Relative basicity of cyclic amines Can anyone explain this trend (the amines are listed in order of increasing basicity)?

Actually, I thought of understanding this by thinking of inductive effect of the alkyl groups attached to the amine group but since they are cyclic I am confused how to apply the inductive effect here or I want to know whether there are other ways of dealing with this problem.

Inductive effect in this sequence in the reason of the increased basicity only in the case of N-methyl piperidine compared to piperidine. Increased basicity in the sequence of azetidine < pyrrolidine < piperidine is connected with the ring strain. Upon the decrease in the C-N-C angle, sp3-hybridized N atom "rehybridize" to increase p-character of orbitals involved in the C-N bonds formation, leading to the increased s-character of orbital involved in the N-H bond and the lone pair of electrons. Since s-electrons are closer to nucleus, this rehybridization leads to the increased stability of the N-H forming orbital and the lone pair in more strained cyclic amines making the lone pair being 'less exposed' to protonation (less basic).


Elevation in boiling point of water $\pu{1060 g}$ of urea aqueous solution is subjected to heating. It starts boiling at $\pu{100.53 °C}.$  It is heated up to $\pu{102.12 °C}.$ Find amount of water vapourised in grams. (Pure water boiling point is $\pu{100 °C},$ $K_\mathrm{b}$ of $\ce{H2O}$ is $\pu{0.53 K kg mol-1}.)$

When water evaporates will the equivalent amount of urea also evaporate i.e. will we calculate final molality with the same amount of solute or a decreased amount?

I first calculated the initial amount of water as (53000/3) g and then taking the same moles of urea calculated the final amount which came out to be (53000/12) g. So the amount of water evaporated should be 13250 g but the given answer is 750 g.

While considering colligative properties we always assume solute to be non-volatile. So, it doesn't contribute to the vapor. The amount of vapor will only be due to solvent particles.
So, in the problem we can use the following relation:
$$ \Delta T = K_\mathrm b\cdot m $$
Initially $ \Delta T  = 0.53\ \mathrm{^\circ C}$ so plugin the value of $K_\mathrm b$ and you end up with the value of initial molality of solvent as $1$. From this you can find the initial moles of solute which remain constant throughout.

Next you put $ \Delta T  = 2.12\ \mathrm{^\circ C}$ and plug in the known values of $K_\mathrm b$ and 'moles of solute' in the above formula to get the final value of solvent amount that remains in the solution.
Subtracting the final amount of solvent left from the initial amount will let you know the amount of water vaporized which is $750\ \mathrm g$.


Why are acetic acid and sodium bicarbonate mixed to unblock a drain? I have always wondered about the rationale behind combining white vinegar and sodium bicarbonate to perform household cleaning jobs, and to unblock drains. I would have expected white vinegar to act well on its own as an acidic cleaner, and sodium bicarbonate to act well on its own as a base cleaner.
Does combining them not simply neutralise the potential of each to perform its job?
The same combination is often recommended for unblocking slow drains. Again, depending on what is blocking the drain, I would have expected that an acid or base would act well in isolation. And given that hair and soap are the likely culprits, I would have anticipated that sodium bicarbonate, a base, would be the more effective.
What is the justification for this? Is it, perhaps, the mechanical action of the carbon dioxide produced that is intended to lift whatever is blocking the drain? Or is there something else at play?
Thank you in advance.

Quoting text from this source under the section "Why could baking soda and vinegar clean clogged drains?" :

Bicarb soda and vinegar react because of the acid-base reaction. Bicarb soda is bicarbonate $(\ce{NaHCO3})$ and vinegar is acetic acid $(\ce{CH3COOH})$.

When bicarb soda and vinegar react, they fizzle and sizzle and they expand. This is why they are able to remove clogs from drains. The pressure from the expanding product shoves the clog down as it moves along. The vinegar solution (vinegar plus water) can remove soap residue that clings to the walls of the pipes.

So it seems like your conjecture is justified, that it is not actually the reactivity of these two substances with the waste that is responsible for majority of the cleansing, but simply the mechanical action of the expansion of the carbon dioxide gas and the forming product that pushes the waste down the drain, and leads to unclogging.
The relevant chemical equation is:
$$\ce{CH3COOH (aq) + NaHCO3 (s) -> CH3COONa (aq) + CO2 (g) + H2O (l)}$$
This is similar to the mechanism by which baking powder makes a cake fluffy on addition of a little vinegar as explained here.


Magnetic and acidic properties of nitric oxide I was faced with the following question:

PART C – CHEMISTRY
Which one of the following properties is not shown by $\ce{NO}$?

It combines with oxygen to form nitrogen dioxide

Its bond order is $2.5$

It is diamagnetic in the gaseous state

It is a neutral oxide

I am confused about (3) & (4). I know $\ce{NO}$ reacts with water & oxygen to produce $\ce{HNO2}$, and then $\ce{NaOH}$ to produce $\ce{NaNO2, N2O, H2O}$. So, obviously it's acidic.
But there is one unpaired electron in the $\ce{N}$ atom. So, I see no reason for $\ce{NO}$ to be diamagnetic in gaseous state.
But, these questions do not have two correct answers. So, one of them has to be wrong. I am not sure which one.

$$\ce{2NO + O2 → 2NO2}$$
Nitric oxide reacts with oxygen to produce nitrogen dioxide. This reaction takes place quickly at room temperature in air. (chemiday)
The bond order in $\ce{NO}$ is 2.5
The oxidation state of $\ce N$ is +2, and this is a colourless, paramagnetic gas (ie. it is a radical, or contains an unpaired electron). (source)
Neutral oxides are those oxides which show neither basic nor acidic properties when they react with water. Examples include carbon monoxide $(\ce{CO})$ and nitrous oxide $(\ce{N2O})$ which are only slightly soluble in water, and nitric oxide $\mathbf{(NO)}$ which is appreciably soluble in cold water. (source)

So, option 3 is the correct answer as it is an inaccurate statement.


Why is the electrode with the lower reduction potential the cathode? Wouldn't it make sense that the electrode with the higher reduction potential be the cathode, since reduction occurs at the cathode?

No you misunderstand the meaning of reduction potential. Reduction potential refers to the voltage required to reduce a material under standard conditions. If a material has a higher reduction potential it takes more energy to reduce it than a lower reduction potential material. Therefore the higher reduction potential material is actually oxidized to reduce the lower reduction potential material.


Ring expansion in cyclo compounds In the following reaction, why does ring expansion not take place?

Mechanism that was the correct one:

In the second step when there is a protonated carbon the ring should expand to become 6 membered as that is more stable. This should be followed by a methyl shift from Quaternary to tertiary and then the $\ce{Cl}$ should get attached. But this doesn't happen. Why is that? According to my textbook's theory, what I described should occur but the mock test says otherwise. Are there some more criteria-besides.

protonated carbon next to ring,

forming of a more stable 6 number ring

for ring expansion that my text book is missing?
Any help would be appreciated.

Simple answer is consider the stability of carbocations. You already got the tertiary carbocation in hand that is much stable than secondary carbocation, which would be resulted by ring expantion. Sure, a methide migration would give you another tertiary carbocation, but before you get that, the reaction progress has to climb another actvation barrier, which would be steep ($3^\circ \rightarrow 2^\circ$). Bottom line is rate of clorination of carbocation is faster than rate of this rearrangement followed by methide migration ($E_\mathrm{a2} \lt E_\mathrm{a3}$):


How is the configuration of an ester enolate double bond assigned according to CIP? Two principal modern methods for enantioselective aldol reactions — the Evans aldol and the Masamune aldol — use a carboxylic-acid derived enolate rather than a ketone enolate as e.g. in the Paterson aldol reaction. For ‘traditional’ ketone enolates, a rule of thumb dictates that (E) enolates give anti-products and (Z) enolates give syn-products, typically rationalised by the Zimmermann-Traxler transition state model. Assigning (E) or (Z) to the enolate of an ethyl ketone poses no problems, as oxygen always has a higher Cahn-Ingold-Prelog priority than carbon, so it is all about the orientation of the enolate oxygen and the methyl group (see figure 1).

An Evans aldol reaction, the enolate of which is also shown in figure 1, can be analysed similarly. Oxygen has a higher priority than nitrogen, so the same basic rules apply.

A Masamune aldol reaction, however, has two oxygens attached to the enolate carbon, formally making it an O,O-ketene acetal. The CIP rules would dictate to follow the connectivities atom by atom. The Masamune auxilliary, like any ester, has a carbon attached to the oxygen. The enolate anion may be strongly attached to a certain atom (e.g. when using a boron- or titanium(IV)-mediated aldol reaction) but also may not be, or the attachment may be ambiguous (e.g. a potassium salt used for enolating in the presence of $\ce{LiCl}$).

Figure 1: The enolates in a Paterson-, Evans- and Masamune-aldol reaction and their double bond configurations where known.

To establish whether the double bond geometry is (E) or (Z), must I know if and which atom is attached to said oxygen? What if two different possibly attaching atoms (e.g. potassium and lithium) would give two different geometries? Is it possible to ignore said attachment altogether and just define the enolate oxygen as a free monoanion?

And do the CIP rules differ from common usage?

Whilst the (E)/(Z) nomenclature works fine for enolates of ketones and aldehydes, there is, as you've discovered an issue with the assignment of enolate geometry for esters (an amides and thioesters) since depending on the nature of the ester and the metal counter-ion used, we end up with a different answer.

Whilst the CIP rules are trivial to apply to any enolate, their use is not always conducive to describing the chemistry going on, especially when we end up with (E) and (Z) enolates both giving syn-adducts (for example) merely because we've changed a counter-ion (the actual spacial configuration of the enolate is identical in both cases).

To avoid this switching of configurations, we usually choose some other way of definition such that at the beginning of any thesis or SI where enolates are used, there is usually a statement that explains the way in which the author is planning to define their enolates (i.e. define them however you wish, but make sure you define them).

This is often not in agreement with formal CIP rules for the reasons outlined above, or at least not consistently in agreement with CIP rules (things may happen to end up named correctly regardless of how we define them).

This statement defining the enolate geometry is generally a re-hash of Prof Evans who stated:

  ...irrespective of the formal CIP criteria, the O-atom bearing the metal (the OM residue) is given a higher priority, and the ipso-substituent X (in enolates) the lower one...

Whilst this is widely accepted, there are some issues, mainly because there are inconsistencies between groups: Evans' calls it (E) but Myers' calls it (Z) with Furstner calling it (E) but for a totally different reason than Evans'.  For this reason alone, its always worth looking at the actual schemes and defining the enolate yourself rather than trusting that enolate X gives Y.

This issue is the source of the major objection to Evans' idea about the nomenclature. The whole reason we use E/Z/R/S is because they are well defined, I can tell you that I have (R)-..... and you'll know what I'm talking about, because we both follow the same rules, and if you don't, you're wrong.

The proposed solution to Evans' solution (Ernest Eliel) is to stop messing with these well defined descriptors, and instead go back to using cis and trans for enolate geometry, as these terms are less well defined and as such abusing them in this way doesn't lead to any significant confusion.


How to obtain reactant lifetime using transition state theory? I've read this paper and the authors calculate life times of an intermediate using Quasi-Classical-Trajectories (QCT), RRKM and transition state theory (TST). Table 3 on page 4755 show the results.

If I understand correctly for the QCT calculations they take all (several hundred in that case) trajectories and just do statistical analysis of the lifetime of that intermediate. That perfectly makes sense and they come up with a value +/- error.

For TST they also calculate a lifetime of that intermediate, assuming it is in equilibrium with it's surrounding. This makes me think that they used the reaction barrier from intermediate to product for their calculation of the lifetime. How to do this? Up to now I've only used TST in combination with Eyring equation for calculation of rate constants.

As far as I understand, there are two things that I should answer for you.

Calculating rate constants with any trajectory methods is different than using TST. In fact, the whole point of using QCT is so we can account for nonstatistical effects. These nonstatistical effects are usually extremely important, as it is also mentioned in the article

  ... This dynamic matching leads to pronounced nonstatistical effects
  on the lifetimes...

Any time you are using TST, you are assuming there are no nonstatistical effects whatsoever. This approximation is probably never true, but it leads to surprisingly good numerical results. The way you calculate reaction rates in QCT is by integrating the reaction probability using a Monte Carlo integration formula, cleverly sampling the initial conditions of the reactants. I usually recommend a good book of W. Miller. The book itself is a bit outdated, but the derivation by Porter and Raff in the first chapter is beautiful.

Now you asked about the determination of the lifetime. The article, as I understand it, talks mostly about the lifetime of the activated complex. This can be done because in QCT we know the coordinates of atoms in every single timestep, so we can compare them to some given values to decide whether we are in the activated complex region or not. From the article:

  Transition zones were defined for the C–C forming (3‡) and the C–N breaking (5‡) steps (see Supporting Information for details). [...] The lifetime of
   intermediate 4b was defined as the elapsed time between departure from the transition zone 3b‡ and entrance in transition zones 5b‡

So, long story short: QCT determines lifetimes differently from TST, this is because TST is inherently statistical, while QCT also considers nonstatistical effects. Instead, QCT calculations rely on defining geometries for different states, and simply calculating how many integration steps the system spends there.


Active mass of solids shouldn't necessarily be unity I've read that active mass of solids and pure liquids is considered unity. I'm absolutely sure there is some misunderstanding. I know that concentration of solids doesn't change during the course of reaction, but it shouldn't necessarily be unity either. Consider graphite for example. It's density is $2260\pu{g/L}$
$$\text{[C]}=\frac{n_C}{V_C}$$ where $n_C$ is the number of moles of the carbon sample and $V_C$ is the volume of that sample. $$\text{[C]}=\frac{w}{M\times V_C} $$ $$\text{[C]}=\frac{\rho_C}{M}=\frac{2260\pu{g/L}}{12\pu{g/mol}}\\=188.3\pu{mol/L}$$This concentration will be constant throughout a reaction if the phase doesn't change and temperature change is quite small. But this clearly is not unity. Then why is the concerntration of solids considered to be unity?

Rather, active mass of solids need not necessarily be unity, but arbitrarily is unity with advantage. You do not have to know the molar mass nor density.
The common thermodynamic expressions like $K = \frac{[C][D]}{[A][B]}$
or kinetic ones like $\frac{\mathrm{d}[C]}{\mathrm{d}t} = k \cdot [A] \cdot [B]$
should be more exactly written as $K = \frac{a_\mathrm{C}] \cdot a_\mathrm{D}}{a_\mathrm{A} \cdot a_\mathrm{B}}$ or $\frac{\mathrm{d}[C]}{\mathrm{d}t} = k \cdot a_\mathrm{A} \cdot \cdot a_\mathrm{B}$
where $a_\mathrm{X}$ is unitless quantity called activity, defined by the chemical potential of the substance $ \mu_\mathrm{A} = \left( \frac {\partial G}{\partial n_\mathrm{A}} \right)_{T,p}$ by formula:
$$\mu_\mathrm{A} = RT \ln {a_\mathrm{A, unitless}}$$
If we implement the standard chemical potential, we can define unitless activity as ratio of the activity with unit and the standard activity with the unit value.
$$\mu_\mathrm{A} = \mu_\mathrm{A}^{\circ} + RT \ln \left( \frac {a_\mathrm{A}}{a_\mathrm{A}^{\circ}} \right)$$
where $a_\mathrm{A}^{\circ}$ is the activity at the standard chemical potential $\mu_\mathrm{A}^{\circ}$. We can consider the activity $a_\mathrm{A} = x\ \text{"whatever unit"}$ and the standard activity $a_\mathrm{A}^{\circ} = 1 \ \text{"whatever unit"}$.
If we arbitrarily choose $a^{\circ} = \pu{1 mol/L}$, then $a$ then more or less closely follow the respective concentration [A].
Chemical potential of the pure solid substance $\mu_\mathrm{A}^{\circ}$ does not depend on the amount of substance, so we arbitrarily choose $a^{\circ} = x_\mathrm{A}^{\circ}=1.0$ and $a$ then more or less follows the  molar fraction of substance $x_\mathrm{A}$.


How are samples prepared for analysis with ICP-OES? Can it analyse gases? In ICP-OES (Wikipedia page), you introduce a sample into a very hot plasma and then run a spectral analysis on the resulting light emissions. Liquid samples are sprayed into the flame. From the Wikipedia article:

  A peristaltic pump delivers an aqueous or organic sample into an analytical nebulizer where it is changed into mist and introduced directly inside the plasma flame.

From that I can imagine that you would probably dissolve solid samples (the technique is widely used in metallurgy and mineralogy) in something suitable to obtain a liquid sample. If that's correct, I'm wondering how the feeding apparatus is constructed to withstand possibly aggresive solvents, like nitric acid or aqua regia. The injection tube is apparently made from fused silica, I guess that's a probable material for the feed as well.

And how about gases? Are they analysed with ICP-OES as well? Are they simply pumped into the injection tube? For example, can we just introduce a stream of air into the plasma and get the emission spectrum for 78% nitrogen, 21% oxygen (I guess the argon will be hidden because the plasma gas is argon)? I guess since every sample will be beyond vaporized anyway by the intense heat, it can't make much difference, but are gas samples too "thin", i. e. do they contain too little atoms per volume for useful analysis?

About the measurement of solids

Injections tubes are indeed often made out of fused quartz. Dealing with aggressive solvents is actually not that much of an issue because of the extremely low amounts of material that can be detected with ICP-OES. Therefore, you can simply dilute your aggressive solvent before use. You can look for example at this research note where they dissolve a sample in $\ce{HCl}$ and $\ce{HNO3}$ and then dilute it before use.

About the measurement of gases

Yes, it is possible to measure gases directly, but it is not recommended and some adaptations to the normal procedure are needed. In this research paper they investigated aerosol analysis by ICP. The main issue was keeping sampling lines at elevated temperatures to allow proper calibration. A secondary problem arose because their target analyte, metallic compound in flue gases, which are volatile at (or close to) room temperature, are typically highly unstable or explosive. Therefore, they recommend dissolving areosols in a liquid solution of water and some organic solvent (1-methoxy-2-propanol in their case) and use this as feed.

For the measurement of 'normal' gases it seems that the problems they encounter will not occur, because you don't need elevated temperatures in your calibration. So in principle that means that you can indeed measure air. It should be noted that in some machines a shear flow of dried air or nitrogen is used to cut the plasma (see wikipedia) which will disallow measurement of air, as this Perkin Elmer instrument (overview table in slide 7 of this presentation). Some other instruments, like the one in the research paper I referenced, do not use such a shear flow and will thus be able to detect air.

Gas samples are not at all 'too thin', because the normal nebulized feed that goes into the plasma is similarly dilute.


Explanation for the trends in nucleophilicity—Orbital interactions or electrostatic attraction The trend of halide nucleophilicity in polar protic solvents is $$\ce{I- > Br- > Cl- > F-}$$

The reasons given by Solomons and Fryhle[1], and by Wade[2] are basically as follows.

Smaller anions are more solvated than bigger ones because of their 'charge to size ratio'. Thus, smaller ions are strongly held by the hydrogen bonds of the solvent.
Larger atoms/ions are more polarizable. Thus, larger nucleophilic atoms/ions can donate a greater degree of electron density to the substrate than a smaller nucleophile whose electrons are more tightly held. This helps to lower the energy of the transition state.

Clayden et al.[3] give the reason that in SN2 reactions, the HOMO–LUMO interaction is more important than the electrostatic attraction. The lone pairs of a bigger anion are higher in energy compared to a smaller anion. So, the lone pairs of a bigger anion interact more effectively with the LUMO σ*. Thus, soft nucleophiles react well with saturated carbon.

What I don't understand is the trend of halide nucleophilicity in polar aprotic solvents.

$$\ce{F- > Cl- > Br- > I-}$$

In aprotic solvents, only the solvation factor is absent. Bigger anions are still more polarizable and can interact in a better way with the LUMO. Then, what is the reason for the trend reversal in aprotic solvents? Solomons and Fryhle[1] say

  In these[aprotic] solvents anions are unencumbered by a layer of solvent molecules and they are therefore poorly stabilized by solvation. These “naked” anions are highly reactive both as bases and nucleophiles. In DMSO, for example, the relative order of reactivity of halide ions is opposite to that in protic solvents, and it follows the same trend as their relative basicity.

Shouldn't the orbital interactions matter more than the electrostatic attractions even in polar aprotic solvents? Why does nucleophilicity follow the basicity trend in polar aprotic solvents?

References:

Solomons, T. W. Graham; Fryhle, C. B. Organic Chemistry, 10th ed.; Wiley: Hoboken, NJ, 2011; pp. 258–260.
Wade, L. G. Organic Chemistry, 8th ed.; Pearson Education: Glenview, IL, 2013; pp. 237–240.
Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed.; Oxford UP: Oxford, U.K., 2012; pp. 355–357.

This is a rather intellectually-stimulating question and one that is also very difficult to answer. You have constructed a very good case for why the nucleophilicity order would not be expected to reverse in polar aprotic solvents, i.e. the order of intrinsic nucleophilicities of the halide ions should be $\ce {I^- > Br^- > Cl^- > F^-}$, based on the concepts of frontier molecular orbital interactions, as well as electronegativity and polarisability of the nucleophilic atom.

Affirming your viewpoint

Admittedly, the saturated carbon atom is a soft centre and based on electronegativity considerations, the energy of the $\ce {HOMO}$ of the halide ions should increase from $\ce {F^-}$ to $\ce {I^-}$, i.e. the hardness of the the halide ions as nucleophiles decreases from $\ce {F^-}$ to $\ce {I^-}$. Based on the hard-soft acid-base (HSAB) principle, we would expect the the strength of the interaction between the halide ion and the carbon centre to increase from $\ce {F^-}$ to $\ce {I^-}$.

We can provide more quantitative justification for this using the Klopman-Salem equation$\ce {^1}$:

$$\Delta E=\frac{Q_{\text{Nu}}Q_{\text{El}}}{\varepsilon R}+\frac{2(c_{\text{Nu}}c_{\text{El}}\beta)^2}{E_{\text{HOMO}(\text{Nu})}\pm E_{\text{LUMO}(\text{El})}}$$

The first term is indicative of the strength of the Coulombic interaction while the second term is indicative of the strength of the orbital interactions. In the reaction with the soft saturated carbon, it is actually expected for the reactivity to increase from $\ce {F^-}$ to $\ce {I^-}$ (ref 1, p. 117).

At the moment, all the evidence presented points us towards the conclusion that you have posited. However, we have overlooked the importance of a thermodynamic factor - the strength of the $\ce {C-X}$ bond.

A not-particularly-relevant add-on

Actually, if you think about it the considerations made in the Klopman-Salem equation are essentially kinetic considerations. We cannot conclude anything about reaction energetics from such considerations. The strength of the $\ce {HOMO-LUMO}$ interaction or that of the Coulombic attraction tells us nothing about how the electrons would organise themselves subsequently, what would be the strength of the resultant bonds etc.

A comment made on this by Olmstead & Brauman (1977)

On this topic of explaining the relative order of intrinsic nucleophilicity of the halide ions, Olmstead & Brauman (1977) had this to say$\ce {^2}$:

  The intrinsic nucleophilicities follow the reverse order of the polarizabilities (e.g., $\ce {CH3O^-}$ > $\ce {CH3S^-}$ and $\ce {F^-}$ > $\ce {Cl^-}$ > $\ce {Br^-}$). This could be due to a stronger interaction between the more concentrated molecular orbitals of the anion with the carbon center. It could also be simply a reflection of the greater thermodynamic methyl cation affinities of the smaller anions.

It is important to note that Olmstead & Brauman are merely postulating the reasons. However, I do feel that the point on thermodynamics leads us on the right track to answering this question. On the other point of "more concentrated MOs", I am unclear of how it can be interpreted.

Let us focus now on this point.

How does thermodynamics even influence kinetics?

Well... It certainly does here. What are we concerned with when dealing with nucleophilicity? As it is a kinetic phenomenon, we are essentially concerned with the rate of the substitution reaction. This makes the consideration of the activation energy ($E_\mathrm{a}$) a particularly important one because the higher the $E_\mathrm{a}$, the lower the rate of the substitution reaction.

Here, the $E_\mathrm{a}$ can be taken to be the difference in energy between the reactants and the transition state. Thus, we can decrease the $E_\mathrm{a}$ by either raising the energy level of reactants or decreasing the energy level of the transition state. The reaction energy profile for an $\ce {S_N2}$ reaction is shown below, taken from ref. 3 (p. 394).

There are two ways to see how the strength of the $\ce {C-X}$ bond can affect the $E_\mathrm{a}$. I am not sure if they may be considered equivalent perspectives. Nonetheless, I will present both.

The first perspective is well-explained by Francis & Sundberg (2007) with regard to the $\ce {S_N2}$ reaction$\ce {^3}$:

  Because the $\ce {S_N2}$ process is concerted, the strength of the partially formed new bond is reflected in the TS. A stronger bond between the nucleophilic atom and carbon atom results in a more stable TS and a reduced activation energy.

Another perspective (which may be somewhat similar to the first, or even possibly the same) is that the transition state structure bears resemblance to the product structure, based on the Hammond's postulate. However, this would be more applicable for an endothermic $\ce {S_N2}$ reaction where the transition state structure does indeed bear more resemblance to the product. When the product is more stable, i.e. has bonds that form more exothermically, the energy of the transition state would thus be lowered, and the $E_\mathrm{a}$ would also decrease, although not necessarily proportionally.

Conclusion

The argument you have presented failed to consider the important thermodynamics factor. Thus, it resulted in the predicted reactivity of the halide ions in the reversed order. The takeaway here is that the application of the $\ce {HSAB}$ concept alone does not always produce the correct prediction.

Additional information

More examples are cited for which the HSAB concept fails. These include:

The favourable combination of $\ce {H^+}$, a hard acid, and $\ce {H^-}$, a soft base (discussed here as well) (ref. 4).
The fact that the rate of reaction between $\ce {Ag^+}$, soft acid, and an alkene, soft base, is slower than that between $\ce {Ag^+}$ and $\ce {OH^-}$, a hard base (ref. 1, p. 110)

References

Fleming, I. Molecular Orbitals and Organic Chemical Reactions (Student Edition). John Wiley & Sons, Ltd. United Kingdom, 2009.
Olmstead, W. N.; Brauman, J. I. Gas-phase nucleophilic displacement reactions. J. Am. Chem. Soc. 1977, 99(13), 4219–4228.
Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer, 2007.
Pearson, R. G.; Songstad, J. Application of the Principle of Hard and Soft Acids and Bases to Organic Chemistry. J. Am. Chem. Soc. 1967, 89(8), 1827-1836.


Why do cookies prevent toothpaste from foaming? If I eat cookies and subsequently brush my teeth such that there is still small amounts of cookie material stuck to my teeth, the toothpaste will not foam as well as usually, making the process less enjoyable. This effect is consistently reproducible with most (sweet) cookies but any not other foods I have tried so far. The toothpaste contains typical foaming agents, such as sodium lauryl sulfate. What could be the mechanism behind this effect?

My first take on this is that the cookies contain fat, which "consumes" the surfactants in the toothpaste (like sodium lauryl sulfate) so it does not foam up the way it normally would. You  might want to see what Wikipedia says about micelles and how they interact with fats. A similar effect can be seen when washing really greasy dishes with an insufficient amount of dishwashing liquid - in which case no suds will form. In your case I don't understand why the effect would not also be observed with other fatty foods such as chocolate or popcorn.


Oxymercuration demercuration, PCC, NaCN/HCl reaction In this reaction of the alkene, the correct answer is below but I don't understand why it goes to that. This is my thought process - reagents 1 and 2 do oxymercuration demercuration and makes a secondary alcohol, then reagent 3 turns the alcohol into a ketone, and finally reagent 4 and 5 performs nucleophilic addition and turns the alcohol into a tetrahedral with CN attached (my steps I drew below). What am I doing wrong here? Thanks

Check the mechanism of the last reaction here.
http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch20/ch20-3-5-1.html

It is not the displacement of nitrile by carboxylic acid(which you do not have, btw). It is the combination of water addition and the displacement of ammonia by second equivalent of water.


Cahn–Ingold–Prelog priority of carboxylic group versus ortho carboxylic acid Should the C=C double bond below be labelled as (E) or (Z)?

When using the Cahn–Ingold–Prelog (CIP) priority system and the ‘Sequence Rules’, double bonds are split into two bonds and the doubly bonded atoms are duplicated.

Thus

is treated as

where (O) and (C) are duplicate representations of the atoms at the other end of the double bond;

whereas

is treated as

Therefore, the name of the structure that is given in the question is (2E)-3-bromo-3-iodo-2-(trihydroxymethyl)prop-2-enoic acid.


What is the most convenient way to prepare ferrous oxide (FeO) in the laboratory? The Wikipedia page for ferrous oxide states that $\ce{FeO}$ can be prepared by the thermal decomposition of iron(II) oxalate, with the following reaction:

$$\ce{FeC2O4 → FeO + CO2 + CO}$$

And that the procedure is conducted under an inert atmosphere to avoid the formation of ferric oxide.

Sicius' Eisengruppe: Elemente der achten Nebengruppe: Eine Reise durch das Periodensystem states that $\ce{FeC2O4}$ should "slowly" be heated in a vacuum and then rapidly cooled down.

Both of these resources, however, omit critical information as to the preparation process. These questions remain:

At which temperature does this reaction really start occurring, and what is the ideal temperature range to carry out this thermal decomposition?
How "fast" should it be cooled down? And what are appropriate methods to do so?

I'd be very thankful for any additional information on the topic.

In appears that the synthetic procedure for preparation of iron(II) oxide from iron(II) oxalate has been first described by Günther et al. [1] and subsequently summarized in Brauer's Handbook of Preparative Inorganic Chemistry [2, p. 1497]:

  Iron (II) Oxide

  I.

  $$\ce{\underset{143.8}{FeC2O4} = \underset{71.8}{FeO} + \underset{28.0}{CO} + \underset{44.0}{CO2}}$$

  Thermal decomposition of $\ce{FeC2O4}$ yields pure $\ce{FeO}$ only under specific conditions.
  The decomposition is carried out in a quartz vessel (Fig. 332) whose lower section is kept at $\pu{850 °C}$ by means of an electric furnace.
  The joint is surrounded by a water-cooled lead coil or a rubber hose.
  The nascent gases should be removed as quickly as possible; for this reason, the reactor is connected to two parallel mercury pumps and a good forepump; the gas is carried into two liquid-nitrogen-cooled traps containing activated charcoal.

  Fig. 332. Preparation of iron (II) oxide.

  The starting $\ce{FeC2O4}$ $(\pu{0.5 - 0.8 g.})$ is placed in the small bulb above the quartz vessel, and the water of crystallization is completely vaporized by heating in vacuum for 12 hours at $\pu{200 °C}.$
  The bulb is turned in the joint, and the $\ce{FeC2O4}$ drops into the heated lower section of the reactor where it is rapidly decomposed to $\ce{FeO},$ $\ce{CO}$ and $\ce{CO2}$ (the decomposition is complete in about 20 seconds). The product $\ce{FeO}$ is retained by a quartz wool plug, which must be loose enough to prevent a buildup of pressure during the decomposition.

  The furnace is now removed and the hot quartz tube is chilled as rapidly as possible in cold water, since $\ce{FeO}$ is unstable in the range of $\pu{300-560 °C}$ and decomposes according to:

  $$\ce{4 FeO = Fe3O4 + Fe}$$

  (this decomposition proceeds most rapidly at about $\pu{480 °C},$ but
  ceases below $\pu{300 °C}).$
  The above procedure yields a jet-black product, readily soluble in dilute acids; it is rapidly oxidized in air, but does not ignite.

Alternatively, iron(III) oxide can be reduced by iron to iron(II) oxide [2, p. 1498]:

  II. The preparation from stoichiometric quantities of commercial $\ce{Fe2O3}$ and reduced iron can also be recommended.
  The mixture and a few drops of water are sealed into a preevacuated quartz tube, heated for about three days at $\pu{900 °C},$ and quenched in cold water.

$$\ce{Fe + Fe2O3 ->[\pu{900 °C}] 3 FeO}$$

Iron(II) oxide can also be produced from iron(III) oxide reduction by carbon monoxide at the temperature range between $\pu{500 °C}$ and $\pu{600 °C}$ (adapted from [3, p. 415]):

$$
\begin{align}
\ce{3 Fe2O3 + CO &->[\pu{400 °C}] 2 Fe3O4 + CO2} \\
\ce{Fe2O3 + CO &->[\pu{500-600 °C}] 2 FeO + CO2} \\
\ce{Fe2O3 + 3 CO &->[\pu{700 °C}] 2 Fe + 3 CO2}
\end{align}
$$

as well as by thermal decomposition of magnetite [3, p. 415]:

$$\ce{2 Fe3O4 ->[>\pu{1538 °C}] 6 FeO + O2}$$

References

Günther, P. L.; Rehaag, H. Über Die Thermische Zersetzung von Oxalaten II. Mitteilung. Darstellung von Reinem Ferrooxyd. Z. Anorg. Allg. Chem. 1939, 243 (1), 60–68. https://doi.org/10/cqzgk8. (in German)
Handbook of Preparative Inorganic Chemistry, 2nd ed.; Brauer, G., Ed.; Academic Press: New York; London, 1965; Vol. 2. (Archive.org)
R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)


How do enzymes affect a reaction's equilibrium? I am not sure to understand something I read in a educational journal (1)
Introduction
Lets consider the following reversible enzyme-catalysed reversible reaction;
$$E + S \leftrightharpoons ES \leftrightharpoons E + P$$
with forward rate constants $k_1$ and $k_2$ and backward rate constants $k_{-1}$ and $k_{-2}$ respectively.
The equilibrium constant of this reaction can be expressed in terms of reaction quotient at equilibrium, in terms of thermodynamic potential and also in terms of rate constants;
$$K_{eq} = (\frac{[P]}{[S]})_{eq} = e^{-\frac{\Delta_r\mu^\circ}{RT}} = \frac{k_1 \cdot k_2}{k_{-1} \cdot k_{-2}}$$
It is possible to experimentally determine the Michaelis constants of this reaction for S and P by studying it in conditions where $[S]>>[P]$ and $[S]<<[P]$ respectively. The Michaelis constants are in fact only combinations of the rate constants;
$$K_M^S = \frac{k_{-1} + k_2}{k_1}$$
$$K_M^P = \frac{k_{-1} + k_2}{k_{-2}}$$
and the maximum forward and backward rate are respectively
$$v^+_{max} = [E] \cdot k_2$$
$$v^-_{max} = [E] \cdot k_{-1}$$
where $[E]$ is the total enzyme concentration (including $[ES]$). Which leads to the Haldane relationship by rewriting $\frac{k_1 \cdot k_2}{k_{-1} \cdot k_{-2}}$;
$$K_{eq} = \frac{v^+_{max} \cdot K_M^P}{v^-_{max} \cdot K_M^S}$$
Problem
I was thinking that enzymes did not affect the position of the equilibrium of a reaction; that position would always be determined by thermodynamics, and the presence of an enzyme would simply affect the speed at which that equilibrium is reached (by lowering the activation energy). However, I stumbled upon a document (1) suggesting that both forward and backward rate expressions ($v^+ = v^+_{max} \frac{[S]}{K_M^S + [S]}$ and $v^- = v^-_{max} \frac{[P]}{K_M^P + [P]}$ respectively) apply on the whole reaction advancement axis and that the point where $v^+ = v^-$ then determines the apparent equilibrium of the reaction. The point of this document was to show that different values of enzyme parameters could correspond to the same $K_{eq}$, therefore an enzyme can set the apparent equilibrium of a reaction to a value totally different from its thermodynamic equilibrium.
I find that point confusing; if the equilibrium of a reaction changes in the presence of an enzyme, to the point that two different enzymes can change the direction of a given reaction (which is also the subject of another article of the same journal (2)), then the equilibrium constant of that chemical reaction cannot be predicted from thermodynamics (ie based on the chemical potential of its reagents). Then, can it be said that an enzyme modifies the chemical potential change of a reaction?
(1) A. Mellors, "The Haldane relationship, enzymes & equilibrium", Biochemical education 1976 vol 4 no. 4
(2) M. L. Uhr, "The influence of an enzyme on the direction of a reaction", Biochemical education 1979 vol 7 no. 1

I believe you have misread the first referenced paper. What it says is that the $V_{max}$ and $K_m$ values for the forward and reverse reactions can be varied to optimize a property such as the maximum velocity (represented by $V_{max}$) in one direction, but are always constrained by the fixed value of $K_{eq}$.
The reason that the $K_m$ values (which are pseudo equilibrium constants) can be adjusted is that they represent a reaction in which the enzyme is a reactant. For the substrate, this reaction is approximately $\ce{E + S <=> ES}$.
Likewise, the second paper also describes how the $V_{max}$ for one direction or the other can be maximized, but again subject to the constraint that the respective $K_m$ values must vary such that the overall $K_{eq}$ maintains its fixed value.


Good sources to understand Stat Mech with concepts I never had a proper course on Statistical Mechanics during my undergrad. I've tried picking up the subject by reading books, but I find that many focus on the mathematical derivations and I get easily lost in the details.

I was wondering if someone could point me to some useful reference materials where I could better understand concepts that are confusing to me. For example, why do chemical systems overcome energy barriers? I know that when a molecule samples a potential energy surface, the different states are occupied with probabilities relating to their energy difference according to the Boltzmann law. But, in classical physics, a ball rolling down a hill would eventually get stuck in the energy minima? I fail to see where do chemical systems get the energy from to be constantly in motion.

Sorry if this is a vague question or ill-posed, but as you can see I have some confusion of concepts in my mind which is not allowing me to develop a more accurate question.

I highly recommend Molecular Driving Forces by Dill & Bromberg.

https://books.google.com/books/about/Molecular_Driving_Forces.html?id=hdeODhjp1bUC

Contains many examples throughout to help with conceptual and mathematical understanding.


re/si face of alkenes How to determine whether the face of the attached alkene is re or si?
If assigning priorities in regard to the left C atom it is re, but for the right C atom it is si. How is it actually determined?

I’m pretty sure that this depends on the atom attacked. Most attacks on double bond are stepwise to a certain extent. And even when they aren’t, the absolute configuration should be discussed on a per-atom basis.

But if you have a concerted reaction and want to label both: (2si,3re).


What is the importance of iron in complexes and in the process of electron transport? What is the importance of the iron in complexes and in electron transport? This question is related to synthesis of ATP. I have read about iron-sulfur proteins, but I don't understand how this works.

The Fe-S proteins are part of the electron-transport chain (together with other complex molecules such as other proteins and coenzymes etc.) In my opinion, I think it would be unfair just to describe the Fe-S complex without understanding bits of the processes and other important molecules involved in this system since it involves a chain of interlinked events.

Components of the Electron-Transport Chain

  Many of the proteins embedded in the inner mitochondrial membrane are
  organized into the four respiratory complexes of the
  electron-transport chain. Each complex consists of several protein
  components that are associated with a variety of redox-active
  prosthetic groups with successively increasing reduction potentials.

The mitochondrial respiratory chain consists of a series of sequentially acting electron carriers, most of which are integral proteins with prosthetic groups capable of accepting and donating either one or two electrons. Three types of electron transfers occur in oxidative phosphorylation:

direct transfer of electrons, as in the reduction of $\ce{Fe^3+}$ to $\ce{Fe^2+}$;
transfer as a hydrogen atom ($\ce{H^+ + e_}$); and
transfer as a hydride ion ($\ce{:H^-}$), which bears two electrons.

In addition to NAD and flavoproteins, three other types of electron-carrying molecules function in the respiratory chain: a hydrophobic quinone (ubiquinone) and two different types of iron-containing proteins (cytochromes and iron-sulfur proteins). Ubiquinone (also called coenzyme Q, or simply Q).

The electron-transport chain involves several different molecular species, including:

Flavoproteins, which contain tightly bound FMN or FAD as prosthetic groups
and which may participate in one- or two-electron transfer events.
Coenzyme Q, also called ubiquinone (and abbreviated CoQ or UQ), which can function in either one- or two-electron transfer reactions.
Several cytochromes (proteins containing heme prosthetic groups), which function by carrying or transferring electrons), including cytochromes b, c, c 1, a, and a3. Cytochromes are one-electron transfer agents in which the heme iron is converted from $\ce{Fe^3+}$ to $\ce{Fe^2+}$ and back.
A number of iron–sulfur proteins, which participate in one-electron transfers involving the $\ce{Fe^3+}$ and $\ce{Fe^2+}$ states.
Protein-bound copper, a one-electron transfer site that converts between $\ce{Cu^+}$ and $\ce{Cu^2+}$.

The Electron-Transport Chain Can Be Isolated in Four Complexes

1. Complex I (NADH Dehydrogenase or NADH–Coenzyme Q Oxidoreductase)

Complex I (also called NADH dehydrogenase) passes electrons from NADH to CoQ. Situated in the inner mitochondrial membrane,e contains one molecule of flavin mononucleotide (FMN; a redox-active prosthetic group that differs from FAD only by the absence of the AMP group) and eight or nine iron–sulfur clusters that participate in the electron transport process.

a.    Iron–Sulfur Clusters

These are redox active. They commonly occur as prosthetic groups in iron–sulfur proteins.
There are four common types of iron–sulfur clusters: Those designated [2Fe–2S] and [4Fe–4S]. Clusters consist of equal numbers of iron and sulfide ions and are both coordinated to four protein Cys sulfhydryl groups. The [3Fe–4S] cluster is essentially a [4Fe–4S] cluster that lacks one Fe. One means of identifying these clusters
utilizes the fact that their sulfide ions are acid labile.
The oxidized and reduced states of all iron–sulfur clusters differ by one formal charge regardless of their number of Fe ions. This is because the Fe ions in each cluster form a conjugated system and thus can have oxidation states between the +2 and +3 values possible for individual Fe ions.

The overall equation for the reaction is:

$$\ce{NADH +  H+ +  CoQ -> NAD+  +  CoQH2}$$

b. The Coenzymes of Complex I

FMN and ubiquinone (CoQ), the coenzymes of Complex I, can each adopt three oxidation states. Although NADH can only participate in a two-electron transfer, both FMN and CoQ are capable of accepting and donating either one or two electrons because their semiquinone forms are stable. In contrast, the cytochromes of Complex III, to which reduced CoQ passes its electrons, are only capable of one-electron reductions.
FMN and CoQ, therefore, provide an electron conduit between the two-electron donor NADH and the one-electron acceptors, the cytochromes.

2. Complex II [Succinate Dehydrogenase or Succinate–Coenzyme Q Oxidoreductase (SQR)]

Complex II, which is also the citric acid cycle enzyme succinate dehydrogenase, passes electrons from succinate to CoQ to yield fumarate and CoQH2. It does so with the participation of a covalently bound FAD, a [2Fe–2S] cluster, a [4Fe–4S] cluster, a [3Fe–4S] cluster, and one heme b560.

Complex II is important because it injects these relatively high-potential electrons into the electron transport chain.  It catalyzes the transfer of electrons to coenzyme Q.

However, its source of electrons (in other words, the substance being oxidized) differs from the oxidizable substrate (NADH) acted on by NADH-CoQ oxidoreductase. In this case, the substrate is succinate from the citric acid cycle, which is oxidized to fumarate by a flavin enzyme;

$$\ce{Succinate + E-FAD -> Fumarate +  E-FADH2}$$

The notation E-FAD indicates that the flavin portion is covalently bonded to the enzyme. The flavin group is reoxidized in the next stage of the reaction as another iron–sulfur protein is reduced:

$$\ce{E-FADH2  +  FeOSoxidized -> E-FAD _ Fe-Sreduced}$$

This reduced iron–sulfur protein then donates its electrons to oxidized coenzyme Q, and coenzyme Q is reduced;

$$\ce{Fe-Sreduced  +  CoQ  + 2 H+ ->  Fe-Soxidized  +  CoQH2}$$

The overall reaction is:

$$\ce{Succinate  + CoQ -> Fumarate  + CoQH2}$$

a. Cytochromes

  Cytochromes are redox-active proteins that contain heme groups that
  reversibly alternate between their Fe(II) and Fe(III) oxidation states
  during electron transport.

3. Complex III (Coenzyme Q–Cytochrome c Oxidoreductase or Cytochrome bc1 Complex)

Complex III passes electrons from reduced CoQ to cytochrome c. It contains four redox cofactors: two b-type hemes, a c-type heme, and one [2Fe–2S] cluster. This complex -CoQH2-cytochrome c oxidoreductase (also called cytochrome reductase), catalyzes the oxidation of reduced coenzymeQ (CoQH2). The electrons produced by this oxidation reaction are passed along to cytochrome c in a multistep process. The overall reaction is;

$$\ce{CoQH2  +  2 Cyt c[Fe(III)]  -> CoQ  + 2 Cyt c[Fe(II)]  +  2 H+}$$

Cytochrome c

Cytochrome c is a peripheral membrane protein that is loosely bound to the outer surface of the inner mitochondrial membrane. It alternately binds to cytochrome c1 (of Complex III) and to cytochrome c oxidase (Complex IV) and thereby functions to shuttle electrons between them.

Complex IV (Cytochrome c Oxidase)

Cytochrome c oxidase (COX or CcO), the terminal enzyme of the electron transport chain, catalyzes the one-electron oxidations of four consecutive reduced cytochrome c molecules and the concomitant four-electron reduction of one  $\ce{O2}$ molecule to yield $\ce{H2O}$:

$$\ce{4cytochrome c^2+  +   4H^ + O2   -> 4 cytochrome c3+ + 2H2O}$$

Overview:

Summarised version:

The mitochondrial electron-transport chain. The pathways of electron transfer (black) and proton pumping (red) are indicated. Electrons are transferred between Complexes I and III by membrane-soluble CoQ (Q) and between Complexes III and IV by the peripheral membrane protein cytochrome c (Cyt c). Complex II (not shown) transfers electrons from succinate to CoQ

These Fe-S clusters are present in the complexes described above and play a big part in the electron transport chain, as they are redox active hence allow the transport of electrons along the chain through a series of redox reaction yielding $\ce{H2O}$ in the final reduction of $\ce{O2}$ molecule (producing the ATP molecules along the way). Although this is not exhaustive of the whole process it gives the "main" idea of molecules (including the so-called iron-sulfur proteins you mentioned in the question) and processes involved in the ETC. It was just impossible for me to explain only iron-sulfur proteins without getting into detailed processes

Hope this helps you

References

Biochemistry (Campbell and Farrel) 5th ed (Chapter 20)

Biochemistry (Voet and Voet) (Chapter 22)

Biochemistry (Grisham) 4th ed(Chapter 20)

Lehninger  Principles of Biochemistry (Chapter 19)


Can a system of 2 consecutive biphenyl systems be optically active , even if any adjacent biphenyl systems aren't optically active? This question came in an online test, where they asked whether or not 22,26-dibromo-12,32-dichloro-23,25-diiodo-16,36-dimethyl-11,21:24,31-terphenyl was optically active.

The answer is that, it is optically active.

According to Solomon and Fryhle:

  "Biphenyl is an asymmetric compound so it must not have a plane of symmetry or center of symmetry to be optically active".
  Hence compounds which have 2 same adjacent groups on a single phenyl in any biphenyl system cant be considered optically active .

Applying that concept above as same groups (Br and I) are attached to the middle phenyl system, the individual biphenyl systems on both sides are optically inactive, hence the compound should be optically inactive.

Please tell me where I am going wrong in my interpretation.

The key point you have to remember is that optical activity isn't decided by a single section of the molecule but rather it is a property of the entire molecule - and while not being able to find a mirror plane is almost a guarantee, the nail in the coffin to decide that a molecule is optically active is if you cannot find an improper rotation axis (reflect and rotate).

There is no mirror plane in the tri-phenyl molecule you show (cut it into left and right the Br and I don't overlap; cut it up and down the methyl and chloride won't overlap; and so on and so forth). Hence, it is optically active.


Why is isoindole unstable? According to this paper, isoindole is unstable because the 1-carbon in the isoindolidine tautomer is very electron-deficient.  I figured that the electron-deficiency of this carbon must be because it is bonded to an aromatic carbon and double bonded to a nitrogen, both of which atoms are electron-withdrawing.

By this logic, I would expect other carbons in this situation (C(aryl)-C=N) to also be very electron-deficient.  However, aromatic Schiff bases seem pretty common, for example salen.  What am I missing?

Kopecký et.al. indeed utter the word unstable, but not in the sense of its recommended definitions, which were established later.

  Although the parent compound, isoindole (I), has resisted isolation, the existence of this unstable species has been shown by trapping with dienophiles (i).

The problem here is that stability always requires a point of reference, which is not given in the quote above.
I would assume that the authors use unstable (or instability) as a way to say the compound cannot be isolated. They go on and justify why this compound behaves in this way, describing what I would say is its reactivity. While their argument is sound and I believe correct, the choice of their wording is a little bit ambiguous (in today's terms).
We clearly need to distinguish a compounds reactivity (towards a certain reaction) from its stability within a common set of parameters. As such the following statements are correct, but neither of them is a justification for the other.

2H-Isoindole is a less stable $\ce{C8NH7}$ isomer than 1H-indole.
2H-Isoindole is more reactive towards dienophiles than 1H-indole.

While statement one is easy enough to prove computationally, it is a lot harder to explain or analyse. Quick calculations on the DF-M06L/def2-TZVPP level of theory gives the following hypothetical reaction energy:
$$\begin{align}
\ce{1$H$-indole &-> 2$H$-isoindole} & \Delta_\mathrm{r}G^\circ &= +38.6~\pu{kJ mol^-1}
\end{align}$$

Unfortunately Kopecký et.al. offer no explanation for this, because they are more or less focused on explaining why it is "unstable", by which they most likely mean highly reactive or not isolatable. They are concerned with the reaction paths that lead away from this species and find that 1H-isoindole plays a big (probably most important) role in this.

  The [...] electron deficiency for position 1 of the isoindolenine [...] leads us to conclude leads us to conclude that the reactive species responsible for the instability of this series of compounds probably is the isoindolenine [...].
  (Note that the numbering differs from the IUPAC name and the position referred to as 1 is actually 3.)

I think this analysis is spot on a holds up to modern standards of computational methodology. The 1H-tautomer indeed opens up a very big variety of decomposition and reaction pathways, and it is readily accessible at room temperature (DF-M06L/def2-TZVPP):
$$\begin{align}
\ce{2$H$-isoindole &-> 1$H$-isoindole} & \Delta_\mathrm{r}G^\circ &= +3.4~\pu{kJ mol^-1}
\end{align}$$

The reason for this is - and this is the point where Joule and Mills get it completely wrong - the retention (not the creation) of the aromatic system in the benzene moiety. The chemical behaviour (reactivity towards cycloadditions) is not determined by the lack of a 'complete' benzene ring. But that is us slipping into statement 2, which we are not quite ready yet.
Let us first dissect the latter statement in the same book (p.447), which is equally as wrong:

  Isoindole, benzo[c]thiophene and isobenzofuran are much less stable than their isomers, indole, and benzo[b]thiophene and benzo[b]furan.

Although they offer no appreciable effort to prove this statement they are probably correct. At least the calculations for the $\ce{C8NH7}$ system confirm that. They even use the correct (or recommended) definition of stable in this case.

  This is undoubtedly associated with their lower aromaticity, which can be appreciated qualitatively by noting that in these [c]-systems, the six-membered ring is not a complete benzenoid unit.

Again, they offer no proof for that other than a hand-wavy argument based on a single Lewis-structure. If we include resonance into the mix then we have an at least as reasonable argument as this to disprove it.

If we have a look at the molecular orbitals of the compounds, we will see that there is more to this than just one resonance description. I have pulled out two characteristic molecular orbitals for example, but the complete set can be found here. On the left side (top) we have 1H-indole and on the right side (bottom) we have 2H-isoindole. Apart from a light shift of the electron density due to the loosened symmetry in indole, the orbitals are pretty much the same.

When we take the 1H-isoinole isomer into the mix, we see that the aromatic system of the benzene ring is retained completely.

Aromaticity is still not a fully understood concept and as a result there is not really a rigorous concept that can be applied (see the definitions below). There will probably be debate about it for the next decades. (Hopefully my bounty will pay off and we can find out more in this question.)
As a result we don't really have a handle on judging whether or not 2H-isoindole has a lower aromaticity than 1H-indole. And according to Kopecký's paper, with available methods at this time, this cannot be backed up with theoretical results.

While I hoped to find some experimental evidence for the existence of isoindole due to the general progress within the last fifty years, I was still unable to find a single source.

I'll skip over the rest of the quote because the stated facts are certainly accurate, just the reasoning with the 'incomplete' benzene ring is false.

Let's look at statement two for a while.
The reactivity of 2H-isoindole towards dienophiles is very easily explained looking at the HOMO of the compound; for comparison the HOMO of butadiene on the right (bottom).

The very easy rationale is that the HOMO is of the right symmetry for a Diels-Alder reaction, which additionally retains the aromatic character of the benzene moiety.
Now a look at the HOMO (left/ top) of 1H-indole will tell us that the only viable attack can be carried out on the 4,7-positions, leaving a much smaller aromatic moiety, i.e. pyrrole, than isoindole. The only way for an inverse electron demand  Diels-Alder reaction would also be in the benzene moiety as the LUMO (right/ bottom) shows.

This is why one molecule reacts readily, while another has to be forced (really, really hard) or reacts as the dienophile via 'the double bond in the pyrrole system'.

After all that information the question still remains, or rather the question has become

"Why is isoindole less stable than indole?"

As I said before, this is really not easy to interpret. In general, lifting symmetry restrictions results in a more stable configuration, as fewer boundary conditions equal more flexibility and that usually gives a lower energy.
One aspect is certainly also that the lone pair of the nitrogen will be better delocalised in indole as there are more carbon atoms in proximity. This is actually something visible in the molecular orbitals.
Apart from that, neither geometries differ much from idealised benzene ($\mathbf{d}(\ce{C-C})=138.7~\pu{pm}$) or pyrrole($\mathbf{d}(\ce{C-N})=136.7~\pu{pm}$, $\mathbf{d}(\ce{C-C'})=137.1~\pu{pm}$, $\mathbf{d}(\ce{C'-C'})=141.4~\pu{pm}$, picture), so that a significant distortion energy is probably not a large contributing factor.

Other than that, I don't know. I also cannot explain, why the other aromatic $\ce{C8NH7}$ isomers are even less stable than isoindole. That would probably require more extensive calculations and analyses.
\begin{array}{lr}\hline
\text{Isomer}& \Delta E_\mathrm{el}/\pu{kJ mol^-1} & \Delta G/\pu{kJ mol^-1}\\\hline
\text{1$H$-indole} & 0.0 & 0.0\\
\text{2$H$-isoindole} & 37.3 & 38.6\\
\text{indolizine} & 52.0 & 50.6\\
\text{1$H$-cyclopenta[$b$]pyridine} & 83.4 & 82.4\\
\text{2$H$-cyclopenta[$c$]pyridine} & 79.7 & 79.1\\\hline
\end{array}

  By this logic, I would expect other carbons in this situation, $\ce{C_{ar}-\color{\red}{C}=N}$, to also be very electron-deficient. However, aromatic Schiff bases seem pretty common, for example salen. What am I missing?

It is true that the carbon atoms in question will be very electron deficient, and they will probably be a good attach point for nucleophiles. In the salen ligand the nitrogens are probably stabilised by intramolecular hydrogen bonds, and while in complexes the entropy factor should make is very stable, too.

On another note, the isoindole moiety is also  very, very common. Just think heme ligands and complexes.

One final thought: Carbonic acid decomposes rapidly in water, however, it is seldom questioned, that the molecule itself is stable under certain conditions.

My personal preferred definition for stable:  An certain arrangement of atoms that forms a local minimum on a potential energy hypersurface and therefore could theoretically probed is stable.

Helpful definitions from the IUPAC gold book

  unstable
  The opposite of stable, i.e. the chemical species concerned has a higher molar Gibbs energy than some assumed standard. The term should not be used in place of reactive or transient, although more reactive or transient species are frequently also more unstable. (Very unstable chemical species tend to undergo exothermic unimolecular decompositions. Variations in the structure of the related chemical species of this kind generally affect the energy of the transition states for these decompositions less than they affect the stability of the decomposing chemical species. Low stability may therefore parallel a relatively high rate of unimolecular decomposition.)

  stable
  As applied to chemical species, the term expresses a thermodynamic property, which is quantitatively measured by relative molar standard Gibbs energies. A chemical species A is more stable than its isomer B if $\Delta_\mathrm{r}G^\circ > 0$ for the (real or hypothetical) reaction $$\ce{A -> B},$$
  under standard conditions. If for the two reactions:
  \begin{align}
\ce{P &-> X + Y} & (\Delta_\mathrm{r}G_1^\circ)\\
\ce{Q &-> X + Z} & (\Delta_\mathrm{r}G_1^\circ)\\
\end{align}
  $\Delta_\mathrm{r}G_1^\circ > \Delta_\mathrm{r}G_2^\circ$, P is more stable relative to the product Y than is Q relative to Z. Both in qualitative and quantitative usage the term stable is therefore always used in reference to some explicitly stated or implicitly assumed standard. The term should not be used as a synonym for unreactive or 'less reactive' since this confuses thermodynamics and kinetics. A relatively more stable chemical species may be more reactive than some reference species towards a given reaction partner.

  aromatic

  In the traditional sense, 'having a chemistry typified by benzene'.
  A cyclically conjugated molecular entity with a stability (due to delocalization ) significantly greater than that of a hypothetical localized structure (e.g. Kekulé structure ) is said to possess aromatic character. If the structure is of higher energy (less stable) than such a hypothetical classical structure, the molecular entity is 'antiaromatic'. The most widely used method for determining aromaticity is the observation of diatropicity in the 1H NMR spectrum.
  See also: Hückel (4n + 2) rule, Möbius aromaticity
  The terms aromatic and antiaromatic have been extended to describe the stabilization or destabilization of transition states of pericyclic reactions The hypothetical reference structure is here less clearly defined, and use of the term is based on application of the Hückel (4n + 2) rule and on consideration of the topology of orbital overlap in the transition state. Reactions of molecules in the ground state involving antiaromatic transition states proceed, if at all, much less easily than those involving aromatic transition states.

  aromaticity
  The concept of spatial and electronic structure of cyclic molecular systems displaying the effects of cyclic electron delocalization which provide for their enhanced thermodynamic stability (relative to acyclic structural analogues) and tendency to retain the structural type in the course of chemical transformations. A quantitative assessment of the degree of aromaticity is given by the value of the resonance energy. It may also be evaluated by the energies of relevant isodesmic and homodesmotic reactions. Along with energetic criteria of aromaticity, important and complementary are also a structural criterion (the lesser the alternation of bond lengths in the rings, the greater is the aromaticity of the molecule) and a magnetic criterion (existence of the diamagnetic ring current induced in a conjugated cyclic molecule by an external magnetic field and manifested by an exaltation and anisotropy of magnetic susceptibility). Although originally introduced for characterization of peculiar properties of cyclic conjugated hydrocarbons and their ions, the concept of aromaticity has been extended to their homoderivatives (see homoaromaticity), conjugated heterocyclic compounds (heteroaromaticity), saturated cyclic compounds (σ-aromaticity) as well as to three-dimensional organic and organometallic compounds (three-dimensional aromaticity). A common feature of the electronic structure inherent in all aromatic molecules is the close nature of their valence electron shells, i.e., double electron occupation of all bonding MOs with all antibonding and delocalized nonbonding MOs unfilled. The notion of aromaticity is applied also to transition states.


Why doesn't the carbon dioxide produced by a fire put out said fire? Carbon dioxide ($\ce{CO2}$) is used in some fire extinguishers to put out fire, but a combustion reaction itself produces $\ce{CO2}$. Why do we get fire then?

$$\ce{CH4 + O2 -> CO2 + H2O}$$

Why doesn't this $\ce{CO2}$ put out the fire?

The fuel for the chemical reaction is the CH4 (in this case) and the oxygen.  Assuming the CH4 isn't going to run out any time soon, the chemical reaction will continue as long as there is enough oxygen accessible. The force behind fire extinguishers can replace the oxygen available to the fire in its immediate area with CO2, and without one of the reactants (oxygen) available, the reaction won't continue.

The CO2 produced by the fire in an open area isn't going to replace the oxygen of the surrounding area.  However, in a closed system (for example, if you put a lid on top of a burning candle), then the air inside of the candle will become deficient in oxygen and rich in CO2 after a certain amount of time and the flame will go out, even without blowing on it or forcing it out.


Concentration of Hydrogen Ions I am slightly confused about one certain task.

  There are 4 different solutions ($\ce{NH3}$, $\ce{HCl}$, $\ce{NaOH}$, $\ce{CH3COOH}$), their concentration is the same $(c = \pu{1 mol L-1}).$ Which solution has pH that is higher than 7, but lower than 14?

It is obvious that $\ce{HCl}$ and organic acid cannot have such concentration of hydrogen ions. Both ammonia and natrium hydroxide have base properties. $\ce{NH3}$ is a weaker base though. I am not sure how to find the concentration of hydroxide in $\ce{NH3}$ and $\ce{NaOH}$  properly.

Thank you for the explanation.

Since $\ce{NaOH}$ is a strong base, the pH can be directly calculated from the analytic concentration (1M):

$\ce{pOH} = -\log\ce{[OH-]} = -\log(1) = 0  $

$\ce{pH} = 14 - \ce{pOH} = 14 - 0 = 14$

For ammonia, you need to consider its reaction in water and the associated $\ce{K_b}$ value,

$\ce{NH3 + H2O -> NH_4+ + OH-}$

The equilibrium concentrations for an unknown $x$ reacted amount (in concentration units) would be

$\ce{[NH3]} = \ce{[NH3]_0} - x$

$\ce{[OH-]} = x$

$\ce{[NH4+]} = x$

$$\ce{K_b = \frac{[NH4+][OH-]}{[NH3]} = \frac{x^2}{[NH3]_0 - x}}$$

By plugging in the proper $\ce{K_b}$ value for ammonia and its analytical concentration ($\ce{[NH3]_0}$), you can calculate $x$, from which

$$\ce{pOH} = -\log(x)$$


How does Fermi's golden rule yield a specific rate? According to Fermi’s golden rule, the rate of an electronic transition is proportional to the magnitude squared of $\langle i|\hat{H}|f\rangle$.  Since energy is always relative to some reference point, I could add an arbitrary constant to the Hamiltonian, and it would do nothing but shift all the energy eigenvalues by that constant.  So I don’t understand how Fermi’s golden rule can yield a specific rate – changing the Hamiltonian by an arbitrary constant will (despite not changing the system) change the calculated rate.  Can anyone explain?

I'm fairly sure the states $\require{\begingroup} \begingroup \newcommand{\ket}[1]{|#1\rangle} \newcommand{\braket}[1]{\langle #1 \rangle} \ket{i}$ and $\ket{f}$ are eigenstates of the unperturbed Hamiltonian $\hat{H}_0$. Therefore, by virtue of the Hermiticity of $\hat{H}_0$, they are necessarily orthonormal

$$\braket{i|f} = \delta_{ij}$$

therefore if we define $\hat{V'} = \hat{V} + k$ where $\hat{V}$ is the perturbation Hamiltonian and  $k$ is some arbitrary constant

$$\begin{align}
\braket{i|\hat{V'}|f} &= \braket{i|\hat{V}|f} + \braket{i|k|f} \\
&= \braket{i|\hat{V}|f} + k\braket{i|f} \\
&= \braket{i|\hat{V}|f}
\end{align}$$

if $\ket{i}$ and $\ket{f} \endgroup$ are different states. If they are the same state, then you are calculating the transition probability from one state to itself - not a very useful concept.


Safe yet strong exothermic reaction? I'm studying chemical engineering at uni (Belgium), and this year (my 2nd) we have to develop and build a stirling engine. It has to be powered by a relatively safe exothermic reaction (not dangerous for bystanders, so no formation of toxic fumes,...) and the whole thing is electrically controlled but that's not important here,

I've come up with this so far:

Sodium acetate crystallization
Hydratation of anhydrous copper sulphate
calcium chloride and water
sodiumhypochlorite oxidizing sodiumsulfite

An 'open flame' reaction is not allowed sadly enough..
I'm hoping to at least reach 70-80 degrees Celsius (158-176° Fahrenheit)

Chances are quite big that there are better reactions but I really can't think of anything else..

Thanks A LOT in advance everyone!

Here are some more options:

Pd-catalyzed oxidation of dilute hydrogen gas in air: $\ce{2H2 + O2 -> 2H2O}$  If you use a dilute hydrogen gas mix (say 95% nitrogen and 5% hydrogen), no flaming combustion can occur in air.  But oxidation still happens, at least when catalyzed by palladium particles. This is how microbiological anaerobic chambers maintain a zero-oxygen environment.  You can probably buy some catalyst fairly cheaply from a microbiology anaerobic chamber company like Coy labs.
Phosphorus pentoxide hydration.  Wikipedia says this reaction, $\ce{P4O10 + 6 H2O -> 4 H3PO4}$, has $\Delta H_{reac}=-177\text{ kJ/mol}$.  There is no flame and no fumes, but the reaction does create phosphoric acid, which is corrosive and can be dangerous.
A modification of your sulfite / bleach idea: use sodium dithionite instead of sodium sulfite.  Sodium dithionite ($\ce{Na2S2O4}$) is air-unstable but if you can work out a way to handle without air exposure it will be more exothermic than sulfite.  Instead of dithionite itself you could use the much easier to handle Rongalite and hypochlorite.  Possibly, using calcium hypochlorite suspensions in water (mostly insoluble) will allow you to use higher concentrations of hypochlorite and Rongalite without the reaction rate and temperature skyrocketing (since the rate will be controlled by $\ce{Ca(OCl)2}$ dissolution).


Existence of Hartree–Fock limit I'm reading McQuarrie's Physical chemistry textbook [1]. On p. 281, the author explains:

If we use a more flexible trial function of the form in which $\psi(\mathbf{r_1},\mathbf{r_2})$ is a product of one-electron functions, or orbitals,
$$\psi(\mathbf{r_1},\mathbf{r_2})= \phi(\mathbf{r_1})\phi(\mathbf{r_2})\tag{8.15}$$
and allow $\phi(r)$ to be completely general, then we reach a limit that is both practical and theoretical. […]
This limiting value is the best value of the energy that can be obtained using trial function of the form of a product of one-electron wave equations (Equation 8.15). This limit is called the Hartree-Fock limit.

But why does Hartree–Fock limit exist? If we increase the number of parameters of the trial function we use, we may expect better calculation results. Nevertheless, why is the wave function multiplied by the two trial functions limited to the accuracy that can be achieved?
My guess is that the functional form does not properly consider the interelectronic correction term by treating the two trial functions independently. But can't this difference be overcome by using more parameters? I wonder if I can theoretically explain the existence of the Hartree–Fock limit.
In addition, if this limit exists, the function will have to be altered to produce better calculation results. I wonder what theory computational chemistry has developed about this.
Reference
McQuarrie, D. A.; Simon, J. D. Physical Chemistry: A Molecular Approach; University Science Books: Sausalito, California, 1997. ISBN 978-0-935702-99-6.

The issue is: what parameters can we introduce? As for changing the form of the trial wavefunction to something else to achieve a lower energy, this is prohibited by the variational principle, which HF theory satisfies.
In short (and within the context of Hartree-Fock), the variational principle guarantees that any given trial function will produce an energy equal to or greater than the solution provided by Hartree-Fock.


Substitute molten salt bath with temperature range from 400–500 °C I'm looking for a suitable substitute for a $\ce{NaNO2/KNO3}$ salt bath (for quite obvious reasons due to the other applications of the latter chemical). The salt mixture needs to have around 400 to 550 °C of molten working temperature (the lower ranges are not interesting), and preferably be reasonably safe to work with (as safe as the aforementioned salt bath, so totally harmless isn't the goal). The purpose is precise annealing of brass alloys in a home shop environment (so no fume hoods and blast cabinets available).

I wasn't sure what other conditions you needed for your salt bath, yet I listed few for your convenience together with the salt bath of $\ce{KNO3/NaNO2}$ (bath 5) for comparison:
$$
\begin{array}{ccc}
\text{Bath#} &\text{Composition} & \text{Approx. melting poing} & \text{Working temp. range} \\\hline
1 & \ce{NaCl}, \ce{KCl}, \ce{BaCl2}, \ce{CaCl2}  & \pu{400 ^\circ C} & \pu{500-800 ^\circ C} \\
 & (10\!-\!15 \!:\!20\!-\!30 \!:\!40\!-\!50 \!:\!15\!-\!20) &  &  \\
2 & \ce{NaCO3}\, (45\!-\!55), \ce{KCl}\, (55\!-\! 45)  & \pu{450 ^\circ C} & \pu{550-900 ^\circ C} \\
3 & \ce{NaCl}\, (20), \ce{KCl}\, (30), \ce{BaCl2}\, (50)  & \pu{540 ^\circ C} & \pu{570-900 ^\circ C} \\
4 & \ce{NaNO3}\, (100)  & \pu{370 ^\circ C} & \pu{400-600 ^\circ C} \\
5 & \ce{KNO3}\, (50\!-\!60), \ce{NaNO2}\, (50\!-\!40) & \pu{135 ^\circ C} & \pu{160-550 ^\circ C} \\
\end{array}
$$

I also listed $\ce{NaNO3}$ bath in case you won't mind using it since it is not potassium salt and working range is ideal for your requirement. As indicated, you can change composition as much as you wanted to to get right melting point and the working temperature range you wanted to.


Why are the "extents of reaction" the same for all reactants and products? Consider the follwoing reaction:
$$\ce{a A + b B -> c C + d D}\tag{1}$$
where $\ce A$ and $\ce B$ are the reactants, $\ce C$ and $\ce D$ are the products, and $a (<0)$, $b(<0)$, $c(>0)$, and $d(>0)$ their respective stoichiometric coefficients. The extents of reaction are respectively defined for these reactants and products as:
$$\begin{align}
\xi_A(t) &= \frac{n_A(t) - n_A(0)}{a}\tag{2.1}\\
\xi_B(t) &= \frac{n_B(t) - n_B(0)}{b}\tag{2.2}\\
\xi_C(t) &= \frac{n_C(t) - n_C(0)}{c}\tag{2.3}\\
\xi_D(t) &= \frac{n_D(t) - n_C(0)}{d}\tag{2.4}
\end{align}$$
However, I have found in many references (e.g. source) the extent of reaction defined for the whole reaction as:
$$\xi(t) = \frac{n_A(t) - n_A(0)}{a}
         = \frac{n_B(t) - n_B(0)}{b}
         = \frac{n_C(t) - n_C(0)}{c}
         = \frac{n_D(t) - n_D(0)}{d}\tag{3}$$
I do not understand how you can justify that the extent of reaction of the different reactants and products are equal.


Azimuthal Quantum Number In the radial equation of hydrogen atom the differential equation is described by

But why is l taken to be integer. I know the principal quantum number n correspond to energy levels so that's why it's taken as integer. Why should azimuthal quantum number be taken as integer though

There are two parts to the answer to your question. (a)  When solving a differential equation the boundary conditions must be specified (this is not always emphasised enough in textbooks), and (b) these conditions are determined by the physics of the problem via the postulates of quantum mechanics, including the nature of the wavefunction. Quantum mechanics (QM) has no derivation; we accept its postulates and QM survives only because experiment has so far always confirmed it.
As an example of quantisation consider the particle in a box. Here there are all sorts of solutions to the differential equation, i.e Schroedinger's equation. However, as the walls are infinitely high the wavefunction must have zero amplitude here and this means that the only solutions that remain and are physically realistic are when quantum number $n$ is quantised. The fact that the wavefunction must be zero at each wall are the boundary conditions.
A similar situation occurs on a particle on a ring where the angular momentum quantum number $l$ is quantised only because we insist, by applying the postulates of QM that the wavefunction repeats itself exactly each $2\pi$ round the circle. This is in effect the boundary condition.
In the H atom the equation is far more complicated but the general conditions imposed on the wavefunction remains the same; the imposition of boundary conditions, as determined by QM, results in the generation of integer quantum numbers so that theory matches experimental data.


Calculating standard enthalpy of reaction, which reactant to choose for the amount of substance For example, this reaction: $$\ce{2NaOH + H2SO4 -> Na2SO4 + H2O}$$

Suppose both concentrations are $1\ \mathrm{mol\ l^{-1}}$ (standard condition). There is $0.02\ \mathrm{mol}$ ($20\ \mathrm{cm^3}$) of $\ce{NaOH}$, and $0.01\ \mathrm{mol}$ ($10\ \mathrm{cm^3}$) of $\ce{H2SO4}$. So neither reactant is the limiting reactant.

Well, I know one way is to find the standard enthalpy of formation of all components and subtract the ones of products from reactants'. My question is if only experiment data is given.

For temperature change of $\Delta T$,
$\Delta H = mc\Delta T = (0.03)\times(4200)\times\Delta T$,
where $m$ is the mass of the solution. Since the density should be similar to water $1000\ \mathrm{kg\ m^{-3}}$,
$20\ \mathrm{cm^3} + 10\ \mathrm{cm^3} = 30\ \mathrm{cm^3} = 0.00003\ \mathrm{m^3}$
which translates to $0.03\ \mathrm{kg}$. $c$ is just the specific heat capacity of water, it was assumed that the specific heat capacity of the mixture is more or less the same as the water's.

Now the standard one $\Delta_\mathrm{r}H^\circ$ requires division by the amount of substance, but which one – $0.02\ \mathrm{mol}$ or $0.01\ \mathrm{mol}$? What if $\ce{NaOH}$ is $0.021\ \mathrm{mol}$? Or $\ce{H2SO4}$ is $0.011\ \mathrm{mol}$?

The difference between the Enthalpy of Reaction $\Delta_{\mathrm{r}} H$ and the Standard Enthalpy of Reaction $\Delta_{\mathrm{r}} H^{\ominus}$

The standard value is not related to a standard temperature (although Standard Enthalpies of Reaction are often given with respect to the standard temperature - they could have been given for any arbitrary temperature). The term "Standard" refers to standard pressure and the reactants and products being in their standard states. The standard states are usually those of the pure compounds. Furthermore gases are usually assumed to behave like ideal gases. This definition of standard states means that Standard Enthalpies of Reaction are often purely calculational quantities since the underlying reactions are practically not feasible. For example, the reactants usually have to be mixed in order to react with one another, meaning that they are not present in their pure form. Similarly, the products of a reaction are often formed as a mixture and not as pure substances. And real gases certainly don't show the same behavior as ideal gases.
So, enthalpically the difference between the Enthalpy of Reaction and the Standard Enthalpy of Reaction is that you would have to add the Enthalpy of Mixing of the reactants to $\Delta_{\mathrm{r}} H$ and subtract the Enthalpy of Mixing of the products from it in order to get $\Delta_{\mathrm{r}} H^{\ominus}$. Furthermore you would have to account for the enthalpic difference between real and ideal gases if there are any gases involved. Furtunately, those differences are usually quite small compared to the often very big Enthalpies of Reaction so that they can be ignored without introducing a large error. But you have to be mindful of the difference between $\Delta_{\mathrm{r}} H^{\ominus}$ and $\Delta_{\mathrm{r}} H$: for example, if $\Delta_{\mathrm{r}} H^{\ominus}$ is small and the reaction is conducted under very non-ideal conditions with respect to the behavior of gases and/or including liquid mixtures with a very large Enthalpy of Mixing, then you have to expect big differences between the (tabulated) standard values and the measured values.

So, in this sense you would have quite a hard time to calculate the exact Standard Enthalpy of Reaction from your measured $\Delta_{\mathrm{R}} H$. But, under the appropriate conditions your measured value is a good approximation.

How to calculate $\Delta_{\mathrm{r}} H$?

The key to solve your calculational problem is to look at the definition of $\Delta_{\mathrm{r}} H$:

\begin{equation}
  \Delta_{\mathrm{r}} H = \biggl(\frac{\partial H}{\partial \xi} \biggl)_{p, T} \ ,
\end{equation}

whereby the extent of reaction $\xi$ is defined as

\begin{equation}
  \mathrm{d} \xi = \frac{\mathrm{d}n_{i}}{\nu_{i}}
\end{equation}

with $n_{i}$ and $\nu_{i}$ being the amount and the stoichiometric coefficient of substance $i$, respectively.
Consider the reaction:

\begin{equation}
  \ce{\nu_{A} A + \nu_{B} B <=> \nu_{C} C + \nu_{D} D}
\end{equation}

From its definition you see that $\Delta_{\mathrm{r}} H$ is the amount of heat that is exchanged with the surroundings in a isothermal and isobaric process when $\ce{\nu_{A}}$ moles of A and $\ce{\nu_{B}}$ moles of B are converted to $\ce{\nu_{C}}$ moles of C and $\ce{\nu_{D}}$ D (= "conversion of one formula unit").
So, using the differential form of your calorimetric formula (under the assumptions you made) and "dividing" it by $\mathrm{d} \xi$ you get

\begin{equation}
  \mathrm{d} H = m C \mathrm{d}T \\
  \frac{\partial H}{\partial \xi} = m C \frac{\partial T}{\partial \xi} \\
  \Delta_{\mathrm{r}} H = m C \frac{\partial T}{\frac{\partial n_{i}}{\nu_{i}}} = m C \nu_{i} \frac{\partial T}{\partial n_{i}} \ ,
\end{equation}

where $m$ is the mass of the stuff you heat in the calorimeter (assumed to be constant) and $C$ its heat capacity.
Considering finite instead of infinitesimal changes of $n_{i}$ leads to

\begin{equation}
  \Delta_{\mathrm{r}} H = m C \nu_{i} \frac{\Delta T}{\Delta n_{i}} \ .
\end{equation}

So, in order to get your Enthalpy of Reaction you simply have to pick one substance from your reaction (it doesn't matter which) and substitute its stoichometric coefficient and the amount that was consumed/produced during the reaction into the equation above. Note that since $\xi$ is always positive the sign of $\Delta n_{i}$ is positive for products and negative for reactants (the same is true for $\nu_{i}$).

Side Note: Standard Enthalpy of Formation $\Delta_{\mathrm{f}} H^{\ominus}$

The Standard Enthalpy of Formation is a special Standard Enthalpy of Reaction for which further conditions are in effect:

the target compound is to be formed by reacting the (pure) elements
it consists of, whereby each element is expected to be in its most
stable modification for the given temperature.
The reaction equation is to be formulated in such a way that $1 \,
   \text{mol}$ of the compound in question is formed.


Silver Nitrate Electrolysis i have a solution of silver nitrate at 1M. the aim of my experiment is to determine in 40 mins, how much Ag will be deposited. Should i use copper or graphite electrodes.

I'd suggest you do it with graphite electrodes.

Because:

Its  more inert than copper
If copper is used, it readily reacts and more copper will be deposited at the cathode compared to silver
the yield of silver will be greatly reduced


Does 11.2 dm3 of nitrogen consist of 0.5 moles at STP? Which of the following statements is true?
$\pu{11.2 dm^3}$ of nitrogen at STP-
A: has a mass of $\pu{14 g}$.
B: consists of $\pu{0.5 mol}$ atoms.

My instinct was that the answer is B because any gas has a volume of $22.4$ liters regardless of what the gas is. If $\pu{0.5 mol}$ of nitrogen has a volume of $\pu{11.2 dm^3}$ at STP, then $\pu{1 mol}$ of nitrogen has a volume of $22.4$ liters at STP (if I am correct).
It turned out the answer was A. Why is B not the correct answer?

You are right that 0.5 mol of $\ce{N2}$ is 11.2 L, but read the question carefully. Option B says that 11.2 L of $\ce{N2}$ contains 0.5 mol of atoms, which is wrong, it contains 11.2 L of $\ce{N2}$ contains 0.5 mol of molecules. It seems to be a trick answer. The first option is correct as 0.5 mol of $\ce{N2}$ has a mass of 14 grams.


Does adsorption violate thermodynamics? My textbook reads as follows:

  When a gas is adsorbed, the freedom of movement of its molecules become restricted. This amounts to decrease in the entropy of the gas after adsorption, i.e. Entropy change is negative.

But now, I am unable to understand why it isn't a violation of the second law of thermodynamics, which states that the entropy change of a system can never be negative. Please guide.

The second law of thermodynamics states that the entropy of the universe always increases.

$$\mathrm{d}S > 0$$

In the case of adsorption the entropy of the system; the gas being adsorbed; decreases but the entropy of the surroundings;the rest of the gas and the surface (and everything else in the universe); increases and this outweighs the decrease in entropy of the system.

$$\Delta S_\mathrm{sys} < 0 \qquad \Delta S_\mathrm{surr} > 0$$
$$|{\Delta S_\mathrm{surr}}| > |{\Delta S_\mathrm{sys}}|$$

This is because adsorption is an exothermic process and so the surroundings are heated up and therefore increase in entropy. If you consider the Gibbs energy change for adsorption it will be negative because the negative $\Delta H$ term is larger than the positive $-T\Delta S$ term. This chem.uic.edu page goes through the maths of this very nicely.


Relationship between rate of reaction and pressure Given the equation

$$\ce{A(g) + B(g) → C(g) + D(g)}$$

with reaction rate given by

$$v = k[\ce{A}]^2$$

How can I prove that reaction rate goes up with pressure? I tried substituting $k$ for $Ae^{-E_A/(RT)}$ which leaves me with

$$v = Ae^{-\frac{E_A}{RT}}[A]^2$$

If I use

$$PV = nRT \implies RT = \frac{1}{n}PV \implies RT = \frac{P}{C}$$

I end up with this:

$$v = Ae^{-E_A\frac{C}{P}}[A]^2$$

This shows there is a relation, but I don't know what to do with the concentration $C$, or whether it was correct to plug in the ideal gas law in the first place.  Any help?

The exponential term in the Arrhenius equation does not imply reaction with other molecules. It applies to ready to react unstable adduct, about to overcome the energy barrier $E_a$ to form the product.
The concentration of this adduct is driven by concentration of reagents and therefore by the pressure.

Therefore the reaction rate equation implicitly involve 2 multiplicative terms:

Concentration of reagent adduct

$$\mathrm{const}\cdot [A]^2 = \mathrm{const}\cdot \left(\frac{p_{\rm A}}{RT}\right)^2$$
Probability the adduct reaches the reaction energy barrier, from the Boltzmann distribution:

$$\exp{\left(-\frac{E_A}{RT}\right)}$$

By other words, the constant $A$ in your equation

$$v = A \exp\left(-\frac{E_A}{RT}\right)$$

implicitly involves concentration, respectively pressure of the reagents.


Does the pre-exponential factor (Arrhenius constant) depend on temperature? I was taught that the the pre-exponential factor $A$ in the Arrhenius equation \eqref{arrhenius} does not increase with temperature.

$$k = A\cdot\mathrm \exp\left(-\frac{E_\mathrm a}{RT}\right)\tag1\label{arrhenius}$$

But, a component of the pre-exponential factor is the number of collisions, which presumably increases with temperature. Should the pre-exponential factor therefore not increase with temperature as well?

Your intuition is correct. The factor $A$ changes with temperature.

This article details how the value of $\ce{A}$ for an elementary, bimolecular reaction between $\ce{P}$ and $\ce{Q}$ can be derived to be:

$$A_{\ce{PQ}}=N_\ce{P}N_\ce{Q}d^2_{\ce{PQ}}\sqrt{\frac{8k_\mathrm{B}T}{\mu}}$$

The RHS is clearly a function of temperature. Without going into the details,[a] it suffices to remember that $A$ is a function of temperature because it is related to molecular collisions, which themselves are a function of temperature.

However, it is worth noting this paragraph on Wikipedia:

  ...under a wide range of practical conditions, the weak temperature dependence of the pre-exponential factor is negligible compared to the temperature dependence of the $\mathrm{e}^{(-E_\mathrm{a}/RT)}$ factor[b] (my emphasis); except in the case of "barrierless" diffusion-limited reactions, in which case the pre-exponential factor is dominant and is directly observable.

Given this, it may be within sufficient experimental errors to make the assumption that $A$ does not vary with temperature. However, it is just that, an assumption. In reality, $A$ does vary with temperature.

[a]: This isn't the exactly correct expression though. As the article itself notes, "Often times however, when the term is determined experimentally, $A$ is the preferred variable and when the constant is determined mathematically, $Z$ is the variable more often used. The derivation for $Z$, while mostly accurate, ignores the steric effect of molecules."
[b]: Of course, here Wikipedia is using the actually computed values of $A$ (and not vague estimated formulae like the one above; as actually $\sqrt{T}$ grows faster than $\mathrm{e}^{-1/T}$)


Why are the total energies of transition states so commonly corrected for zero-point vibrational energy? So often I see total energies of transition states corrected for zero-point vibrational energy which always confuses me.  Zero point energy is the lowest energy that a ground state minimum energy configuration can have at 0K in a vacuum.  Transition states are indeed not minima, can never be isolated, have one imaginary mode of vibration, etc.

So why do people constantly factor in zero-point vibrational energy to the total energy for a transition state?  Obviously this is done to have 'comparable' minima and TS energetics (if you are applying ZPE to the minima) but is this even physical?  If zero-point correcting the TS, should the zero-point energy of the imaginary mode be included in the summation or excluded (or should both quantities be provided)?

At the transition state, you have exactly one vibration mode with imaginary frequency (which corresponds locally to the reaction coordinate), and $3N-1$ other “normal” vibration modes. In the transition state theory, you separate out the reaction coordinate (consider it infinitely slow compared to the other movements). Thus, when you consider the energy of the transition state, you have to include the contributions of ground state energy of all the other vibration modes.

The more rigorous way of looking at it is to use statistical physics, and to consider the partition function of the transition state. See here or there for such treatments.


What is a "Strong" Acid? If a reaction requires a "strong" acid - what does this mean? Does it mean any acid of a specific concentration or pH? Are only certain acids classed as "strong"? I have read on Wikipedia that it needs to ionise completely in an aqueous solution - in this case, is a strong acid (for example, HCl) still classed as a strong acid if it is in a 0.00001M solution?

The strength or weakness of an acid is independent of its concentration. A strong acid dissociates completely in solution, so $\ce{HCl}$ at any molarity would be considered strong, and $\ce{HCOOH}$ (for example) at any molarity would be weak. It is more enlightening to consider $K_a$ values, which are essentially the equilibrium constants for the ionization of an acid $\ce{HA}$.
$$K_a=\frac{[\ce{H}^+][\ce{A}^-]}{[\ce{HA}]}$$
This value will either be very small (<<1) or very large (>>1000). Acids with very small $K_a$ are considered weak, and acids with very large $K_a$ are considered strong (the $K_a$ of strong acids is usually not stated and assumed to be infinite).

Addendum: The $K_a$ value is concentration-independent because you are only interested in how well the acid dissociates from its molecular form. If you had $0.0000001\ \text{mol}\cdot\text{dm}^{-3}\ \ce{HCl}$, practically every molecule of $\ce{H-Cl}$ would dissociate into $\ce{H^+}$ and $\ce{Cl^-}$ no matter how many there are. But if you had $12 \ \text{mol}\cdot\text{dm}^{-3}\ \ce{HCOOH}$ (methanoic acid), only a small proportion of the $\ce{HCOOH}$ would dissociate into  $\ce{H^+}$ and $\ce{HCOO}^-$, irrespective of how many you had to start.


IUPAC nomenclature of cycloalkadienes with substituents and alkenes on the same carbons Consider the following organic compounds:

Compound $(1)$ has two double bonds, which I'll label $C1$ and $C4$ such that the methyl group is attached to $C1$ and thereby assign lowest numbers to all functional groups and substituents. Officially its name is 1-methylcyclohexa-1,4-diene.
Generally we try to give just enough information to be able to reconstruct the original compound, but not too much. With that in mind, I feel that compound $(1)$ should be named simply 1-methylcyclohexa-4-diene:

Since it ends in -diene, it's understood that there are two double bonds.
Since it ends in -4-diene, it's understood that one of them begins at $C4$.

If the second double bond were located after $C4$, the numbering is backwards; instead of, say, -cyclohexa-4,6-diene, it should've been named -cyclohexa-1,3-diene.
If the second double bond were located at $C2$ or $C3$, then priority was given to the methyl substituent over the alkene functional groups. Rather than calling it 1-methylcyclohexa-2,4-diene, it should've been named 5-methylcyclohexa-1,3-diene.
Therefore, the only option left is for the second double bond to be at $C1$, and therefore specifying 1-methylcyclohexa-1,4-diene is extraneous.

Following a similar line of thought, compound $(2)$ ought to be named just 1-methylcyclohexa-3-diene, rather than its proper name of 1-methylcyclohexa-1,3-diene.
Taking this logic to an extreme, the third compound should be named 1,4-dimethylcyclohexadiene, instead of its proper name of 1,4-dimethylcyclohexa-1,4-diene, because the same logic used above to determine that the second double bond is located at $C1$ should be sufficient to determine the locations of both double bonds in this case, where they both have substituents!
Why in all of these cases, and others like them, do we specify both locants of the alkenes, when one (for compounds $(1)$ and $(2)$) or none (for compound $(3)$) is sufficient?

The question is based on a false premise:

Generally we try to give just enough information to be able to reconstruct the original compound, but not too much.

This isn’t true. The prime aim of IUPAC nomenclature is clarity and the construction of unambiguous names.

Three logicians walk into a bar
The bartender asks: “Do you all want a drink?”
The first logician says: “I don’t know.”
The second logician says: “I don’t know.”
The third logician says: “Yes.”

IUPAC nomenclature doesn’t work like that. Generally, the complete information about a structure is explicitly given by the name and does not rely on any implied information or hidden logic involving other structures.
The fact that “1-methylcyclohexa-2,4-diene” would actually be called 5-methylcyclohexa-1,3-diene does not change the preferred name for 1-methylcyclohexa-1,4-diene.
Nevertheless, there are a few exceptional cases where locants are omitted, but only when there is no ambiguity. For preferred IUPAC names, the nomenclature rules explicitly stipulate when locants are omitted. In most cases, if any locants are essential for defining the structure, then all locants must be cited for the structure.
Also note that your third example is clearly wrong even when following your unorthodox logic. Your name “1,4-dimethylcyclohexadiene” could mean 1,4-dimethylcyclohexa-1,4-diene as well as 1,4-dimethylcyclohexa-1,3-diene or even 1,4-dimethylcyclohexa-1,2-diene, which are all correct IUPAC names.


Is atom-clean aluminum active enough to react with alkali salts? Thermally evaportated $\ce{Al}$ under high vacuum is free of oxidized species and highly active. But is it active enough to even substitute the alkali cationic metal in alkali salts, like $\ce{LiOH, NaOH, Cs2CO3}$.
Here $\ce{Al}$ is directly depositely onto the salt substrate.

Even normal metallic aluminum (with its spontaneously-formed corrosion-resistant oxide coating) reacts with concentrated sodium or potassium hydroxide to give the aluminate (e.g. NaAl(OH)4), plus hydrogen gas. Lithium hydroxide should also react (lithium is likely functioning as only a spectator ion) whereas alkali metal carbonates may not be sufficiently reactive.

This reaction is not what you refer to in your question as "substitution" reactions, where the alkali metal ion would be reduced to metallic form. Note that aluminum chloride is reduced by sodium or potassium metal to give aluminum metal, so the reverse reaction is unlikely to occur, but the energetics may be different for the hydroxides or carbonates in a non-aqueous environment.