnetization of the magnet as a function of the applied magnetic field, you see these hysteresis loops. Each applied magnetic field corresponds, therefore, to two possible system states, and which state the system exists in will depend on its past history.

Desorption hysteresis is hysteresis applied not to magnetization but to adsorption and desorption, and your excerpt then goes on to describe the physical interactions that drive this phenomenon.


SN2 reaction with alkyl bromide and pentanenitrile? Is it possible to do a Friedel-Crafts alkylation on a plain alkane such as the one I've attached below, or is it only reserved for EArS mechanisms?
Also, I was wondering what would happen if I tried this reaction with another halogen on the starting substrate (if I had 1-bromoethane for example)...

Edit 1: I realize now that the top reaction doesn't make sense. Rather than a Friedel-Crafts Alkylation, I think I found a mechanism to make this type of reaction work (SN2) with Br as a leaving group and CN as a strong nucleophile, but I'm not sure...

Edit 2: I miscounted the carbons.

Edit 3: This is what I understand, from the suggestions given...

It appears that you are having some difficulty with this synthesis. The suggestion in the Comments that you employ lithium diisopropylamide (LDA) to generate the anion of a nitrile is a wise one. The anion is usually formed a -78oC in tetrahydrofuran (THF). The bromide is added and the reaction mixture is allowed to warm to ambient temperature. There are two possible bonds that can be made in what is termed a retrosynthetic analysis. One by method A and another by method B. The bonds formed are shown in red in structures 2. Method A is better than method B because an SN2 reaction is being conducted on a primary halide, ethyl bromide. Method B requires an SN2 reaction on a secondary halide, 2-bromopropane, a process that is hindered and in competition with an E2 elimination that would produce propene 4 and the starting nitrile 3. Nitrile 2 must be a racemate given its method of formation. It is comprised of equal amounts of the two enantiomers, (R)-2 and (S)-2. By the methods employed here, a single enantiomer cannot be formed. Thus, a "normal bond" (not dashed nor bold) is shown in nitriles 2.


Strength of Dipoles, Covalent bonds and H-Bonds My textbook says:

  Dipole-dipole forces are 1% as effective as covalent bonds and the strength of a hydrogen bond is generally twenty times less than that of a covalent bond.

What do these statements mean?

It means that of the three forces, dipole-dipole forces are the weakest, hydrogen bonding is the second weakest, and covalent bonds are the strongest. Now, dipole-dipole attraction happens between an ion and a polar molecule. Due to a net dipole movement, this is weak. In comparison to a covalent bond, it has only $\frac{1}{100}$ the efficacy. A hydrogen bond is a force between H and either N, O, or F, the three most electronegative elements. These are classified as d-d forces, but due to the differing electronegativities, they are a tad stronger. Covalent bonds are much stronger than either, $20$ times more than the H-bond and $2000$ times stronger than the d-d force. A covalent bond involves the sharing of electron pairs between atoms, generally two non-metals. There are also ionic bonds and metallic bonds, an ionic bond involving the transfer of electrons, such as $\ce{NaCl}$ and a metallic bond being between metals.


Thermodynamic stability of benzene derivatives Q1. How can we comment on the thermodynamic stability of the following benzene derivatives?

Q2. What does thermodynamic stability mean, in general? What does it mean to compare the thermodynamic stability of more than compound?

Q3. How do we compare the thermodynamic stability of benzene derivatives? Is there a specific rule, or way to go about it? Would MOT (frost circles, to be specific) help?

I looked at the extent of resonance, and other such stabilizing phenomena/interactions. However, that only tells the extent of stabilisation of the compound due to those phenomena, not as a whole.

Edit: I wish to predict the order of stability using knowledge of chemistry, and not check enthalpy of formation data and then establish the order.

Wikipedia defines thermal stability as .

  Thermodynamic stability occurs when a system is in its lowest energy state, or chemical equilibrium with its environment. This may be a dynamic equilibrium, where individual atoms or molecules change form, but their overall number in a particular form is conserved.

For the particular problem in hand the lowest energy state definition is more relevant for us .

Enthalpy of formation can be used for judging the energy states.

A  negative enthalpy change implies that the product has released heat thus lowering the amount of energy present in the product

This implies.

  More negative the enthalpy of formation, more stable the compound.

Enthalpies of formation

Phenol
$\ce{ -> -165.0 kJ/mol}$.
Hydroquinone
$\ce{-> -371.1 ± 1.3 kJ/mol}$
P-Nitrophenol
$\ce{-> -207.1 ± 1.1 kJ/mol}$

All data has been taken from http://webbook.nist.gov , for solid state.

The order can be calculated now as

Hydroquinone $>$ paranitrophenol $>$ phenol