Effect of magnetism on dissolved Na+ and Cl- ions in water Suppose we have put some salt in water. My state of knowledge is, that a hydration envelope is formed from the oxygen side of the water molecules on the Na+-Ions and one another from the hydrogen side of the water molecules on the (Cl-)-Ions to drag these out of the crystal.
My questions now are:

When the hydration envelope is formed on the ions, does it (partially) change the charge of the ions?

(Now the main question) If I would put two magnets with opposite poles on both sides of the container, in which the salt solution is, would the ions be pulled to the side of the opposite pole, (if yes or no, then why)?

I made a little sketch an let the hydration envelope out for the sake of the little place.
And one another thing. There is no reason to downvote my question, because I am not playing around here, but the question just jumped in my head

There are a couple of issues: first, a uniform magnetic field doesn't exert a force on stationary charges and secondly, when the charges are moving the force is at right angles to the magnetic field. There are a few ways to get the charges moving. You could have an electric current flow from the bottom of your figure to the top. The $\text{Na}^+$ ions would move with the current, up the page and since the magnetic field points to the left in your figure, the $\text{Na}^+$ ions would be pushed towards the reader by the right hand rule resulting in higher electric potential near the reader. This is called a positive Hall effect. The $\text{Cl}^-$ ions would move against the current, down the page and paradoxically would be pushed towards the reader again by the right hand rule, this time resulting in a lower electric potential near the reader: a negative Hall effect.
Another way is to have the electric current directed out of the page. Positive $\text{Na}^+$ ions move towards the reader and then are pushed down the page by the magnetic field according to the right hand rule. Negative $\text{Cl}^-$ ions move away from the reader and are also pushed down the page by the magnetic field via the right hand rule. This is the principle of magnetohydrodynamic propulsion.
Another way is to have the brine flowing up the page which forces $\text{Na}^+$ charges towards the reader and $\text{Cl}^-$ charges away from the reader by the right hand rule, both resulting in higher electrostatic potential on the near side of the stream. This would be magnetohydrodynamic propulsion in reverse operating as a generator.


Why is sulfur dioxide more stable to UV radiation than ozone? Ozone readily decomposes when impacted by UV radiation.

Sulfur dioxide, having the same bond structure and valence electrons in each atom as ozone, is much more stable.

Would someone please help me understand what makes sulfur dioxide a more stable compound?

I'm going to attempt to answer this from three points of view.
Qualitative considerations and MO theory:
First, why isn't ozone linear? You can imagine $\ce{O3}$ like a $\ce{CO2}$ molecule that lacks two electrons. This lack of electrons will result in less $s$-character, and less bonding "attitude". That's why ozone is bent.
Still, the repulsion in its electrons ($\ce{O}$ is a small central atom, very small) will want it to be free. Oxygen's van der Waals radius is 1.5 $\mathrm{\overset{\circ}A}$. Thus, $\ce{O3}$ has a bond length of $\approx 1.2\ \mathrm{\overset{\circ}A}$. That's really cramped. Electrons aren't happy with this and will try to get free as soon as $E_a$ and other conditions are provided.
Now compare that with sulfur's 1.8 $\mathrm{\overset{\circ}A}$ radius; which, results in a $\approx 1.43\ \mathrm{\overset{\circ}A}$ bond length. Less cramped, the electrons aren't as much complaining as they do in $\ce{O3}$'s case. We can't get really further as these two molecules aren't isoelectronic.
_____________________________________________
Approach using classic high school knowledge:
It's because oxygen has an arbitrarily positive formal charge. Take a look at $\ce{O3}$'s electron-dot diagram:
$\hspace{18ex}$
$\hspace{45ex}$ $\small{Source}$
The oxygen atom, which has an electronegativity of $3.6$ in Allen scale, has formed $sp^2$ orbitals. This, a very electronegative atom with a positive formal charge, is highly unfavorable for the molecule, and acts very "non-bonding".
The sulfur, on the other hand, has an electronegativity of $\approx 2.5$. It will also get a positive formal charge. But since it's less electronegative, it bears it.
_____________________________________________
Thermodynamics and kinetics:
Ozone decays into diatomic oxygen under certain conditions:
$$\ce{2O3 -> 3O2}$$
Three moles of oxygen will be produced per two moles of ozone. The measured enthalpy of formation for ozone is $142~ \mathrm{kJ/mol}$, while it's 0 for diatomic oxygen. So
$$\mathsf{\Delta H_f~of ~the~ products - \Delta H_f ~of ~the ~reactants} = \mathsf{reaction~ enthalpy}$$
$$ 0 - 142 = \fbox{-142} ~\mathrm{kJ/mol}$$
Furthermore, this reaction will increase entropy, as two moles of a gas produce another three moles. Thus, it's thermodynamically highly favorable.
Above that, its energy of activation is a measly $\approx 32 \mathrm{~kJ/mol}$, which is a very small kinetic barrier.
Thus, you can also conclude that $\ce{O3}$ is exceptionally "inclined to decompose". Also, note that ozone is a very strong oxidant; while $\ce{SO2}$ is a weaker one.


Origin of the name of barbituric acid Recently, I have encountered two different accounts explaining the origins of the name "barbituric acid". Both accounts do mention that the name of this compound was given by the famous organic chemist Adolf von Baeyer. Li mentions that von Baeyer named the compound after his girlfriend at the time, with the name Barbara$^{1}$. Quin & Tyrell, on the other hand, informs that von Baeyer named the compound after the patron saint of artilleryman, St. Barbara$^{2}$. This is a seemingly trivial question but I'm genuinely just curious.

References

Li, J. J. Name Reactions, 3rd ed.; Springer: Heidelberg 2006; p. 15.
Quin L. D.; Tyrell, J. A.; Fundamentals of Heterocyclic Chemistry: Importance in Nature and in the Synthesis of Pharmaceuticals. John Wiley & Sons: Hoboken, New Jersey 2010; p. 119.

Interesting question but the sad answer is we will never know the true story. Here is the original text (thanks to Google Books) dated 1863, right from the horse's mouth Adolf von Baeyer.

Dieser Gedanke hat mich bei der folgenden Arbeit geeitet (spelling?), und ich habe
mich zunächst bemüht, die Kenntnis des Materials zu vervollständigen.
Man wird sehen, wie sich dasselbe in einfachster Weise um die Substanz
$\ce{N2CO3H4}$, die ich Barbitursäure nennen will, gruppieren lässt und wie
also die Frage nach der Konstitution der Harnsäure und ihrer Derivate
auf die Untersuchung dieser Substanz zurückgeführt ist.

[DeepL Translation] This thought led (?) me to the following work, and I first tried to complete the knowledge of the material. One will see how the same can be grouped in the simplest way around the substance $\ce{N2CO3H4}$, which I want to call barbituric acid, and how the question of the constitution of uric acid and its derivatives can be traced back to the investigation of this substance.
He never explained his choice- no footnotes anywhere. However the unabridged Oxford English Dictionary (not open access) attributes it to a woman's name Barbara.
< French barbiturique (Ann. de Chimie et de Physique, 1865), < German barbitur in barbitursäure (Baeyer 1863, in Ann. d. Chemie u. Pharm.), < Barbara, a woman's name.
Cited from Adolf von Baeyer.


How can oxygen have three bonds? I was reading an article about a chemical reaction, and I came across the phrase:

  The oxygen atom at this point has three bonds and has a net positive charge

How can this happen? Oxygen has 2 missing electrons in the valence shell. Therefore it can only form 2 bonds at the most, if both are sigma bonds.

Does it mean the 3rd bond is not covalent? Can it happen with a hydrogen or an ionic bond?

Consider the auto-ionization of water :

$\ce{ 2H_2O->H_3O+ + OH-}$

The first oxygen has three bonds, the second only has one.

You can think of the reaction taking place by a lone pair on the oxygen of one water molecule ripping off the proton only of the hydrogen of another water molecule to form a covalent bond between them using just the lone pair. The electron of the hydrogen is left behind and stays with the oxygen of the other molecule.

If you calculate the formal charges on each oxygen you will see the first one has a positive charge and the second one has a negative. The formal charge is just the valence number of electrons minus the number of bonds minus non-bonding electrons (using the lewis structure) and is a useful book keeping method to think about where the electrons go/are and what are the most stable structures.

Formal Charge Calculation


Why BiCl5 isn't stable? I read this in a textbook:

  $\ce{Bi(V)}$ is very unstable and is a good oxidizing agent.

Why does it happen that way? Is it because in $\pu{+5}$ oxidation state $\ce{Bi}$ pulls in more electrons and hence gets reduced fast or there's a different concept?

Also then $\ce{BiCl5}$ should be more stable than $\ce{BiCl3}$ because it's getting the electrons that it needs, which is not true.

The reason behind this is the mainly inert pair effect .

In $\ce{BiCl3}$, due to the much electronegativity difference between $\ce{Bi}$ and $\ce{Cl}$, Chlorine atom forms bonds with almost pure $p$ orbitals of $\ce{Bi}$, and the lone pairs on $\ce{Bi}$ are of almost pure $s$ character. Thus, $\ce{Bi}$ atom doesn't actually utilise much of its $s$ orbital electrons in forming the bonds which is energetically much more preferable. ( If it seems non-obvious, recall Bent's rule which says that more electronegative atoms prefer to form bonds with orbitals with more $p$ - character.)

On the other hand, if $\ce{Bi}$ is present in $\ce{BiCl5}$, it has no other option other than utilising its $s$ orbitals for bonding. The equatorial bonds will mainly consist $s$ character and $p_x, p_y$ orbitals will also participate in forming the equatorial bonds. So, $s$ character in the equatorial bonds will be slightly lesser than $33$% (again due to Bent's rule). And the axial bonds will consist of mainly $p_z $ and $d_{z^2}$ orbitals and little bit of $s$ mixing may also be there as it is spherically symmetric. So, overall central atom $\ce{Bi}$ has to use its $s$ electrons in bonding which is difficult energetically.

Now, the participation of $s$ electrons is difficult in bonding is due to the relativistic contraction of $s$ orbitals. It becomes closer to the nucleus and more stable in case of heavy elements like $\ce{Au, Hg, Tl, Pb, Bi}$ etc. Hence, the central atom needs to pay more energy to involve those stable electrons in bonding. That's why, the $s$ electrons become kind of inert and thus these heavy elements don't prefer higher oxidation states to avoid participation of these inner $s$ orbitals in bonding.


Is there a chemical reaction that speeds up when the temprature gets lower? We were talking about chemical reaction speed on our chemistry class, and I asked my teacher if there is a chemical reaction that speeds up when the temprature gets lower?
I think this is not everyday chemistry and my teacher didn't have an example for one.

That's a good question.

A reaction slowing down on a rise in temperature seems counter intuitive but it's not so always.

The Arrhenius equation works for only a few limited elementary reactions and the transition state theory is a much better explanation of rate laws and their variation with temperature.

The dependence on temperature of the reaction rate cannot always be analysed purely using the Arrhenius equation. It can be better understood by knowing the mechanism of the reaction.

For instance
$$\ce{2NO + O2-> 2NO2}$$

Has its first step as
$$\ce{2NO <=> N2O2}$$

And the second step as:
$$\ce{N2O2 + O2->2NO2}$$

The first step is exothermic in nature.
And the equilibrium shifts to the left on increasing temperature based on the Van't Hoff equation of gaseous equilibrium.

So the effective rate constant of the overall reaction decreases on increasing the temperature.

Thus, we see a reaction whose rate slows down with the rise in temperature.

There are in fact many such reactions.

To give another example, I could take an enzymatic reaction in a living cell and show that the rate of reaction falls down drastically with a rise in temperature. Ever wondered why your body rises its temperature during a fever? The aim there is to slow down the reactions in the cells of the invading germ and hence kill it.

From these examples, one could conclude that the famous Arrhenius equation works in only a few limited(and ideal) cases.


When to use 1/Ka vs Kb I'm really unsure what the difference is. We're using these when calculating $K$ of combined reactions. In general, I am also confused what $K$ is in comparison to $K_\mathrm{a}$ and $K_\mathrm{b}$. For example here is a calculation where for the reaction involving $\ce{NO2-}$, the equilibrium constant is $1/K_\mathrm{a}$ even though it shows $\ce{NO2-}$ acting as a base.

  Use the following four aqueous species to write an acid-base reaction with $K > 1$. (An acid-base reaction is one that involves the exchange of a proton. Your reaction will involve all four species.)

  $\ce{NO2-(aq)}$, $\ce{F-(aq)}$, $\ce{HF(aq)}$, $\ce{HNO2(aq)}$

  $\ce{HNO2}$ $(K_\mathrm{a} = \pu{4.6e-4})$ is a weaker acid than $\ce{HF}$ $(K_\mathrm{a} = \pu{6.6e-4})$, so $\ce{HNO2}$ holds onto its proton more tightly than
  $\ce{HF}$. So if we write a reaction when $\ce{NO2-}$ takes a proton from $\ce{F-}$, we'll have a reaction with $K > 1$:

  $$\ce{HF + NO2- → HNO2 + F-}$$

  We can get the $K$ for this reaction that adding the two acid dissocation reactions together as:

  $$
\begin{align}
\ce{HF &→ H+ + F-} &\quad &K_\mathrm{a}(\ce{HF})\\
\ce{H+ + NO2- &→ HNO2} &\quad &1/K_\mathrm{a}(\ce{HNO2})\\
\hline
\ce{HF + NO2- &→ HNO2 + F-} &\quad &\frac{K_\mathrm{a}(\ce{HF})}{K_\mathrm{a}(\ce{HNO2})} = \frac{\pu{6.6e-4}}{\pu{4.6e-4}} = 1.4 > 1
\end{align}
$$

$$
\begin{align}
\ce{HNO2 + H2O &<=> NO2- + H3O+} &\quad K &= K_\mathrm{a} \tag{1}\\
\ce{NO2- + H2O &<=> HNO2 + OH-} &\quad K &= K_\mathrm{b} \tag{2}\\
\ce{H2O + H2O &<=> H3O+ + OH-} &\quad K &= K_\mathrm{w} \tag{3}
\end{align}
$$

Above are the reactions associated with equilibrium constants commonly called $K_\mathrm{a}$, $K_\mathrm{b}$, and $K_\mathrm{w}$. If you reverse the first reaction, you get:

$$\ce{NO2- + H3O+ <=> HNO2 + H2O} \qquad K = 1/K_\mathrm{a}\tag{1a}$$
(This looks a bit different from your equation because it shows the hydronium ion instead of the hydrogen ion, but it refers to the same reaction.)

As you might know, the three equilibrium constants are related:

$$K_\mathrm{a} \times K_\mathrm{b} = K_\mathrm{w}$$

This is because when adding up the first and the second reaction, you get the third.

  When to use $1/K_\mathrm{a}$ vs $K_\mathrm{b}$

Use the equilibrium constant that matches the reaction you are working on.


Why is the electrode potential for a cathode positive? Electrode potential is the potential difference between the electrode and the surrounding electrolyte.
If the potential of the cathode is positive this means that the potential of the electrode is greater than the potential of the electrolyte, which means that the electrons will flow from the electrolyte to the cathode.
Shouldn't it be the other way around? shouldn't electrons flow from the cathode to electrolyte meaning the electrode potential should be negative?
Similarly, we can say that the electrode potential for an anode is positive.
Can someone point out the faults in this reasoning?

The question: Why is the electrode potential for a cathode positive?

Since question didn't directed to what kind of cell, the answer is it is not always positive:

As M. Farooq clarified, cathode is where reduction is happening in a cell. Thus, in Galvanic cell it is positive as the reduction reaction, $\ce{Y+ + e- -> Y}$, happens. The electrons needed for this purpose came from the anode. However, in an electrolytic cell, electrons from an outer source (e.g., a battery), flow to the negative end where the reduction reaction, $\ce{M+ + e- -> M}$, happens. Thus, the cathode is negative here.

Edit: I agree with  Poutnik's comment below. Accordingly, it may be better to say cathodes are more negative than anodes in electrolytic cells and more positive than anodes in galvanic cells. As more positive does not mean positive and more negative does not mean negative. The convention of marking positive and negative cell contacts is good enough for every day life, but it says nothing about the sign of electrode potentials either with respect to free electron or SHE (e.g., relevant half-cell reduction reaction's potential may carry positive or negative sign, based on the corresponding reaction).


How is a carboxylic acid acidic? Carboxyl groups are acids that comprise a carbon double bonded to an oxygen and single bonded to an oxygen-hydrogen. The one oxygen gets the two electrons it wants (so to speak) from the double bond with the carbon, the other gets the two it wants from the bond with the carbon and the hydrogen. The hydrogen is also satisfied by its bond with the oxygen. The only electron stealing opportunity I can see is in the polarity of the hydrogen. Is that what the acidic property of carboxy groups consists in?

There is a lot in chemistry that cannot simply be explained by Lewis structures, usually requiring a detailed molecular orbital description to rationalise observations.

For example considering a carboxyl group versus a simple hydroxy group. The carboxyl group has a $\mathrm{p}K_\mathrm{a}$ of approximately $5$ while the hydroxy group has a $\mathrm{p}K_\mathrm{a}$ of approximately $15$. In Lewis structures, these two look identical save the neighbouring carbonyl group.

Why can this hydrogen be abstracted easily at all? The hydrogen on a neighbouring carbon (e.g. $\ce{C\mathbf{H}3COOH}$) is not easily abstracted ($\mathrm{p}K_\mathrm{a} \approx 25$ if I recall correctly). The Lewis structure in itself provides no information. You need to additionally consider the concept of polar bonds to explain it: The $\ce{O-H}$ bond is polarised towards oxygen, that means the hydrogen already feels as if it were somewhat like a $\ce{H+}$ and therefore the transition to the true $\ce{H+}$ is easier.

While that explains the acidity per se, it doesn’t explain why one is more acidic than the other. The theory says, that an acid is stronger if its conjugate base is more stable. Since we cannot rationalise the different acidities from looking at the acids, let’s look at the conjugate bases. In one, we just generate an alcoholate much like a hydroxide ($\ce{R-O-}$). Nothing special-looking. The same for a carboxylic acid. No distinction. (However, this clearly shows why $\ce{C-H}$ bonds are usually not acidic: the carbanion is very unstable when compared to an oxyanion.)

Since we cannot find the origin of acidity, we need to expand our theory. Specifically, this is where resonance comes into play as a partial representation of some features of molecular orbital theory in Lewis structures. The carboxylate can resonate between two different structures as shown in the scheme below. Therefore, every oxygen atom only has to take the burden or half a negative charge ($\ce{O^{\frac{1}{2}-}}$). This is much more stable than having a full negative charge — even for an electronegative element such as oxygen. The third structure of the scheme simply displays a way of drawing ‘something in-between’: Both bonds are somewhere between a single and a double one and the charge is located on both atoms simultaneously (but that is not easily drawn with chemical drawing programs).


Why is [PdCl4]2- square planar whereas [NiCl4]2- is tetrahedral? The molecule $\ce{[PdCl4]^2-}$ is diamagnetic, which indicates a square planar geometry as all eight d electrons are paired in the lower-energy orbitals.

However, $\ce{[NiCl4]^2-}$ is also $\mathrm{d^8}$ but has two unpaired electrons, indicating a tetrahedral geometry.

Why is $\ce{[PdCl4]^2-}$ square planar if $\ce{Cl}$ is not a strong-field ligand?

The geometry of the complex changes going from $\ce{[NiCl4]^2-}$ to $\ce{[PdCl4]^2-}$. Clearly this cannot be due to any change in the ligand since it is the same in both cases. It is the other factor, the metal, that leads to the difference.

Consider the splitting of the $\mathrm{d}$ orbitals in a generic $\mathrm{d^8}$ complex. If it were to adopt a square planar geometry, the electrons will be stabilised (with respect to a tetrahedral complex) as they are placed in orbitals of lower energy. However, this comes at a cost: two of the electrons, which were originally unpaired, are now paired:

$\hspace{30 mm}$

(source: Wikipedia)

and on top of that, there are additional steric repulsions introduced by moving the ligands closer together. (This is, to some extent, a throwback to VSEPR theory, which simply says that the ligands would rather adopt a tetrahedral arrangement because of their mutual repulsion.)

We can label these two factors as $\Delta E$ (stabilisation derived from occupation of lower-energy orbitals) and $P + S$ (pairing energy + steric repulsions) respectively. One can see that:

If $\Delta E > P + S$, then the complex will be square planar;
If $\Delta E < P + S$, then the complex will be tetrahedral.

This is analogous to deciding whether an octahedral complex adopts a high- or low-spin configuration; where the crystal field splitting parameter $\Delta_\mathrm{O}$, also called $10~\mathrm{Dq}$ in older literature, plays the same role as $\Delta E$ does above. (However, unlike $\Delta_\mathrm{O}$, there is no simple graphical way to represent $\Delta E$ on a diagram since multiple orbitals are changed in energy.)

The two main factors that differentiate the Ni(II) complex from the Pd(II) complex are therefore:

The $\mathrm{4d}$ orbitals of $\ce{Pd^2+}$ are more radially diffuse (i.e. bigger) and therefore form stronger overlap with the orbitals on $\ce{Cl-}$, as compared to the $\mathrm{3d}$ orbitals of $\ce{Ni^2+}$. This leads to larger splitting of the $\mathrm{d}$ orbitals - or the relevant MOs (I'm too lazy to check their symmetry labels) - and hence a larger $\Delta E$.
Again because the $\mathrm{d}$ orbitals are more diffuse, the pairing energy $P$ is smaller in the palladium complex (it basically costs less energy to stuff them into the same orbital).

Together, these two factors ensure that practically all $\mathrm{4d}$ and $\mathrm{5d}$ $\mathrm{d^8}$ $\ce{ML4}$ complexes adopt a square planar geometry, even if the ligand is not a strong-field ligand. Other examples of such square planar complexes are $\ce{[PtCl4]^2-}$ and $\ce{[AuCl4]^-}$.

In fact, I am digressing here, but the same factors also cause the octahedral complexes to be almost invariably low-spin. For example, the $\mathrm{d^5}$ complex $\ce{[Fe(ox)3]^3-}$ has five unpaired electrons, indicating a high-spin configuration, $\mathrm{(t_{2g})^3(e_g)^2}$. However, $\ce{[Ru(ox)3]^3-}$ has one unpaired electron, which implies a low-spin configuration $\mathrm{(t_{2g})^5}$.


Cinnamaldehyde listed as (2E)-3-phenylprop-2-enal , but is the first 2 redundant? Wikipedia lists the molecule as being named "(2E)-3-phenylprop-2-enal", however it would appear that, as there is only 1 double bond which has a possibility of stereoisomerism, the 2 is redundant.

Would the correct IUPAC name not in fact be E-3-phenylprop-2-enal?

The preferred IUPAC name (PIN) for cinnamaldehyde is (2E)-3-phenylprop-2-enal.

With regard to numbering of locants that describe the position of stereogenic units, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-91.3 NAMING OF STEREOISOMERS

  (…)
  In preferred IUPAC names, stereodescriptors are placed immediately at the front of the part of the name to which they relate. They are placed at the front of the complete name when related to the parent structure; they are cited in parentheses followed by a hyphen. When they relate to substituent groups, they are cited at the front of the corresponding prefix. They are preceded by a numerical or letter locant to describe the position of the stereogenic unit when such locants are present; general rules of numbering are applied (see P-14.4).
  (…)

Accordingly, all stereodiscriptors have to be preceded by a locant.
In case of the PIN for cinnamaldehyde, the stereodiscriptor ‘E’ is preceded by the numerical locant ‘2’.

In general nomenclature, however, locants are usually omitted when there is no ambiguity.

Nevertheless, the current IUPAC recommendations are prescriptive about when omission of locants is permissible in PINs. Locants are omitted in PINs only in the exceptional cases that are explicitly mentioned in the IUPAC recommendations. None of these cases applies to the locant of the stereodiscriptor in (2E)-3-phenylprop-2-enal.

As a rule of thumb, if any locants are essential for defining the structure, then all locants must be cited in the PIN.

Note that the current recommendations given in the 2013 Blue Book incorporate a change from earlier recommendations given in the 1993 Guide. By way of comparison, the following examples are explicitly mentioned in the 1993 Recommendations as well as in the 2013 Recommendations:

(2E)-but-2-ene (PIN, 2013 Recommendations)
(E)-but-2-ene (1993 Recommendations)

(1Z)-1,2-dibromo-1-chloro-2-iodoethene (PIN, 2013 Recommendations)
(Z)-1,2-dibromo-1-chloro-2-iodoethene (1993 Recommendations)

(2Z)-2-methylbut-2-enoic acid (PIN, 2013 Recommendations)
(Z)-2-methylbut-2-enoic acid (1993 Recommendations)

(2E)-but-2-enedioic acid (PIN, 2013 Recommendations)
(E)-but-2-enedioic acid (1993 Recommendations)


Thermodynamics and Coffee Heyo, I'm sure you all know of the experiment where you take a mixture of ice and water, plop some salt in there and the temperature of the solution progressively drops. The salt increases the entropy of the liquid state, shifting somewhat the ice/water equilibrium towards water. As ice is converted to liquid water, the endothermicity of this process leads to a decrease in the temperature of the solution.

Now, I was wondering if the reverse would ever be possible, say you were to have a thermos flask which would be sealed. It would contain Hot Tea and have a slot where one could pour in salt grains without releasing steam. By the logic in the previous question, won't some of the water vapor move to the liquid state (exothermic), resulting in the heatin up of your already hot tea. If possible would this change even be measurable or useful in any sort of way?

You're exactly right: the phenomenon is called boiling-point elevation:

  Boiling-point elevation describes the phenomenon that the boiling point of a liquid (a solvent) will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent. [...] It is an effect of the dilution of the solvent in the presence of a solute. It is a phenomenon that happens for all solutes in all solutions, even in ideal solutions, and does not depend on any specific solute-solvent interactions. The boiling point elevation happens both when the solute is an electrolyte, such as various salts, and a nonelectrolyte. In thermodynamic terms, the origin of the boiling point elevation is entropy and can be explained in terms of the vapor pressure or chemical potential of the solvent. In both cases, the explanation depends on the fact that many solutes are only present in the liquid phase and do not enter into the gas phase (except at extremely high temperatures).

The empirical method for computing boiling point elevation for a given solute is described here. In short:

  $\Delta T_b = \sum_i{n_iK_{b,i}m_i}$

Here, $\Delta T_b$ is the observed increase in boiling point, and $K_{b,i}$ and $m_i$ are the boiling point elevation constant and the molality (moles of solute per kilogram of solvent), respectively, of solute $i$.  The $n_i$ factor is the number of ions into which solute $i$ dissociates upon dissolution (e.g., $n_{sucrose} = 1$ and $n_{\ce{NaCl}} = 2$).

Your best bet for a full thermodynamic description is a college or graduate level textbook on the topic. I don't know offhand of any that specifically address boiling-point elevation, though.  The Wikipedia page on colligative properties actually has a bit of thermodynamic detail, providing the expression for computing the boiling point elevation constant from the enthalpy of vaporization $(\Delta H_{\mathrm{vap}})$ and other physical parameters:

  $K_{b,i} = \frac{RM_iT_b^2}{\Delta H_{\mathrm{vap}}}$

where $R$, $M$, and $T_b$ are the ideal gas constant, the formula weight of solute $i$, and the boiling point of pure solvent, respectively.


Difference between Coupled Cluster and Full CI I'm looking at some high precision quantum chemistry methods like Coupled Cluster (CC) and FULL CI (FCI).

It seems both CC and FCI start from Hartree-Fock and then excite the electron to higher states.

If I understand correctly, CC uses an exponential operator to raise the states of the electron while FCI explicitly calculates the coefficients of each determinants.

What makes the difference between CC and FCI? Can someone give an intuitive explanation? Thanks.

Full Coupled Cluster (FCC) vs. Full Configuration Interaction (FCI)

The main theoretical difference is the way excitations are used. "Excitation" refers to putting one or more electrons in higher orbitals than the "reference" calculation, which is often a Hartree-Fock calculation, would. CI uses a "simple", linear excitation operator, whereas CC uses a more complicated, exponential one. One can think of CI as a sum of excitations while CC results in a product of excitations.
Symbolic equations for the wave function ($\Psi$) ansatz of either method read:
$$
\Psi^{\text{CI}} = (1 +C_1+C_2+C_3+...) \Psi^{\text{HF}} \\
\Psi^{\text{CC}} = \exp(T_1+T_2+T_3+...) \Psi^{\text{HF}} \\
= (1 +T_1+T_2+T_3+\frac{1}{2}T_1^2+\frac{1}{2}T_2^2+T_2T_1+\frac{1}{2}T_3^2+...+\frac{1}{6}T_1^3+...) \Psi^{\text{HF}}
$$
In CI, the excitation operator is often denoted $C$ and the prefactor of determinants are called CI coefficients, whereas CC uses $T$ and "amplitudes". In either, the subscript refers to the number of excited electrons.
For "full" calculations, that is, calculations that take into account every possible excitation, the results of CC and CI will be identical.

Full calculations are too expensive for all but the smallest systems, because the cost rises as $O(N!)$, where $N$ is some measure of the system size involving electron number and basis set size. For this reason, one typically decides to "truncate" the excitations to lower the cost.

Single-reference Truncated CC vs. Single-reference Truncated CI (this is what CC and CI commonly refer to)

Truncation first involves collecting the excitations into "classes" of excitations. Within a class, one gets similar cost and hopes for similar contributions. Thus, one may sort of gauge the cost and accuracy of the calculation. Standard ways of doing this have evolved, and this is what CCSD, CCSD(T), CISD etc. refer to. CASSCF/RASSCF can be categorized like this, as well.

With truncation, another difference between CI and CC arises: CC is no longer variational and CI is no longer size-extensize. Variationality is the property that the resulting energy has the true energy (of the given basis set) as a lower bound - one approaches the energy monotonically from above as more excitations become involved. Size-extensivity is the property that two identical systems at infinite separation from each other yield twice the energy of a single system. The latter property is thought of as the more important one, mainly because of its implications for closer, interacting systems. Variationality is also less of a concern because finite basis sets must be used, thus, one is typically far above the true energy anyway.

Another technical difference is that truncated CC and CI lead to different equations, for which different methods of solving have been developed. Overall, CC implementation is, or at least used to be, more difficult. Properties other than the total energy can also be more difficult to implement.

Scaling of CC and CI

Basically, when truncation is performed at similar levels, the two methods scale the same, but the prefactor may be different - this also depends on the quality of the implementation. For reasons stated above, CC has been worked on more intensively since, say, the mid 1990s. So, CISD and CCSD scale similarly, CISDT and CCSDT would as well. Bottom line: CCSD(T) will still be affordable for molecules where full CI would require specialized implementation on a dedicated supercomputer.


Why does pOH increase when pH decreases? Let's say you add $\ce{HCl}$ to water. The $\ce{H+}$ ion concentration increases and that causes a decrease in $\mathrm{pH}$. But why would the $\mathrm{pOH}$ increase?

I can't see why added $\ce{H+}$ ions will decrease the $\ce{OH-}$ concentration? Is this because of the water auto-ionization? Is it because, according to Le Chatelier principle, when you add an amount of a component, the reaction will go in the other direction to counteract the disturbance, thus decreasing the concentration of $\ce{OH-}$ to form water?

At a particular temperature, the $K_{\text{eq}}$ for the following reaction (yes, it's the auto-protolysis of water):

$$\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$

will be constant. Note that $K_\mathrm{w}=K_{\text{eq}}\times[\ce{H2O}]=\ce{[H+(aq)][OH-(aq)]}$ is called the auto-protolysis constant of water. It is considered a constant because concentration of water $([\ce{H2O}])$ is assumed to be constant.

So, you can directly observe from here, that since $\ce{[H+(aq)][OH-(aq)]}=K_\mathrm{w}=\text{constant}$ at a particular temperature, if $\ce{[H+(aq)]}$ increases (i.e. $\mathrm{pH}$ decreases), $[\ce{OH-(aq)}]$ must decrease (i.e. $\mathrm{pOH}$ must increase).

Your interpretation of the Le Chatelier's principle is also correct. In fact, it is actually what is happening behind the scenes. When you add more acid (assuming a 100% ionized acid), the $Q$ value of the auto-protolysis of water increases, causing the reaction to shift backward. However, the backward shift is unable to decrease the concentration of $\ce{H+}$ ions as much as it decreases the concentration of $\ce{OH-}$ ions. See explanation below:

Consider adding $\pu{0.01M}$ $\ce{HCl}$ to pure water (at $\pu{25^\circ C}$) at $t=0$. This is what happens till achieving equilibrium at $t=t_0$:

$$
\begin{array}{ccc}
&\ce{H2O(l) &<=>& H+(aq)&+&OH-(aq)}\\\hline
-&-&&\pu{10^-7M}&&\pu{10^-7M}\\
t=0&-&&\pu{(10^-7+10^-2)M}&&\pu{10^-7M}\\
t=t_0&-&&(10^{-7}+10^{-2}-x)~\pu{M}&&(10^{-7}-x)~\pu{M}\\
\end{array}
$$

Now, if you actually solve for $x$ at $t=t_0$, you'll find $x$ to be of the order $10^{-8}$ or below. Notice that, for the $\ce{H+}$ ions, this means a decreases in concentration from $\approx\pu{0.01M}$ at $t=0$ to $\pu{0.0099999M}$ at $t=t_0$. This a hardly noticeable change. However, it is a catastrophe for the $\ce{OH-}$ ions, whose concentration would probably be reduced, by a factor of more than ten thousand, from the initial $\pu{10^-7 M}$! The change in $\mathrm{pOH}$ is, hence, much more pronounced.


Why did my beer slowly froth up after removing from freezer? I got a beer out of the freezer last night and opened it, only to realize that it was almost frozen, so I left it to warm up. 10mins later I hear a dripping noise and realize that the beer had overflowed and was starting to spill off the table.

What causes this slow expansion of beer after it is warmed from a near 0°C temperature?

Melting takes time.
As your beer is thawing, temperature remains roughly constant. Then the beverage will slowly assume ambient temperature.
Solubility of gases (e.g., $\ce{CO2}$) in common solvents usually decreases with warming. This is because the breaking of solvent$-$solvent interactions are not compensated by new interactions between solute and solvent. In other words, solubility of gases is often exothermic.

The bottom of the beer might have warmed more quickly, causing a buildup of carbon dioxide under a layer of still-frozen beer. Reaching some critical value for pressure, the top popped. According to Henry's law, immediately after even more gas bubbles are formed.

Effects due to positive thermal expansion of beer are probably rather small but these would only aggravate any overflow.

For further reading on similar subjects, see

Does shaking increase the pressure inside a bottle of champagne?,
Plateau–Rayleigh instability.


Can the eigenvalue for a quantum mechanical operator be zero? Can the eigenvalue for a quantum mechanical operator be zero?

$S_z$ (spin along the z-axis) is a QM operator, and it frequently is zero - such as in closed-shell molecules.

Also note that the zero energy point of quantum mechanics can be defined in more than one way: some quantum chemistry programs use "all electrons and nuclei at infinite separation", whereas some solid state programs used "neutral atoms at infinite separation". Nevertheless, they evaluate the total energy.


What is happening at the various boundaries of a phase diagram? Hoping to clarify a few questions I have that the textbook doesn't address.
Lets just say I have a piston (at equilibrium) with 100 moles of gaseous compound that looks like this:

And here is the phase diagram for this compound:
Now lets say I start increasing the pressure on the piston so that the gas starts to approach the gas-liquid boundary.
1.) What would start to happen as you approach the boundary? For example: Would a few moles of gas begin to condense to liquid? Is this process exponential as you approach boundary?
2.) What is happening at the boundary? For example: Are the gas and liquid in perfect equilibrium (50 moles of each)? By that logic, at the triple point, are there 33 moles of each?
3.) Now let's say we approach the critical point along the liquid-vapor boundary. I read online that its when the gas and liquid become "indistinguishable." If so, that means it has a uniform density. Does that make it another state of matter? If not, why is it special?

I got that you are trying to visualize what is actually happening. I have not done the compression myself, but this is what I think will be observed. It will be good if someone with real experience can confirm the answer.

At the beginning you are in the region marked 'vapour'. When you are approaching the liquid-vapour boundary from below, before you touch the boundary, you will still observe only vapour, no liquid at all. You are still in the region marked 'vapour'. Nothing will happen.

When you just reached the boundary, you will see some liquid droplets formed. It should look just like some dew formed at the wall. Regarding the relative amount of liquid and vapour at equilibrium, the pressure of vapour present must be equal to pressure exerted by you (or else the piston will move!) and equal to pressure on the boundary (determined by temperature). Say if you try to push the piston further inside, the volume occupied by sample will decrease, and more liquid formed, but you will still stay on the boundary. You will notice that the pressure needed to push piston inside further does not change, because the pressure of vapour does not change (push in, vapour pressure should increase, but some vapour condensed, lower back the pressure). Only when all vapour has been converted into liquid, you can depart from the boundary, move into the region marked 'liquid'. However, before all vapour turned into liquid, if you stop pushing in and hold your piston still, the liquid and vapour amount won't change. So there is no unique relative amount of vapour and liquid at equilibrium as written in your question. Above liquid-vapour boundary you have only liquid and the piston surface is resting on the liquid surface.

That state is known as supercritical fluid.


How do you determine the hybridisation state of a coordinate complex? If an exercise asks for the hybridisation state of a coordinate complex (ion in solution) consisting of a central cobalt atom surrounded by $\ce{NH_3}$ ligands, and neither coordination number or complex ion is given, how do you decide which configuration will occur?
The answer is that in a solution this will form a $\ce{[Co(NH_3)_6]^{3+}}$ complex ion. I know that this means the hybridisation is $\mathrm{d^2sp^3}$ but can't figure out how to get there since cobalt is found in different oxidation states.

Again, I feel a bit like a broken record. You should not use hybridisation to describe transition metal complexes.

Instead, you should discuss transition metal complexes using molecular orbital theory; a simplified version of it such as the crystal field theory may also suffice for simple complexes. Below, I have attached the orbital scheme of an octahedral coordination complex of a random transition metal not including π effects.

Figure 1: molecular orbital scheme of an octahedral $\ce{[ML6]}$ complex. Originally taken from Professor Klüfers’ coordination chemistry course’s web scriptum (in German).

Forcing hybridisation onto this system is very bad and very wrong; look at the huge energy differences between the 3d, 4s and 4p orbitals. Hybridisation would also lead us to assume six equal bonds, which is not the case: the six $\ce{M\bond{->}L}$ bonds transform as $\mathrm{a_{1g} + e_g + t_{1u}}$ which leads to three different energies. And finally, hybridisation is not able to explain crystal field split, and why $\mathrm{e_g}$ — the orbitals that should take part in hybridisation and thus form bonds — are raised in energy (they are $\mathrm{e_g^*}$ in MO terms).

In most cases, coordination complexes would default to octahedral or derived. The main catch case is low-spin $\mathrm{d^8}$ in which case square-planar should be assumed. This is relatively independent of the type and number of ligands and most exceptions are due to sterics. Thus, octahedral complex is a safe first guess. Note that the only difference between two complexes of the same metal and ligands but with different metal oxidation states is typically whether a Jahn-Teller distortion is relevant and how strong it is.


Causing a metal or glass marble to float in an aqueous solution I am trying to find a way that you could cause metal (Aluminum) or glass to float in a mixture of water and some water-soluble material. I am aware that salt can be used to heighten the density of water, but I doubt it could go to that extent because both aluminum and glass are more dense than salt (possibly flawed logic, I have limited knowledge of the subject). So basically: if there is any way to cause Aluminum or glass to float in water, possibly through adding a water soluble material to the water, please let me know.

Edit: Basically what I'm doing is trying to think of creative ways to sort materials by density, for example a set of marbles. With the example of solutions with salt, you would vary the amount of salt in the mixture to alter what materials would float.

The best current non-toxic alternative to Clerici Solution appears to be Sodium Polytungstate (SPT).  It also has a neutral (~7) pH, which is important when working with aluminum. By dilution/evaporation, density covers the range $1.1\!\!-\!\!3.1$g/cm$^2$, but above $2.5$g/cm$^2$ it can develop a harmless reversible blue tint when exposed to reduced metals. Aluminum has an oxide patina, so I am not sure if it qualifies as a reduced metal in this setting. SPT is rather expensive though.
Aluminum is $2.7$ g/cm$^2$ and Common glass is $2.52$ g/cm$^2$. So both of those materials, many gemstones and some minerals, will float in SPT.


MO diagram of Zundel cation Here is the structure of the Zundel cation (courtesy Wikipedia):

The outermost shell electron count is $16$.

May I have the MO diagram of this cation?

This seems very complicated so I don't know where to start.

Thanks.

Strictly speaking, drawing a MO diagram is only possible for two entities.
That being said, I will attempt to show you the MOs and order them according to their energy.

At PBE0-D3/def2-TZVPD, using NWChem 6.6, I found the following minimum geometry (vacuum):

Figure 1: Minimum geometry found for Zundel cation ($\ce{[H2O...H...OH2]^{+1}}$) with natural atomic charges, angles (in degrees) and distances (in angstrom)

MOs 1 and 2 correspond to $\ce{1s^2}$ core electrons on oxygen and orbitals 3-10 give us the states for the 8 remaining valence electron pairs.
MO 11 is the lowest unoccupied molecular orbital.

Figure 2: Molecular orbital diagram for structure from Figure 1 at PBE0-D3/def2-TZVPD

Since the structure has no symmetry, no symmetry was assigned to any MO and the diagram above has simple labels only.
Energies (in hartree) can be found further below.

Below are images of each MO:

Figure 3: MO #3 (cutoff 0.04)

Figure 4: MO #4 (cutoff 0.04)

Figure 5: MO #5 (cutoff 0.04)

Figure 6: MO #6 (cutoff 0.04)

Figure 7: MO #7 (cutoff 0.04)

Figure 8: MO #8 (cutoff 0.04)

Figure 9: MO #9 (cutoff 0.04)

Figure 10: MO #10 (cutoff 0.04)

Figure 11: MO #11 (cutoff 0.4)

As a side note, we may obtain a finer view of the MOs by representing them as linear combinations of natural bonding orbitals (NBOs):

CMO: NBO Analysis of Canonical Molecular Orbitals

Leading (> 5%) NBO Contributions to Molecular Orbitals
======================================================
 MO   3 (occ): orbital energy =  -1.356442 a.u.
               0.426*[  5]: BD ( 1) O 4- H 7
               0.400*[  1]: BD ( 1) O 1- H 2
              -0.399*[  2]: BD ( 1) O 1- H 3
               0.396*[  4]: BD ( 1) O 4- H 6
               0.396*[  3]: BD ( 1) O 4- H 5
               0.336*[  9]: LP ( 2) O 1(lp)
 MO   4 (occ): orbital energy =  -1.327547 a.u.
               0.444*[  1]: BD ( 1) O 1- H 2
              -0.440*[  2]: BD ( 1) O 1- H 3
              -0.437*[  3]: BD ( 1) O 4- H 5
              -0.434*[  4]: BD ( 1) O 4- H 6
               0.280*[  9]: LP ( 2) O 1(lp)
              -0.272*[  5]: BD ( 1) O 4- H 7
 MO   5 (occ): orbital energy =  -0.859519 a.u.
              -0.606*[  5]: BD ( 1) O 4- H 7
              -0.502*[  9]: LP ( 2) O 1(lp)
               0.369*[  3]: BD ( 1) O 4- H 5
               0.359*[  1]: BD ( 1) O 1- H 2
 MO   6 (occ): orbital energy =  -0.839700 a.u.
              -0.562*[  4]: BD ( 1) O 4- H 6
               0.511*[  2]: BD ( 1) O 1- H 3
               0.471*[  3]: BD ( 1) O 4- H 5
               0.425*[  1]: BD ( 1) O 1- H 2
 MO   7 (occ): orbital energy =  -0.838909 a.u.
               0.533*[  2]: BD ( 1) O 1- H 3
               0.514*[  1]: BD ( 1) O 1- H 2
               0.481*[  4]: BD ( 1) O 4- H 6
              -0.469*[  3]: BD ( 1) O 4- H 5
 MO   8 (occ): orbital energy =  -0.696009 a.u.
              -0.549*[  9]: LP ( 2) O 1(lp)
               0.502*[  5]: BD ( 1) O 4- H 7
               0.302*[ 10]: LP ( 1) O 4(lp)
              -0.301*[  8]: LP ( 1) O 1(lp)
              -0.253*[  3]: BD ( 1) O 4- H 5
               0.244*[  1]: BD ( 1) O 1- H 2
              -0.227*[  4]: BD ( 1) O 4- H 6
 MO   9 (occ): orbital energy =  -0.636884 a.u.
               0.692*[ 10]: LP ( 1) O 4(lp)
               0.687*[  8]: LP ( 1) O 1(lp)
 MO  10 (occ): orbital energy =  -0.623778 a.u.
              -0.620*[  8]: LP ( 1) O 1(lp)
               0.614*[ 10]: LP ( 1) O 4(lp)
               0.344*[  9]: LP ( 2) O 1(lp)
              -0.319*[  5]: BD ( 1) O 4- H 7
 MO  11 (vir): orbital energy =  -0.201773 a.u.
               0.263*[121]: BD*( 1) O 1- H 2*
               0.262*[123]: BD*( 1) O 4- H 5*
              -0.257*[122]: BD*( 1) O 1- H 3*
               0.257*[124]: BD*( 1) O 4- H 6*
              -0.232*[113]: RY*( 1) H 7(ry*)

Figure 12: Atomic numbering for the NBO Analysis of Canonical Molecular Orbitals above


Formation enthalpy vs. bond enthalpy I have been working with thermochemistry at school for a while. I have been introduced to the two terms: formation enthalpy and bond enthalpy. However, I have not quite understood the difference between the two and would be very glad if someone could explain it!

OK, here is the difference: formation enthalpy refers to the formation of a compound from elementary compounds, while bond energy is applicable when considering the formation of a molecule from isolated atoms. The relation between the two is pretty much explained by the Hess's law: the enthalpy of a reaction does not depend on its pathway, but only on the initial and final states.

Now consider the following two pathways for your example.

Pathway 1: we burn graphite in oxygen, just like $\ce{C + O2 -> CO2}$. There is some change of enthalpy associated with this reaction.

Pathway 2: we vaporize graphite to get atomic carbon (see how many C-C bonds are broken, take that into account), atomize $\ce{O2}$ to get elementary oxygen (again, some bonds are broken), and finally produce $\ce{CO2}$ (this time some bonds are formed). Now if you'll look up the energies of all bonds involved and carefully add/subtract them all together, you'll have a decent approximation of the reaction enthalpy obtained in the pathway 1.

(Normally it works the other way around, that is, we use reaction enthalpies to calculate bond energies, but... whatever.)


Boric Acid and Brass Alloy Can someone explain why boric acid prevents zinc from vapourizing out of the brass alloy when it is being melted?  What is the actual chemistry occurring and why?
Thanks

It depends on the temperature, specifically the 1700 F +/- boiling point of zinc , not the flux. At 1700 F the zinc will vaporize regardless of any flux . I never saw a single component flux, likely the the flux also contains soda ash, silica, etc. and it does protect the molten metal from atmosphere by forming a physical layer. Footnote: many years age I worked with copper alloy fluxes and found that for some alloys, the boron promoted the formation of spinels in the metal ; These are exceptionally hard materials that essentially prevent machining of the bronze/ brass.


Nitration of pyrrole with sulfuric and nitric acids Would 2,5-dinitropyrrole be a product of nitration of pyrrole with $\ce{H2SO4/HNO3}$? It seems that 2,5-dinitropyrrole is not formed when using $\ce{Ac2O/HNO3}$ (ref 1). Is this just because of the mild conditions used, or does this reflect the pyrrole ring being more deactivated by the first nitration than a benzene ring would be?

1 Cooksey, A. R.; Morgan, K. J.; Morrey, D. P. Nitropyrroles—II : The nitration of pyrrole. Tetrahedron 1970, 26 (21), 5101–5111. DOI:10.1016/S0040-4020(01)93163-1

Would 2,5-dinitropyrrole be a product of nitration of pyrrole with $\ce{H2SO4/HNO3}$?

I doubt it and would rather expect a lot of tar to be formed here. Pyrrole tends to undergo acid-catalyzed polymerization!

For decades, acetyl nitrate, formed by addition of fuming nitric acid to acetic anhydride, is the reagent of choice.

Mononitration of pyrrole predominantly gives 2-nitropyrrole, accompanied by minor amounts of 3-nitropyrrole. (K.J. Morgan and D.P. Morrey, Tetrahedron, 1966, 22, 57–62. DOI).

In the case of 1-methylpyrrole, the yield of the 3-nitropyrole is somewhat larger (H.J. Anderson, Canad. J. Chem, 1957, 35, 23–29). In you want even more of the 3-nitro isomer, a more bulky alkyl substituent (t-Bu) or N-silylation with $\ce{((CH3)CH)2SiCl}$ is a viable approach.

Nitration of 2-nitropyrrole at −15 °C with nitric acid in acetic anhydride  furnishes a 4:1 mixture of 2,4- and 2,5-dinitropyrrole with a total yield of 61%. (R. Alan Jones, Gerritt P. Bean, The Chemistry of Pyrroles: Organic Chemistry: A Series of Monographs, Vol. 34, Academic Press, p. 124)


Is it possible to predict the lines in the atomic emission spectrum of Na? The atomic emission spectrum for sodium ($\ce{Na}$) is completely dominated by a line in the range of yellow, about $590~\mathrm{nm}$ (to be more precise, it's a doublet). Here is how it looks like:

This line is due to the transition $3\mathrm{p} \to 3\mathrm{s}$.
I know the Rydberg equation which can be used to predict the transitions in hydrogen. I would like to know if there is a way to predict the transitions for other elements, such as the alkali metal $\ce{Na}$.

The origin of the Rydberg equation is the Bohr model's equation for the energy of a hydrogenic state:
$E_\text{B} = -\dfrac{R_{H}hc}{n^2}$

This is accurate for other hydrogenic atoms, with only one electron too (bearing in mind that $R_{X}$ is proportional to the atomic number squared).

When you get to non-hydrogenic atoms you need to model inter-electron repulsion as well. This can be done by introducing an empirical parameter $\delta$, the quantum defect: $E_\text{B} = -\dfrac{R_{X}hc}{(n-\delta_X)^2}$. For sodium the quantum defect is 1.37, allowing good approximation of the transition energies between n-states in the electronic spectrum.

I don't think there is a similar back-of-the-envelope method to calculate the differences in energies between multi-electron atom l-energy levels (like the 3s-3p). The theory of angular momentum tends to mathematical complexity. There are likely good computationally derived approximations though.


Why caffeine is easily oxidized to uric acid? I want to understand how caffeine (1,3,7-trimethylxanthine) can be easily oxidized to uric acid.

Is it safe to conclude that caffeine may increase the uric acid level in blood?

I doubt caffeine can be so easily oxidized to uric acid. I'd propose a way by demethylation via cytochrome P450 to the corresponding xanthines and oxidation via xanthine oxidase (but these are enzymatic reactions).

From a synthetic perspective, one will start with guanine (-> guanine deaminase followed by oxidation; or oxidative deamination of Guanine followed by another oxidation)

It's not safe to conclude that caffeine increase uric acid blood level.

Long story short:
Coffee consumption seems to lower uric acid blood levels.
Decaf coffee consumption also seems to lower uric acid blood levels.
Tea consumption seems to have no effect on uric acid blood levels.

  Serum uric acid level tended to decrease with increasing coffee intake
  (Figure 1). After adjusting for age and sex, serum uric acid level in
  individuals with coffee intake 6 cups daily was lower than in those
  with no use by
  0.43 mg/dl (95% CI 0.22, 0.65; P for trend  0.001). After adjusting for other covariates, the differences were attenuated but remained
  significant (P for trend  0.001) (Table 2).

"Coffee, Tea, and Caffeine Consumption and Serum Uric Acid Level: The Third National Health and Nutrition Examination Survey", Choi HK, Curhan G., 2007
DOI 10.1002/art.22762

It has to be noted this was the first study I found on the fly and there may be other results out there


Do the groups involved in geometrical isomerism have to be in similar planes to show geometrical isomerism? In my textbook, under the topic "Conditions for geometrical isomerism", the following has been given:

  Groups responsible to show geometrical isomerism must be nearly in the same plane.

Is this statement really true? Because, we are well aware that disubstituted cycloalkanes which show geometrical isomerism are not planar. Eg: 1,2-Dimethylcyclopropane

Kindly explain with examples.

To have geometrical isomers you need a structurally rigid bond. This bond can be double and this is a case where groups are in the same plane:

Or can be a cyclic compound where the  carbon-carbon single bond is restricted(2 examples below):

1st example:

2nd example:

Your textbook definition is incomplete and I wouldn't use the word "must" there.


What temperature is needed for carbon monoxide to reduce iron oxide? What temperature is needed for carbon monoxide to reduce iron oxide?
That is at what temperature the following reaction will take place:
Fe2O3 + CO -> Fe + CO2

Based on an experiment, carbon monoxide dose reduce iron oxide at 1200°C (2192°F).

But probably, this is not the minimal necessary temperature.
A better answer is still welcome.


Are ortho, meta and para used in IUPAC nomenclature? I was presented with the following question in an online test:

To avoid any confusion, here are the arranged options:

A. 2-Chlorotoluene
B. 1-Chloro-2-methylbenzene
C. B and D
D. O-Chlorotoluene
Correct answer: (C) B and D.
Explanation: Chloro is given priority so numbered as 1

My question is: Are the ortho, meta and para prefixes part of the official IUPAC nomenclature or not?
I chose B, what should be the correct answer to this question?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book),

The locants o, m, p are no longer recommended; the numerical locants '1,2-', '1,3-', and '1,4-' must be used in substitutive names. However, as an exception, the three isomers of xylene are still recognized as o-, m-, and p-xylene in general IUPAC nomenclature (see P-22.1.3).

and

The letters o, m, and p have been used in place of ortho, meta, and para, respectively, to designate the 1,2-, 1,3-, and 1,4- isomers of disubstituted benzene. This usage is strongly discouraged and is not used in preferred IUPAC names.

The preferred IUPAC name for the compound that is given in the question is 1-chloro-2-methylbenzene.

The name 2-chlorotoluene may be used in general nomenclature.


Subtracting standard electrode potentials As far as I'm aware, if we construct a half cell ($\ce{A + e- <=> A-}$) under standard conditions then the standard electrode potential $E^⦵$ is the potential of the $A/A^{+}$ couple wrt. SHE, which doesn't change depending on whether the reaction is an oxidation or reduction in this half cell. This is because potential is a physical, electrostatic concept.

Consequently, the cell EMF is then also defined to be the potential of the cathode subtract that of the anode, $E^⦵_{cell} = E^⦵_{cat} - E^⦵_{an}$. All of the above I have obtained from IUPAC recommendations.

However, some people say that reversing a half equation reverses the sign of $E^⦵$, and then proceed to add the electrode potentials to find the cell EMF. Although this method is algebraically equivalent, it doesn't make any sense - since the electrode potential has nothing to do with the chemical concept of a reaction direction!

The only justification I have seen is that if you write $\Delta G = -nFE^⦵$ for each half equation, and sum them, then the EMF comes out as expected only if we do change the sign. However this seems inherently flawed as well, since although Gibbs free energies are additive, $\Delta G = -nFE^⦵$ only applies to a cell as a whole (because it is derived by considering the energy change if charge flows from one terminal to the other) and is meaningless in the context of a half equation.

Consequently, why is the notion that reversing the half cell equation negates the electrode potential still taught - since to me at least it seems fundamentally incorrect!

Thanks a bunch!

Let's take a look at what you're doing specifically for computing the potential difference. The potentials at the cathode and the anode are all relative to some standard. Importantly, it must be the same standard. Let's refer to this as ground.

$$E_{\mathrm{cathode}} = V_{\mathrm{cathode}} - V_{\mathrm{ground}}$$
$$E_{\mathrm{anode}} = V_{\mathrm{anode}} - V_{\mathrm{ground}}$$

The $V$'s here represent absolute potentials on some arbitrary scale.

What's the change in potential when I move a test charge from the anode to the cathode? Since potential differences are path independent, we can look at this via states: moving the charge from anode to cathode is the same as moving the charge from anode to ground and then from ground to cathode.

The potential difference is expressed as:

$$E = (V_{\mathrm{cathode}} - V_{\mathrm{ground}}) - (V_{\mathrm{anode}} - V_{\mathrm{ground}}) = E_{\mathrm{cathode}} - E_{\mathrm{anode}}$$

The key is the cancellation of the ground potential, so that we have a potential difference.

This is exactly what are doing when working with half-reactions. And it works because both the reduction and oxidation half reactions are referenced to the standard hydrogen electrode.

Let's look at two half-reactions:
$$\ce{A -> A+ + e-}\ ,\ \ E^{⦵} = -E_{A}$$
$$\ce{B+ + e- -> B}\ ,\ \ E^{⦵} = E_{B}$$

As a general rule, we tabulate reduction potentials, so let's treat the $\ce{A}$ reaction with a reduction potential:

$$\ce{A+ + e- -> A}\ ,\ \ E^{⦵} = E_{A}$$

Now, both of these half reactions are relative to the standard hydrogen electrode.

In the reaction
$$\ce{A + B+ -> A+ + B}$$
we can look at the equivalent process of moving an electron from $\ce{A}$ to a proton in a standard hydrogen electrode, and then taking a electron from dihydrogen in the standard hydrogen electrode and moving it to $\ce{B}$.

The potential difference is:
$$E^{⦵}_{\mathrm{cell}} = E_{B} + (-E_{A}) = E_{B}-E_{A}$$

This is exactly the same process because the anode potential relative to ground is the reduction potential of the oxidation half-reaction. The potential difference is computed by subtracting the potential from the other one.

Alternatively, if you already expressed the anode potential as a oxidation potential (inverse of the reduction potential), then the minus sign is implicitly included, and you can just add the two potentials.


Production of Succinic Acid from Crude Glycerol using Fermentation So I am writing a short paper on how one can produce Succinic acid from glycerol. I have to find a microorganism which can do that. I have found the following microorganism, which are traditionally being used for fermentation: Anaerobiospirillum succiniciproducens, Actinobacillus succinogenes, Mannheimia succiniciproducens and recombinant Escherichia coli strains. But the problem is that whatever article I find, they are written with respect to Glucose.

Does anyone know any article/paper/hints or some information they can share?

I just need to understand which organism can be used to do the fermentation, and how much succinic acid can be made from it.

Glycerol has 3 carbon atoms and succinic acid has 4. So you need (apart from some redox reactions) a way to add another carbon atom.

Also, succinic acid appears in the citric acid cycle. What you have to research are anaplerotic reactions. When you remove any of the metabolites (like succinic acid) from the cycle, you have to replenish them. One way is to turn pyruvate (3 carbons) into oxaloacetate (4 carbons), as shown in the wikipedia article on anaplerotic reactions:

$$\ce{pyruvate + HCO3− + ATP -> oxaloacetate + ADP + P_i + H2O}$$

Then, you have to find a pathway from glycerol to pyruvate. Here is a schematic of glycerol metabolism:

Source: https://www.researchgate.net/publication/224950286_Effect_of_impurities_in_biodiesel-derived_waste_glycerol_on_the_performance_and_feasibility_of_biotechnological_processes/figures?lo=1

As you can see, it is possible to turn glycerol into pyruvate fairly directly (using reactions of glycolysis).

To put everything together, you just need an organism that secretes succinic acid, and make sure that these pathways are turned on sufficiently that most of the provided glycerol gets metabolized to succinic acid. As this involved oxidation, the organism has to have access to oxygen or some other oxidation agent.

UPDATE

Prompted by Andrew's comment, I found a paper where they optimize succinate production from glucose. As you can see from the scheme above, glycerol can be funneled into glycolysis by two enzymes (using one ATP for phosphorylation and one NAD+ for oxidation). The enzymes used for the remainder of the reaction (and to provide NADPH via the pentosephosphate pathway) are shown in the figure from the paper below. You would need to provide some glucose to feed in the pentosephosphate pathway to have sufficient NADPH, or invoke some other pathways:

Source: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4983090/figure/Fig1/?report=objectonly


Why are excess nuclei required to produce an NMR signal? In NMR, an excess population of nuclei is required to produce a signal. If saturation is achieved (same number of nuclei in the α and β states) no net signal can be produced. A signal is produced when a radio frequency causes one state (could be either α or β) to invert. During the process of inverting, the net oscillating electric field of the nuclei is tipped to the xy plane until they undergo relaxation back to the inverted state. Even if a particular set of nuclei are saturated, they will still undergo resonance, thus producing an oscillating electric field in the xy plane. This should be detectable; however, it is not. I am guessing that this is because the oscillating electric fields cancel each other out (since they are occurring for both the α and β states). This would result in canceling out the net electric field; however, I am not sure of my guess.

"A signal is produced when a radio frequency causes one state (could be either α or β) to invert"

This isn't correct. It really isn't, even though it's often incorrectly described that way. The reality is more complicated, and it involves understanding QM as well as the fact that single spins do not necessarily exist in $\alpha$ or $\beta$ states (any superposition of them is allowed). See my answer here for more details.

Without going into too much detail, population differences* can be used to generate coherences, which are responsible for generating the detected signal. If we construct a generalised wavefunction, with two different coefficients of the two states, viz.

$$|\psi\rangle = c_\alpha|\alpha\rangle + c_\beta |\beta\rangle$$

then coherences can be (loosely) understood as an oscillation of the coefficients $c_\alpha$ and $c_\beta$. For example we could have

$$\begin{align}
c_\alpha &= \cos(\omega t + \phi) \\
c_\beta &= \sin(\omega t + \phi)
\end{align}$$

which mean that the coefficients, and hence the wavefunction $|\psi\rangle$, is oscillating with an associated frequency $\omega$ and a phase $\phi$. This generates a signal in the final spectrum at the frequency $\omega$. This sort of oscillation has nothing to do with individual spins flipping from $|\alpha\rangle$ to $|\beta\rangle$. Indeed, in NMR, individual spins have no meaning at all. Only the bulk behaviour of the $\sim10^{20}$ spins in your sample matters.

How does this relate to population differences? For a coherence to be generated, there is an additional requirement that across the entire ensemble of spins, there is a correlation between spins in the phase $\phi$.

A decent analogy would be having a collection of ($\sim10^{20}$) analog clocks on a wall. Each of these is oscillating at a rate of one revolution per 12 hours, since each clock returns to the same time every 12 hours. If you started all the clocks at the same time, then they would (largely) remain in sync with each other, and you could have a pretty good idea of what time it is right now. On the other hand, if they all started out of sync with each other, then you would have zero hope of telling what time it was.

This is the real key to understanding why there is a need for population differences. Without a population difference, you can generate oscillating spins, but each of them has a completely random phase and so there is no correlation between them. Consequently, the net signal sums to zero.

I don't want to go deeper because it will get very, very long, and also because it has been covered in a way that is better than I can ever do. See: James Keeler's book at Resources for learning Chemistry.

* As I've said before, populations should not be interpreted as the actual number of particles in a specific eigenstate, but rather the ensemble average of the coefficient corresponding to the specific eigenstate in the wavefunction. In other words, if each spin has a state $|\psi\rangle = c_\alpha|\alpha\rangle + c_\beta|\beta\rangle$, then we have $n_\alpha = \overline{c_\alpha^* c_\alpha}$ where the bar indicates an average over all spins in the ensemble.


Difference in work done by a gas in reversible and irreversible isothermal expansion Why there is difference in work done by gas in isothermal expansion when gas expands through reversible and irreversible process?

For a massless piston, the force per unit area exerted by the gas on the inner face of the piston is exactly equal to the external pressure exerted by the surroundings on the outer face of the piston, irrespective of whether the expansion is reversible or irreversible (by Newton's 2nd law).  The work done on the surroundings is thus always equal to the integral of the external pressure integrated over the volume change.

However, for an irreversible expansion, the force per unit area exerted by the gas is comprised of two components: (1) the local gas pressure at the piston face (as determined by applying the ideal gas law locally) and (2) viscous stresses that are proportional to the rate at which the gas is expanding.  Furthermore, in an irreversible expansion, the gas pressure and temperature are typically not even uniform within the cylinder.

In a reversible expansion, which is carried out very slowly, the viscous stresses are negligible (since the rate of expansion is negligible), and the gas pressure and temperature are uniform throughout the cylinder. Thus the external pressure is equal to the uniform pressure of the gas calculated from the ideal gas law.  Consequently, for a reversible expansion, one can use the pressure calculated from the ideal gas law in place of the external pressure to calculate the work done.

To get a better feel for how this all plays out, particularly with regard to viscous stresses and viscous dissipation, please see the following link:  https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/.  This article points out the close analogy between the transient behavior of an ideal gas experiencing an irreversible expansion or compression, and that of a spring and damper connected in parallel. It demonstrates that the physical mechanism responsible for the difference between the irreversible work and the irreversible work is viscous dissipation of mechanical energy to thermal energy (yes, even in the ideal gas region, gases exhibit non-zero viscosity).


What you understand by "Order of reaction"? When studying Kinetic chemistry (and most of the Physical chemistry topics), we often find definitions based upon the mathematical formalism. For example, Order of reaction is often defined as the sum of the partial order of a Rate law.
But is there a way to define it without using the mathematical formalism? I mean, if we think microscopically, how would we define Order of reaction?

Historically, the concept of reaction order came from Ostwald's famous Lehrbuch of 1890s. An abridged translation is available in Internet Archive as "Outlines of General Chemistry". In this book, surprisingly, when a single reactant undergoes a transformation he does not call it a first order reaction. He designated rate of reactions controlled by two species as reaction of the second order.

  Reaction of the Second Order. We come to a second kind of chemical process, regulated by a new law, when in its course two substances
  undergo change of quantity. We must again assume that the action is
  proportional to the concentration of each present.

  Reactions of Higher Order. We must now consider reactions in which three or more substances take part.

So roughly, you can think of reaction order as the number of species in a chemical reaction which control the rate of reaction.

Coming to modern times, you have fractional reaction orders, and zero order and so on. For understanding you will need to see the concept of elementary and overall reaction orders.
A reaction may be occurring in several steps. In such cases, you might end up in fractional exponents. However, once you are able to know the steps of that reaction, you will see that those elementary reactions don't have fractional exponents. You can then interpret, each elementary reaction in the original Ostwald's way.


Determining the rate of stratospheric ozone destruction I am a sophomore. A friend and I ran across a kinetic chemistry exercise. We solved it in a slightly different way and we can't come to an agreement on who is right. We would appreciate some help:

  Destruction of stratospheric ozone as determined by using the steady-state approximation.

  The balance of ozone in the stratosphere is of critical concern
  because this molecule absorbs ultraviolet light that would be harmful
  to life at Earth's surface. The principal production mechanism for
  ozone is recombination of $\ce O$ atoms with $\ce{O_2}$. The principal
  destruction mechanism is that given below. There is increasing concern
  over alternative destruction mechanisms involving molecules introduced
  into the stratosphere by human activity.

  Determine the destruction rate of ozone in the following mechanism.
  $$\begin{align}\ce{O3 + M &<=>[k_1][k_{-1}] O2 + O  + M}\\
\\
\ce{O3 + O &->[k_2] 2 O2}\end{align}$$

Our answers:

We agree on the beginning, so we have :

$$\begin{align}
-\frac{\mathrm d[\ce{O_3}]}{\mathrm dt} &= k_1[\ce{O_3}][\ce{M}] - k_{-1}[\ce{O_2}][\ce O][\ce M] + k_2[\ce{O_3}][\ce O]\\
-\frac{\mathrm d[\ce{M}]}{\mathrm dt} &= k_1[\ce{O_3}][\ce M] - k_{-1}[\ce{O_2}][\ce O][\ce M]
\\
-\frac{\mathrm d[\ce O]}{\mathrm dt} &= -k_1[\ce{O_3}][\ce M] + k_{-1}[\ce{O_2}][\ce O][\ce M] +k_2[\ce{O_3}][\ce O]
\end{align}$$

Now here comes the disagreement, we need to use the steady-state approximation to say that one of the above speed is null:

My friend argues that $-\frac{\mathrm d[\ce O]}{\mathrm dt}$ should be approximated to null because it is a transient compound.

Even if in most other exercise, it would be correct, I disagree with him because in the proposed mechanism, $\ce{O_3}$ and $\ce O$ play a symmetrical part. So if we approximate $-\frac{\mathrm d[\ce O]}{\mathrm dt}$ to $0$ we should also approximate $-\frac{\mathrm d[\ce{O_3}]}{\mathrm dt}$ to $0$. Since we want to calculate this specific value, we don't want to approximate it.
In my demonstration, I approximate $-\frac{\mathrm d[\ce{M}]}{\mathrm dt}$ to $0$, arguing that the compound $\ce M$ is on both side of the equilibrium equation and therefore it's concentration will remain constant.

We both manage to find a result:
My friend’s calculation:

$$
- \frac{\mathrm d[\ce O]}{\mathrm dt} = -k_1[\ce{O_3}][\ce M] + k_{-1}[\ce{O_2}][\ce O][\ce M] + k_2[\ce{O_3}][\ce O] = 0
$$
$$
k_1[\ce{O_3}][\ce M] - k_{-1}[\ce{O_2}][\ce O][\ce M] = k_2[\ce{O_3}][\ce O]
$$
We substitute the left member in the speed of ozone:
$$
-\frac{\mathrm d[\ce{O_3}]}{\mathrm dt} = 2\times k_2[\ce{O_3}][\ce O]
$$
Because $[\ce O]$ is transient, it is difficult to mesure experimentally, so we expresse it in function of $[\ce{O_3}], [\ce{O_2}]$ and $[\ce M]$ :
$$
k_1[\ce{O_3}][\ce M] = [\ce O] \times ( k_2[\ce{O_3}] + k_{-1}[\ce{O_2}][\ce M] )
$$
$$
[\ce O] = \frac{k_1[\ce{O_3}][\ce M]}{k_2[\ce{O_3}] + k_{-1}[\ce{O_2}][\ce M]}
$$
And finally we have my friend’s result:
$$
-\frac{\mathrm d[\ce{O_3}]}{\mathrm dt} = \frac{2k_1k_2[\ce{O_3}]^2[\ce M]}{k_2[\ce{O_3}] + k_{-1}[\ce{O_2}][\ce M]}
$$

Now my calculations:
$$
-\frac{\mathrm d[\ce M]}{\mathrm dt} = k_1[\ce{O_3}][\ce M] - k_{-1}[\ce{O_2}][\ce O][\ce M] = 0
$$
Which allow us to simplify the destruction rate of ozone:
$$
-\frac{\mathrm d[\ce{O_3}]}{\mathrm dt} = k_2[\ce{O_3}][\ce O]
$$
We already disagree by a factor 2 on the rate. But if we replace $[\ce O]$ like before, we obtain a much simpler result:
$$\begin{align}
k_1[\ce{O_3}][\ce M] &= k_{-1}[\ce{O_2}][\ce{O}][\ce M]
\\
[\ce{O}] &= \frac{k_1[\ce{O_3}][\ce M]}{k_{-1}[\ce{O_2}][\ce M]}
\\
[\ce O] &= \frac{k_1[\ce{O_3}]}{k_{-1}[\ce{O_2}]}
\end{align}$$
Finally I have:
$$
-\frac{\mathrm d[\ce{O_3}]}{\mathrm dt} = \frac{k_1k_2[\ce{O_3}]^2}{k_{-1}[\ce{O_2}]}
$$

Which one of us is wrong and why?

Dioxygen is split in the stratosphere by UV light into oxygen atoms. This is the main source of free oxygen atoms. Their production rate depends on the intensity of the UV light.

In steady-state we can assume that the concentration of oxygen atoms is constant.

$$\frac{d[\ce{O}]}{dt} = 0  \tag{1}$$

For the above destruction mechanism we can formulate the rate laws

$$\begin{align}
\frac{d[\ce{O3}]}{dt} &= -k_1 [\ce{O3}] [\ce{M}] + k_{-1} [\ce{O}] [\ce{O2}] [\ce{M}] - k_2 [\ce{O}] [\ce{O3}] \tag{2}\\
\\
\frac{d[\ce{O}]}{dt} &= k_1 [\ce{O3}] [\ce{M}] - k_{-1} [\ce{O}] [\ce{O2}] [\ce{M}] - k_2 [\ce{O}] [\ce{O3}]  \tag{3}
\end{align}$$

Adding equation $(2)$ and $(3)$ which by $(1)$ is $0$ we get

$$\frac{d[\ce{O3}]}{dt} = -2~k_2 [\ce{O}] [\ce{O3}] \tag{4}$$

With equations $(1)$ and $(3)$ we can express the concentration of oxygen atoms as

$$[\ce{O}] = \frac{k_1 [\ce{O3}] [\ce{M}]}{k_{-1} [\ce{O2}] [\ce{M}] + k_2 [\ce{O3}]} \tag{5}$$

Substituting $(5)$ into equation $(4)$ we get

$$\frac{d[\ce{O3}]}{dt} = \frac{-2~k_1~k_2 [\ce{O3}]^2 [\ce{M}]}{k_{-1} [\ce{O2}] [\ce{M}] + k_2 [\ce{O3}]}  \tag{6}$$

So your friend seems to be right.


What is the non-ideal equivalent of mass action kinetics? Starting from the equations of mass action kinetics, one may show that (for suitably chosen numbers $A_i$), the function $\sum_i x_i(A_i + \log x_i)$ is a Lyapunov function, i.e. it decreases over time.

Apart from an irrelevant factor of $RT$, this is the expression for the free energy of an ideal solution (or ideal gas, etc). So it seems that mass action kinetics carries with it an assumption that the reactions are taking place in an ideal solution.

I'm interested in how to think about kinetics in non-ideal solutions, but I haven't been able to find any information about it. Is there a standard set of kinetic equations that remain valid when the chemical potential is given not by $\mu_i^0 + RT\log x_i$ but by $\mu_i^0 + RT\log \gamma_i x_i$, for activity coefficients $\gamma_i$ that depend on the concentrations?

There are several alternatives to mass action kinetics that I know about, such as Michaelis Menton kinetics for enzymes - but these are actually derived from mass action kinetics applied to elementary steps, so they don't seem to be able to escape from the assumption of ideality.

According to Catalytic reaction rates in thermodynamically non-ideal systems Journal of Molecular Catalysis A: Chemical 163 (2000) 189–204 (citing to kinetics textbooks) non-ideal kinetics needs to consider the activity of the transition state in addition to the activities of the reactants:

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

rate $=\frac{k_BT}{h}K^{\ddagger}\frac{a_Aa_B}{\gamma_\ddagger}=k_0\frac{\gamma_A\gamma_B}{\gamma_\ddagger}C_AC_B $

Where $K^{\ddagger}$ is the equilibrium constant for the formation of the transition state, $a$ is activity, $\gamma$ is activity coefficient, $C$ is concentration, $k_B$ is Blotzmann constant, $T$ is temperature, $h$ is Planck's constant and $k_0$ is the thermodynamically ideal rate constant.

Another reference to look at for more info is The status of transition-state theory in non-ideal solutions and application of Kirkwood–Buff theory to the transition state J. Chem. Soc., Faraday Trans. 2, 1986,82, 1297-1303.


Solubility of metal halides Are LiBr and LiCl ionic or covalent? Are they soluble in water? Some texts say they are partially soluble in water and in organic solvents also. Or can it be believed that they have more of a covalent character than ionic character ?

$\ce{LiCl}$ and $\ce{LiBr}$ are ionic compounds with a significant covalent character. This is due to the strong polarising power of $\ce{Li^{+}}$ cation. This polarising power is due to its large charge to mass ratio.

$\ce{Li^{+}}$ pulls the electron cloud of the halogen, hence resulting in a covalent character. The covalent character is more in $\ce{LiBr}$ than $\ce{LiCl}$ as $\ce{Br^-}$ is larger in size, hence valence shell is held comparatively loosely.

This is what essentially happens.


Effect of pH on electrolysis I understand (more or less) that acidification of water helps in the dissociation of water into ions. Acids easily dissociate into ions in aqueous solution. So, accordingly, the hydrogen from the acid goes to the cathode and the hydrogen from the water makes a compound with the negative radical from the acid. In the process, oxygen is left alone and hence goes to the anode. I want to confirm whether this is the correct explanation.

I have been trying to find out why is copper sulphate acidified during its electrolytic refining.  I interpret  that acidifying copper sulphate makes it dissociate easily into ions but I want to know what the exact process is. It doesn't seem to me that copper sulphate may behave in the same way as water does above. Please help.

In general, I would like to understand the effect of decreasing or increasing pH on the process of electrolysis.

I'd rather not put it in such terms. Dissociation of water is determined by the constant $\ce{K_w=[H+][OH-]=10^{-14}}$, which you can't change simply by adding acid (or anything else, for that matter). Acids dissociate easily into ions, just like you said, therefore adding acid to pure water (which is a poor conductor by itself) makes the latter more conductive and thus more prone to electrolysis.

Electrolytic refining of copper is a whole different story. Indeed, once dissolved in water, $\ce{CuSO4}$ would dissociate into ions completely, no matter whether or not you acidify it. The role of acid is just to prevent hydrolysis.

In general, as you change pH, some of your ions may undergo certain transformations (like $\ce{CO3^2- -> HCO3- -> CO2\uparrow}$, or $\ce{Fe^2+ -> Fe(OH)2\downarrow}$), which in turn might or might not affect the reactions on one or both electrodes. As it happens, being too general prevents us from drawing more specific conclusions.


Chain transfer agents and rate of free radical polymerization In free radical addition polymerizations sometimes chain transfer agents are added. From what I understand chain transfer agents (e.g. XR' below) reduce the average molecular weight by transferring the free radical  from the growing polymer chain (P•) to another molecule (R'•).

P• + XR' → PX + R'•

My question is what effect would this have on the overall "rate" of polymerization itself, i.e say the consumption of monomer.

In other words the R'• species is then free to combine with more Monomer, right? And being a smaller molecule (in general) would it be expected to have a higher rate constant for the reaction:

R'• + M--> P•

Not really.

As long as you are in solution (=unreacted monomer or solvent), it makes no difference if the reactive chain end is diffusing slow or fast. The monomer molecules come by at the same rate.

The mobile chain end would help against a local depletion of monomer, but once your mixture gets so viscous that this could happen, neither a long or a short active chain can still diffuse well (and neither can the transfer agent).

The only general difference you will see with transfer agent is that the Trommsdorff-Norish or gel effect sets in a bit later, because the overall viscosity of short chains is lower. In that sense, the transfer agent even slows down the reaction a bit. Which in that case is very welcome.

Of course the transfer agent radical can be either more reactive with the given monomer or less. (See "copolymerisation parameter".) So the chain restart can either be fast (in which case it makes practically no difference), or it can eventually be much slower. That depends on the specific transfer agent / monomer pair.


Chemical compounds responsible for the colors in flowers? I know that anthocyanins are a class of compounds responsible for the purple colors found in flower petals.

  Anthocyanins (also anthocyans; from Greek: ἀνθός (anthos) = flower + κυανός (kyanos) = blue) are water-soluble vacuolar pigments that may appear red, purple, or blue depending on the pH. (same reference as above)

The anthocyanins, when extracted from flower petals with water, can be used as acid/base indicators, producing red colors in acid and blue in neutral solutions (in basic solutions they turn green, yellow, and then become colorless). When I've done the extractions, some red petals produce a red color in acid and maintain the red color at higher pH than purple flowers (see plate below), but some red flowers produce colors that seem to follow the normal progression.

Row 2 from the top was a purple flower, row 3 a red flower and row 4 also a red flower. (pH values start at 2 on the left and increase by 1 pH unit as you move right by adding the water extracts to buffer samples in the wells.)

So, are there other classes of compounds responsible for the red and yellow colors found in flowers? Can the red colors that give a "normal" anthocyanin pH response arise from pigments held in an acidic environment in the plant?   (That may be a biology question, but I'm interested in the chemistry side of where the color comes from.)

are there other classes of compounds responsible for the red and yellow colors found in flowers?

Interest in colors, specifically dyes, was a real motivator for serious chemical work. Some of the early efforts were aiming to mimic Tyrian purple, also known as Royal Purple, because in antiquity it was worth its weight in gold.

Two classes of molecules come to mind other than anthocyanins in phytochemistry, such as the closely related flavones (its root means yellow) and the distinct carotenes (its root means carrot, but also responsible for flamingo color).

  Can the red colors that give a "normal" anthocyanin pH response arise from pigments held in an acidic environment in the plant?

A real sharp organic chemist by the name of Richard Willstatter proposed this  in the early 1900s. He was working on determining the structure of the pigment responsible for the blue color in cornflower. He extracted the same substance coming from a rose and proposed that all flower colors were due to a different pH in the sap. As your experiment clearly points out, however, alkaline conditions are typically required for the blue anion. A Japanese group argued that alkaline conditions in the sap were unlikely and that instead these different colors can be explained by the formation of metal-organic complexes.

It wasn't until 2005 that a definitive crystal structure was determined from the cornflower pigment called protocyanin. It turns out that the blue color comes from a supermolecule containing one $\ce{Fe^{3+}}$, one $\ce{Mg^{2+}}$, two $\ce{Ca^{2+}}$, six anthocyanin and six flavone molecules packaged together in a complex structure. This color, as I vaguely recall, remains blue from pH 7 until pH 1! This agreed with some earlier evidence which suggested a complex species by the temperature dependence of the pigment.

These are not easy studies to conduct and I'm not familiar with how in-depth red colors have been studied, but it is certainly possible a complex species is actually how the red pigments form in vivo. Keeping in mind that plant vacuoles have a pH on the order of 5, we'll generally notice a loss in red above pH 3 in solution for a pure anthrocyanin in solution as hydration of the cation occurs. Stability of the color, and thus cation, obviously must be obtained either via copigmentation or with metals in a plant.


What is the pH of starch? We need to identify an unknown for a chemistry lab. It's either starch or zinc sulfate.

We know that zinc sulfate is acidic, but what is the pH of starch?

Starch is a weak acid, but a stronger acid than water.

If you make a slurry of starch and add sodium hydroxide the starch will react with the sodium hydroxide, the starch slurry will then change in viscosity. It normally becomes a lot more viscous in my experience. This is part of the method by which starch is alkylated. I have made moderate sized batches of benzyl starch from corn starch by first stirring the starch with sodium hydroxide and then treating the resulting paste with benzyl chloride and then heating it up.

If you were to get a very senstive pH meter you would see that a slurry of starch in water is slightly more acidic than water would be. An alternative would be to use a dilute sodium hydroxide solution to make the measurement, I will warn you that it can get messy as the starch can be transformed into a paste / slime.

MaxW is right in part, a pure starch will give no ash but zinc sulfate will dehydrate to anhydrous zinc sulfate unless you heat it very very hard.

Another method is to add iodine, many starches will form a deep blue / black colour with a small amount of iodine. Zinc salts will not.


Why do metals form cations? metals that have a low number of occupied shells such as lithium and potassium should have a stronger electrostatic attraction to their nuclei, so what causes them to always lose their electrons during ionic bonding with a non metal?

Potassium is in the 4th period and therefore has at least 4 shells occupied with electrons while lithium is in period 2 and thus only has 2 shells occupied with electrons. Perhaps you mean they both have 1 valence electron that must be ionised before they become a cation? Anyhow, ionic bonds are formed due to the strong electrostatic attraction between oppositely charged ions. All elements would like to achieve a noble gas electron arrangement due to its stability. However, to be able to achieve this, elements must either gain or lose electrons depending on how close their electron arrangement is to their closest noble gas, e.g. Fluorine would like to gain 1 electron while Magnesium is more likely to lose 2 electrons to form its closest noble gas.

For instance, NaCl: Na would like to lose an electron and Cl would like to gain an electron. Together this is possible. The nucleus of the Cl atom electrostatically attracts the valence electron from the Na atom– this attraction is strong enough the overcome the the attraction this valence electron feels from its own nucleus. As a result, the Cl gains a negative charge due to it gaining a negatively charged electron and the Na atom now has a positive charge as it has one less negatively charged electron. The Cl- and Na+ are oppositely charged and therefore attract each other. This is known as ionic bonding and results in the formation of an ionic lattice.

Now to your question. All elements in group 1 have approximately the same effective nuclear charge, i.e. the attraction the valence electrons feel from the nucleus of their atom after the shielding of the inner electrons is considered. Hence, the fact that they are in group 1 should not significantly affect the ability to lose their valence electrons. However, as one goes down periods, the number of occupied outer shells increases, which means that the valence shells experience a weaker attraction from their nucleus. This is confirmed by the lower first ionisation energy trends. To answer your question frankly (I perhaps overcomplicated with my verbose writing style), it is the strong affinity for electrons of the non-metals that result in the loss of the valence electrons from the metals. Even if the the metal has a smaller atomic radius due to occupying less shells, the affinity for the electrons from the non-metals is strong enough to take the metal's valence electrons and make it into a cation.

References:

Pearson Higher Level Chemistry Textbook, 2nd Edition. By Catrin Brown and Mike Ford.


electronegativity and how it relates to bond strength my question is, if the strength of the bond between two atoms increases as the difference in electronegativity increases as well.

so say, a H-F bond would be stronger than a O-H bond.

I'm a little confused because I know that acidity increases from left to right across the periodic table. So H-F is a stronger acid than H-CH3. Wouldn't this mean that H-F has weaker bonds than H-CH3?

Bond strength usually refers to the energy required to break a bond homolytically--with one electron going to one of the atoms and the other electron going to the other atom. In the case of HF this requires more energy than in CH4. Acidity is the tendency for the bond to break heterolytically--with two electrons going to one atom and none going to the other atom. In the case of HF, we get the two ions H+ and F-. This still requires a lot of energy, but in water solution, the ions are stabilized by hydrogen bonding, so ionization is much more favorable. Methane on the other hand would ionize to give a CH3- anion, which is a lot worse than F- due to carbon being less electronegative.


Which is more amphoteric: zinc or copper hydroxide? $\ce{K_{a} ~Cu(H2O)_6^2+}{=~5*10^{-7}} $

$\ce{K_{a} ~Zn(H2O)_6^2+}{=~2.5*10^{-10}} $

Given this information, which forms a more amphoteric hydroxide?

My reasoning is that copper hydroxide is more amphoteric than zinc hydroxide. My reasoning is that because $\ce{K_{a} ~Cu(H2O)_6^2+}$ > $\ce{K_{a} ~Zn(H2O)_6^2+} $, $\ce{Cu(OH)_2}$ should be more amphoteric because it will probably have the bigger $\ce{K_{a}}$. This fulfills half of the definition of amphoteric - behaving as both an acid and base.

But how do I compare the base strengths of $\ce{Zn(OH)_2}$ and $\ce{Cu(OH)_2}$? I need to, because being amphoteric means the substance acts as both an acid and a base!

Consider the following:

$1. ~{K_{a1}:} ~\ce{Cu(H2O)_6^2+ + H_2O -> Cu(H2O)_5(OH)^+ + H_3O^+}$

$2. ~{K_{a2}:} ~\ce{Cu(H2O)_5(OH)^+  + H_2O -> Cu(H2O)_4(OH)_2 + H_3O^+}$

$3. ~{K_{a3}:} ~\ce{Cu(H2O)_4(OH)_2  + H_2O -> Cu(H2O)_3(OH)_3^- + H_3O^+}$

Basic character of copper hydroxide is illustrated by the below equation:

$4. ~\ce{Cu(H2O)_4(OH)_2  + H_2O -> Cu(H2O)_5(OH)^+ + HO^-}$

The above equation, reversed, is:

$5. ~\ce{Cu(H2O)_5(OH)^+ + HO^- -> H_2O + Cu(H2O)_4(OH)_2 }$

This equation is very similar to $~{K_{a2}}$ except that the base is hydroxide ion instead of water.

We can write a similar equation for zinc hydroxide:

$6. ~\ce{Zn(H2O)_5(OH)^+ + HO^- -> H_2O + Zn(H2O)_4(OH)_2}$

Let's compare equations $5$ and $6$. We can expect $5$ to be a larger extent reaction than $6$ because hydrated copper ion is a stronger acid than hydrated zinc ion to start with. So:

${K_{rxn5} > K_{rxn6}}$

And therefore,

${K_{rxn4} < K_{rxn7}}$

$7. ~\ce{H_2O + Zn(H2O)_4(OH)_2 -> Zn(H2O)_5(OH)^+ + HO^-}$

Zinc hydroxide is more amphoteric as quantitatively investigated in Effect of pH, Concentration and Temperature on Copper and
Zinc Hydroxide Formation/Precipitation in Solution

In summary, in the case of Cu(OH)2, only at extremely low concentrations and very high pH can a third OH- be coordinated.

  At higher pH values, copper
  hydroxide, Cu(OH)2, is the dominant species up to pH 12.3 where the copper ion
  Cu(OH)3
  - forms according to Equation 3. At higher copper concentrations, solid
  Cu(OH)2 is formed and precipitates out of solution at copper concentrations above the
  solubility product of copper hydroxide at 1×10-8 M. It is important to note that the
  domain of stability of solid Cu(OH)2 is expanding to lower and higher pH values with
  increasing copper concentration.

On the other hand:

  Above pH 11.4, Zn(OH)3
  - forms according to Equation 7. At
  higher zinc concentrations, solid Zn(OH)2 forms and precipitates out of solution at zinc
  concentrations above the solubility product of zinc hydroxide of 1×10-5 M.

So qualitatively both display similar amphoteric behavior in the limit of infinite dilution, but at a appreciably non-zero concentration zinc is more amphoteric.

The article says it is getting the equilibrium constant values from Mineral Equilibria, Low Temperature and Pressure.

An alternative source of this information is Hydrolysis of cations. Formation constants and standard free energies of formation of hydroxy complexes Inorg. Chem., 1983, 22 (16), pp 2297–2305, which gives quantitative values for OP equations 1, 2 and 3 as well as the corresponding equations for zinc, in terms of Gibbs energy of formation and in terms of cumulative equilibrium constants.

Simplifying the general equation in the paper to only mononuclear (one metal atom) complexes:

$$K_y$$
$$\ce{M^{n+} + yH2O <=> M(OH)_y^{n-y} + yH+}$$

$\ce{Cu^{2+}}$ :
$pK_1 = 7.96$
$pK_2 = 16.26$
$pK_3 = 26.7$
$pK_4 = 39.6$

$\ce{Zn^{2+}}$ :
$pK_1 = 8.96$
$pK_2 = 16.9$
$pK_3 = 28.4$
$pK_4 = 41.2$

(These are all experimental values, calculated values are also given)

So really, without considering the solid phase (solubility), Cu2+ and Zn2+ seem very similar, with Cu2+ showing slightly more acidity.

Another study, which does consider the solid phases, is Zinc Hydroxide: Solubility Product and Hydroxy-complex Stability
Constants from 12.5-75 [degrees] C Can. J. Chem. 53, 3841.

This study approaches the problem with 5 equilibria involving the solid phase. (These authors use the symbol "c" for the solid phase.)

$\ce{Zn(OH)2(c) <=> Zn(OH)+ + OH-}$
$K_1 = [\ce{Zn(OH)+}][\ce{OH-}] = 2.54 \times 10^{-11}$

$\ce{Zn(OH)2(c) <=> Zn(OH)2(aq)}$
$K_2 = [\ce{Zn(OH)2}] = 2.62 \times 10^{-6}$

$\ce{Zn(OH)2(c) +OH- <=> Zn(OH)3-}$
$K_3 = [\ce{Zn(OH)3-}]/[\ce{OH-}] = 1.32 \times 10^{-3}$

$\ce{Zn(OH)2(c) +2OH- <=> Zn(OH)4^{2-}}$
$K_4 = [\ce{Zn(OH)4^{2-}}]/[\ce{OH-}]^2 = 6.47 \times 10^{-2}$

$\ce{Zn(OH)2(c) <=> Zn^{2+}} + \ce{2OH-}$
$K_{sp} = [\ce{Zn^{2+}}][\ce{OH-}]^2 = 1.74 \times 10^{-17}$

This study (see Fig. 1) shows that aqueous $\ce{Zn(OH)2}$ is only the major aqueous species in the pH range 9-11, having acted as an acid or base outside this range.

Instead, if I were just given the information in the OP, meaning the two supposed Ka values, as some kind of test or homework question,  I would say copper is more amphoteric.  My reasoning would be, for zinc, just to remove one proton, you need to get up to almost pH 10.  For two more protons to be removed, as required to get to Zn(OH)2 and lose yet another proton to act as an acid, seems impossible because I would expect pKas to be spaced apart and if you would need to approach or go beyond pH 14 to observe the acidic behavior, that isn't reasonably considered amphoteric.

Note that EXAFS Investigations of Zn(II) in Concentrated Aqueous Hydroxide Solutions J. Phys. Chem. 1995, 99, 11967-11973 finds that rather than being 6-coordinate as indicated in the OP, 4 hydroxides coordinate Zn2+ tetrahedrally with no water ligands.


Reduction of amide with LiAlH4 In reduction of carboxylic acid with $\ce{LiAlH_4}$ the first step is deprotonation.

But in the mechanism of reduction of amide with $\ce{LiAlH_4}$ (here), the first step is addition of hydride on carbonyl group. Why doesn't the hydride ion abstract a proton from $\ce{-NH2}$? Since acid base reactions are faster than nucleophilic addition.

As compared to the carboxylic acid, deprotonation of the amide is negligible.

I don't recall any data for THF or 1,4-dioxane as the solvent, but F.G. Bordwell has determined a lot of $pK_a$ values in DMSO.

In DMSO, the $pK_a$ of acetic acid is 12.6 [1], while the $pK_a$ of acetamide is 25.5 [2].