What happens to the mobility of Group 1 elements going down the periodic table? Whilst reading the book, So you want to go to Oxbridge (Third Edition), I noticed that it states that a past question given during an interview for the place on the undergraduate chemistry course (M.Chem) was:

  What happens to the mobility of Group 1 elements going down the
  periodic table?

I do not understand the question. And so, I would both like to understand the question and know and understand its answer.

Preemptively, Thank you

When you have charged particles in a liquid acted upon by a electric field, the experience a force, and said force accelerates them to a terminal, drift speed.

Note: The physical treatment of the problem, and the notion of terminal speed is quite similar to say a ball falling from the top of a sky scraper through air (a fluid medium)

Now, let's go back to our charged particles in fluids. Let's say we have applied a voltage $\Delta\phi$ between two large, planar electrodes, immersed in the liquid and separated by a distance $l$

The  magnitude of the electric field, and the electric force experienced by an ion of charge $ze$ are given by the following relations:

$$ E = \frac{\Delta\phi}{l}$$
$$F = zeE = (ze).\frac{\Delta\phi}{l}$$

A cation responds by accelerating towards the negative electrode, and an anion responds by accelerating towards the positive electrode. This acceleration, however, is short lived because the fluid medium offers a retarding force $F_{\text{friction}}$.

$$F_{\text{friction}}\propto s\text{, where s := speed} $$

Assuming we have spherical particles of radius $a$, let's say our retarding force is given by Stokes' Law

$$F_{\text{friction}} = (6\pi\eta a)s$$ $\eta$) is the viscosity of the medium.

When the electric force is exactly balaced by the viscous drag, the acceleration drops to zero and this final speed of the particle is called the drift speed ($s_d$).
$$F_{\text{friction}} = F_{\text{electric}}$$ and solving for the drift speed we get,
$$s_d = \frac{zeE}{6\pi\eta a}$$

At this point we note that the drift speed is proportional to the applied electric field and introduce the following definition:

$$s_d = \mu E$$ where, $$\mu = \frac{ze}{6\pi\eta a} $$

and we call $\mu$ the mobility.

From the aforementioned equations/definitions one can deduce that $\mu \propto \frac{1}{a}$. Thus, bigger/bulky ions ought to have lower mobilities, and lower drift speeds in solution.

However, $a$ is not the ionic radius, but the Stokes radius (or the hydrodyanmic radius), i.e the "effective radius" of an ion in solution taking into account its solvation shell (i.e the solvent molecules surrounding the ion). I believe the following illustration is instructive,

The size of this solvation shell depends on how "attractive" an ion appears to a solvent molecule, and that in turn depends on the electric field produced by the ions.

For spherical ions, we consider the electric field on the surface of a charged sphere: $$ E _\text{surface of a charged sphere} \propto \frac{ze}{r^2} $$

here $r$ is the ionic radius. So smaller ions have a larger solvation shell than larger ions (assuming they carry the same charge).

This is the reason why mobilities of alkali metal ions increase from $\ce{Li^+}$ to $\ce{Cs^+}$ (and not the other way around); the lithium ions are effectively dragging a large amount of solvent molecules with them and thus, move slower.


What are range and domain of Hamiltonian? What are range and domain of Hamiltonian? Hamiltonian is function so it should have ones. What are they?

Hamiltonian is not a function, but an operator (a function in a broader sense, if you'd like):
$${\hat {H}}=-{\frac {\hbar ^2}{2m}}\nabla ^2+V(\bar{r},t)$$
It is applied to some function (our supposed $\psi$) and yields some other function. So its domain is the set of all doubly differentiable functions on our underlying space (the space itself can be 1D, 3D, or whatever else, depending on the nature of the problem), and the range is even broader than that, because the derivatives (or $V$, for that matter) do not have to be differentiable.


Removing trace amounts of water from agricultural grade anhydrous ammonia Agricultural grade anhydrous ammonia (liquefied gas) contains 2000 ppm $\ce{H2O}$. What would be the process to remove this water to obtain refrigeration grade (<100 ppm $\ce{H2O}$) anhydrous ammonia?

Calcium oxide can get water down to 4ppm in anhydrous ammonia according to http://www.google.com/patents/US4075306.


What's the point of solubility products? What is the advantage of solubility products, e.g.

LiF(s) ⇄ Li+(aq) + F-(aq)  |  K = [Li+(aq)] * [F-(aq)] = 0.00184 M2 at 25°C (source)

compared to just stating that

Solubility in water = 0.134 g/100 mL (25 °C) (source)

(There's a little difference between (0.00184 M2)½ = 1.11 g/L and the solubility of 1.34 g/L above; I take it that's because of the difference between activity and concentration?)

It seems to me that solubility products are just a roundabout, complicated way of expressing the concentration of a saturated solution. I suspect that I'm missing something though: otherwise why would people publish long lists of solubility products?

What I read: Wikipedia; all my high school textbooks.

Solubility products figure prominently in equilibria involving precipitated species.  Following is an example that may arise in a practical situation.

Problem:  We have 0.01 M ferrous ion in water and we propose adding a base to drive out the iron as hydroxide.  We do not want excess base dissolving into the water so we will try magnesium hydroxide.  How well will it work?

The proposed reaction is then

$\ce{Fe^{2+} + Mg(OH)2(s) <=> Mg^{2+} + Fe(OH)2(s)}$

$K=\frac{[\ce{Mg^{2+}}]}{[\ce{Fe^{2+}}]}$

Compare this equilibrium constant with:

$K_{sp}(\ce{Mg(OH)2})=[\ce{Mg^{2+}}][\ce{OH^-}]^2=5.61×10^{-12}$ (source)

$K_{sp}(\ce{Fe(OH)2})=[\ce{Fe^{2+}}][\ce{OH^-}]^2=4.87×10^{-17}$ (Ibid.)

So then

$K=\frac{[\ce{Mg^{2+}}]}{[\ce{Fe^{2+}}]}=\frac{K_{sp}(\ce{Mg(OH)2})}{K_{sp}(\ce{Fe(OH)2})}=1.15×10^5$

Since the reaction stoichionmetry implies that the sum $[\ce{Mg^{2+}}]+[\ce{Fe^{2+}}]$ will be constant at the original iron concentration of 0.01 M, we find that the iron level remaining in the water may be reduced by five orders of magnitude, the equilibrium concentration is below $10^{-7}$ molar!

Magnesium hydroxide, despite being supposedly "weak" because of its limited solubility, is actually like a strong base here, displacing the weak iron hydroxide base almost quantitatively -- but, in effect, self-regulating because of that limited intrinsic solubility.  The same property applies with respect to most heavy metals commonly found in water; their hydroxide solubilities, like that of iron, are much less than magnesium's.  Thus magnesium hydroxide (or, in practice, magnesium oxide which becomes the hydroxide in situ) is a good agent for treating water to remove heavy metals.  See here, under "Applications".


Temperature changes in exo/endothermic reactions In exothermic reactions, heat is released, whereas heat is absorbed in endothermic reactions. But exothermic reactions are often described as being associated with an increase in temperature, and the opposite is the case for endothermic reactions.
How does heat release result in a higher temperature of the system, and heat absorption result in lower temperature? Shouldn't it be the other way around?

Your confusion is understandable.  I looked at the answers linked in the comment on your OP, and I'd like to explain it differently from how they do.  Hopefully this explanation will get to the heart of your confusion.
You are thinking (incorrectly) that an exothermic reaction works something like this:
The system has a certain amount of thermal energy.  In an exothermic reaction, some of the system's thermal energy is relased to the environment, and thus the system must lose thermal energy and thus decrease in temperature.  I.e., you're imagining something like what happens when a warm block of metal is placed in a cool cup of water.  The metal releases some of its thermal energy to the water, and thus cools.
But that's not what happens in an exothermic reaction.  In an exothermic reaction, the system internally generates extra thermal energy. This is generated through a chemical reaction in which chemical energy (essentially, the energy in chemical bonds) is converted to thermal energy.  Now, if the system is internally generating extra thermal energy, any of a number of things can be done with this energy.  The two most common pictures are as follows:

The system is in a large (relative to the system) constant-temperature heat bath, and is surrounded by diathermal (heat-conducting) walls.  In this case, because the temperature of the system is kept constant, the thermal energy flows out (i.e., is released) into the bath.

The system is adiabatic, i.e., is surrounded by a perfect thermal insulator.  In this case, the extra thermal energy stays within the system, causing its temperature to increase.

The reverse of what I wrote above applies to an endothermic reaction.


Nomenclature of lithiated dithianes Lithiated dithianes have a few uses in organic synthesis (albeit fairly niche ones) such as the Corey-Seebach alkylation / Corey-Seebach reaction, which uses dithianes as masked acyl anions (d1 synthons) as shown below.

Searching through the 2013 IUPAC recommendations for organic nomenclature for dithian, dithiane didn't come up with any useful results, and the usual technique of looking for the corresponding oxygen analogue seems redundant in this case as the corresponding molecule with oxygen is impossible!

In most papers, the authors avoid explicitally naming the lithiated intermediate (certainly one way to get around the issue):

  Lithiation of dithiane X with t-BuLi followed by alkylation with epoxide Y resulted in Z upon quenching.

In others, a derivative name is used based on the 1,3-dithiane itself:

  2-lithio-1,3-dithiane derivatives are usually prepared by deprotonation with n-BuLi in THF at low temperatures2 and the anionic species obtained are able to react with many types of electrophiles.

  Taken from Tetrahedron 2003, 59, 6147

ChemDraw outputs something more in-keeping with other nomenclature:

The exact nomenclature of alkyllithiums is an inorganic problem (cf. 2005 Red Book) and in there one finds as an example (IR-10.3.2, p 229; repeated in P-69.3, p 1043 of the 2013 Blue Book)

$\ce{LiMe}$ lithium methanide (compositional name)
$\ce{[LiMe]}$ methyllithium
$\ce{[(LiMe)4]}$ tetra-µ3-methyl-tetralithium
$\ce{(LiMe)_n}$ poly(methyllithium)

so, the name you choose will depend on what structural features you want to emphasise. Note that the Red Book does not prescribe "preferred" IUPAC names, and the Blue Book explicitly refuses to assign PINs (p 1036):

Neither preferred IUPAC names or preselected names (see P-12.2) for organometallic compounds involving the transition elements (including the Group 3 elements) and Groups 1 and 2 elements, excpet for 'ocene' compounds, are noted. This determination will await consideration by a task group on inorganic and/or organometallic nomenclature.

Obviously, from a practical point of view, the name "methyllithium" is more commonly used and more easily recognised. In line with this, we will therefore look for a name of the form "_yllithium". The use of "methyl" instead of "methanide" is permissible because (p 1037, 2013 Blue Book)

Because of a long established tradition, names for substituent groups as defined in Section P-29 can also be used as ligand names.

and hence the question becomes: what is the preferred prefix for the following group?

This is a purely organic question, so can be answered with the 2013 Blue Book. The relevant rules are as follows:

P-29.2 [...] (2) The suffixes 'yl', 'ylidene', and 'ylidyne' are added to the name of the parent hydride with elision of the terminal letter 'e', if present, when followed immediately by the letter 'y'. The locants for the atoms of free valences are as low as is consistent with any established numbering of the parent hydride and, except for mononuclear parent hydrides or the suffix 'ylidyne', the locant '1' must be cited. This method is used to generate names of 'alkanyl-type substituent groups' that are simple substituent groups with free valences at positions other than '1'.
P-57.1.5.1 Preferred prefixes derived from cycloalkanes are of the 'cycloalkyl type' (see P-29.2); preferred prefixes derived from cyclic compounds other than cycloalkanes are all of the 'alkanyl type'.

The preferred IUPAC name of the parent hydride, according to Hantzsch–Widman nomenclature (P-22.2.2.1), is 1,3-dithiane. Correspondingly, the preferred prefix is 1,3-dithian-2-yl:

Finally, returning to the Red Book:

IR-2.2.3.2 Parentheses are used in names in the following ways:
(a) To enclose substituent group or ligand names in order to avoid ambiguity, for example if the substituent group or ligand names contain multiplicative prefixes, such as (dioxido) or (triphosphato), or if substitution patterns would otherwise not be unambiguously specified, or if the substituent group or ligand name contains numerical or letter descriptors [...]

we can conclude that a suitable name would be (1,3-dithian-2-yl)lithium.


Cis-trans isomerism with electron pairs Is it possible to name compounds with lone pairs on the double bonded species using the cis-trans nomenclature? For instance, can't you name $\ce{Ph-N=N-Ph}$ by the relative position of the nitrogen atoms' electron pairs?

Actually, most of linear imines and exocyclic imines display $E/Z$-isomerism (commonly known as $cis/trans$-isomerism) with a lone pair on the nitrogen, which is a part of the double bond. The isomerism follows various mechanisms. For instances, $cis/trans$-isomerism of azobenzene a noted elsewhere is operated by UV/vis-irradiation is a good example. The mechanism of this transformation has been depicted as follows (Ref.1):

Similar to cis/trans-isomerism in alkenes, extensive studies has been done on azo-compounds and imines. For example, acid-catalyzed $E/Z$-isomerization of imines has been reported frequently (Ref.2,3). The abstract of Ref.2 states that:

The kinetics and mechanism of acid-catalyzed Z/E isomerization of O-methylbenzohydroximoyl chloride (1Za and 1Ea), methyl O-methylbenzohydroximate (1Zb and 1Eb), ethyl O-methylbenzohydroximate (1Zc and 1Ec and five para and meta substituted derivatives), O-methylcinnamohydroximoyl chloride (2Za and 2Ea), and methyl O-methylcinnamohydroximate (2Zb and 2Zb) have been investigated. The kinetics of Z/E isomerization of these imines have been studied in glacial acetic acid (1Ea and 1Zc) and in dioxane solutions containing $\ce{HCl}$, trifluoromethanesulfonic acid, or tetrafluoroboric acid (1Ea, 1Zb, 2Ea, and 2Zb). The isomerization takes place by either (a) rotation about the carbon−nitrogen double bond of the protonated imine (iminium ion rotation) or (b) nucleophilic attack on the protonated imine to form a tetrahedral intermediate that undergoes stereomutation and loss of the nucleophile (nucleophilic catalysis). The hydroximoyl chlorides 1Ea and 2Ea only isomerize by the nucleophilic catalysis mechanism. The hydroximate 1Zb appears to be capable of isomerizing by either mechanism. The hydroximate 2Zb may be isomerizing only by iminium ion rotation. Theoretical calculations support the notion that increased conjugation in the protonated imine increases the rate of iminium ion rotation.

Photochromic properties of a series of novel pyrrolidene imines bearing functionalized aryl or naphthyl moieties was studied by UV spectroscopy (Ref.4):

UV irradiation of these heterocyclic Schiff bases at room temperature promotes the trans–cis photoisomerization of the $\ce{C=N}$ double bond with formation of a variable amount of the cis-isomer that, in the absence of light, returns thermally to the original form in few seconds. The thermal cis–trans re-isomerization of these molecular switches is much more slower than for the common benzylidene aniline-type imines, allowing the observation of the photochromic phenomena at room temperature.

Another interesting studies have been done with indigo derivatives (Ref.5) noting that:

Substituted indigo derivatives undergo photoisomerization of the central double bond if both nitrogen atoms are functionalized. Indigo itself however does not photoisomerize because of a competing and highly efficient excited‐state proton transfer...

References:

H. M. Dhammika Bandara, Shawn C. Burdette, "Photoisomerization in different classes of azobenzene," Chem. Soc. Rev. 2012, 41(5), 1809-1825 (https://doi.org/10.1039/C1CS15179G).
James E. Johnson, Nora M. Morales, Andrea M. Gorczyca, Debra D. Dolliver, Michael A. McAllister, "Mechanisms of Acid-Catalyzed Z/E Isomerization of Imines," J. Org. Chem. 2001, 66(24), 7979–7985 (https://doi.org/10.1021/jo010067k).
James E. Johnson, Nancy M. Silk, Elizabeth Ann Nalley, Mohammed Arfan, "Mechanism of an acid-catalyzed geometric isomerization about a carbon-nitrogen double bond," J. Org. Chem. 1981, 46(3), 546–552 (https://doi.org/10.1021/jo00316a013).
Paulo J. Coelho, M. Cidália R. Castro, M. Manuela M. Raposo, "Reversible trans–cis photoisomerization of new pyrrolidene heterocyclic imines," Journal of Photochemistry and Photobiology A: Chemistry 2013, 259, 59-65 (https://doi.org/10.1016/j.jphotochem.2013.03.004).
Ludwig A. Huber, Peter Mayer, Henry Dube, "Photoisomerization of Mono‐Arylated Indigo and Water‐Induced Acceleration of Thermal cis‐to‐trans Isomerization," ChemPhotoChem 2018, 2(6), 458-464 (https://doi.org/10.1002/cptc.201700228).


Do the π-electrons from the C=O bond in tropone contribute to Hückel 's rule? This is a continuation of Is tropone aromatic?

I'm looking at the resonance form that does have the carbonyl double-bond, 1a. Can this resonance structure also be considered aromatic? Would the electrons from the C=O double bond also be delocalized through extended conjugation?

To me, this looks like a very similar situation to styrene, when we have an alkenyl sidechain attached to a benzene ring. Apparently, for alkenyl side-chains, Huckel's number is not affected, leaving an already aromatic compound to stay aromatic even though more electrons come into the conjugation.

In the case of tropone, would the C=O double bond interfere with electron counting for Hückel's rule?

In addition to Philipp's answer I'd like to share some orbital pictures which shall make the case rather obvious.

Let me shortly remind you of the key part of the IUPAC definition for aromaticity:

  In the traditional sense, 'having a chemistry typified by benzene'.

I will therefore follow the same approach as in my answer to "Is phenoxide aromatic?" in comparing the orbital pictures.

The following orbitals are for the π system of benzene, in increasing order of energy (DF-BP86/def2SVP). The two orbitals on the right are degenerate:

Analogously, the π system for cyclohepta-2,4,6-trien-1-one is extensively delocalised and holds eight electrons. Note that the symmetry of this molecule is only $C_\mathrm{2v}$ removing a lot of constraint from the electronic structure. This is also why the Hückel rule breaks down, since there are no more degeneracies observed.

Also note, that the LUMO of this molecule (pictured below) still has considerable bonding character and that reduction would most likely not destroy aromaticity. (The same statement also applies to oxidation.)


metal salt solutions g/l means hydrate or anhydrous I want to redo an experiment. They've used

Nickel chloride: 40 g/L

Nickel chloride can be bought in two forms: anhydrous ($\ce{NiCl2}$) and hydrate ($\ce{NiCl2.6H2O}$). The paper doesn't say which one they've used. Which one did they mean in the paper? In one liter of water should I dissolve 40 grams of $\ce{NiCl2}$ or 40 grams of $\ce{NiCl2.6H2O}$? Is there a convention?

Unless the experimental part of the paper provides additional details, such as equivalents or moles or the colour of the nickel chloride used ($\ce{NiCl2*6H2O}$ is green, while the anhydrous chloride is yellowish) you're up to guessing.

However, if the experiment you're trying to repeat is conducted in aqueous solution, it is safe to assume that you can use the hexahydrate.

Btw, gloves and proper waste management are a good idea here.


Units of the rate constant for reactions whose order is greater than 1 In many textbooks, it is written that:

  The value of rate constant depends on the nature of the reactants, temperature and catalyst. It is independent of the concentration of the reactants.

However, the unit of the rate constant is

$$\left(\frac{\mathrm{mol}}{\mathrm{l}}\right)^{1-n} \mathrm{s}^{-1}$$

where $n$ is the order of reaction.

Therefore, except for first order reactions, the unit of the rate constant will have a concentration term $\displaystyle\frac{\textrm{mol}}{\textrm{l}}$ for all other reactions. Does this contradict the statement given in books?

As an example if the reaction is $\ce{A + B -> C}$ the rate expression for the appearance of $\ce{C}$ is

$$\frac{\mathrm d[\ce{C}]}{\mathrm dt} = +k[\ce{A}][\ce{B}].$$

The units have to be the same on both sides of this equation, these are concentration/time on the left and to be the same on the right the rate constant $k$ has to have units $(1/\text{concentration})(1/\text{time})$ usually expressed as $\mathrm{dm^3\,mol^{-1}\,s^{-1}}.$

It is true that the numerical value of the rate constant usually depends of several things such as temperature, but the units remain the same for each type of rate constant, 0th,  1st, 2nd, 3rd simply because the rate equation units have to balance.


Alternatives to Ring Stands I'm trying to find an alternative to simple ring stands to hold a simple electrochemical cell.

The issue is that it is difficult to adjust the height to the right position to within even a couple inches. If you try to screw the stand in it can drop or lean, and I need to be able to adjust it to within at least an inch without it falling down.

I use a micrometer to adjust an electrode coming in from the top, and I don't have a lot of play to push down.

It is also not easy to adjust the stand from under a few hood, due to the nature of the screw.

It would also be nice if the clamp could be adjusted along the y-axis to move it closer or farther from the stand for different situations.

On the more improvised/DIY side, you may be able to get away with something like a gooseneck clamp of the sort commonly used for microphones; something like this:

(from here)

Obviously they're not really designed to be exposed to hazardous chemicals, though, and you'll want to make sure that it can handle the weight you need to put on it without bending further.


Can you determine whether a compound can form given particular elements? Say you are given a set of particular elements, such as Dy, C, I, Er, P. Is it possible to determine whether a compound exists with all of the given elements and no more?

If not, is it possible to find this out if given the number of a particular element, such as; two hydrogen and one oxygen where you could come to the conclusion that $\ce{H2O}$ is possible.

Answer on first question is no, on second is yes.

Consider this short list C, H, O, N this elements can produce unlimited number of compounds.
If you have short formula like C6H6 and you assuming only covalent bonds in molecule, define valence 4 for Carbon and 1 for Hydrogen you can make mathematical calculation according to graph theory and you will get 217 structures. But most of them doesn't exist due to bonds strain and intersection. If you want to obtain existing molecules you can filter this 217 models using for instance quantum mechanics calculation.


Is the Hydrogen spectrum only an emission spectrum? In a statement question it was stated that hydrogen spectrum is an emission spectrum and the scheme considered it to be ture. But my argument is that it can be both emission and abosorption spectrum. Please tell me whether I am correct.

Emission and absorption are inverse processes.
So, naturally, if an atom exhibits some emission spectrum, it will also give an absorption spectrum. So you were right in assuming that.
However, the processes by which these two spectra are generated are completely the opposite of each other.
An atom emits a photon when an electron goes from a higher energy state to a lower energy state, thereby releasing energy in the form of light. When a photon is absorbed, the electron uses this energy to excite to a higher energy state.
What kind of spectrum you have at hand depends on which instrument and detector you are using. When atoms releases photons (light), the detector catches the production of the photon and gives out emission spectra. For absorption spectroscopy, varying wavelengths of photons are incident on the atom and the corresponding photon is absorbed by the atom resulting in no photon of that particular wavelength being detected.
I am attaching the images of the emission and absorption spectra of hydrogen atom. Note how they are complete inverses of each other.

Image source: https://chem.libretexts.org/Courses/Solano_Community_College/Chem_160/Chapter_07%3A_Atomic_Structure_and_Periodicity/7.03_The_Atomic_Spectrum_of_Hydrogen


Accuracy of pH meters in MgSO4 solution I bought a 2 pH meters several months back from Amazon for roughly $10 a piece.  I couldn't tell you how they work, but here's the link on Amazon:  http://www.amazon.com/gp/product/B0087LOWIA/ref=wms_ohs_product?ie=UTF8&psc=1

I also got a neutral buffer solution for calibration.  It worked fine initially but I'm thinking it's off.  It's now being used for pH testing in the salt water of an floatation tank (147 gallons of water + 750 lbs pharmaceutical grade epsom salts and 50 pounds of dead sea salt).  One meter shows a pH of 6 and another of the same kind at about 8.

My question is, does the salt effect the accuracy of the meter?  Are the meters inherently bad, or do they need periodic recalibration?  I didn't want to keep buying litmus paper.

The answer is yes, ionic force does influence real pH signal. Because pH meter is sensitive to activities (instead of concentration of H+ only) of ions that modifying it junction potential you must take in account ionic force when it is strength enough to significantly impact activities. There is a lot of limitation with pH-meter: above pH>12 or bellow pH<2 they are not precise at all (see for acidic and basic errors). Or if they are used in different solvents than water. Organic chemists should tell you what pH could mean in hexane. Assessing pH is a complex task that must be addressed with the care it deserves.

Anyway, you should find the user-guide of your pH meter on internet. I am sure the manufacturer will provide limitation and use tips.

There is a detailed chapter on pHmetry in Skoog and West Analytical Chemistry. It may help you to deeper understand those phenomena.


Law of mass action & Collision Theory I don't understand why the stoichiometric coefficients of reactants and products are expotentialized in law of mass action.

So I checked on Wikipedia. And this popped out:

  In the 1879 paper[9] the assumption that reaction rate was proportional to the product of concentrations was justified microscopically in terms of collision theory, as had been developed for gas reactions. It was also proposed that the original theory of the equilibrium condition could be generalized to apply to any arbitrary chemical equilibrium.

How does collision theory explain this?

The law of mass action only rigorously applies to elementary reactions (in that otherwise the exponents do not match stoichiometric coefficients). These are reactions whose equations match which particles are colliding on the microscopic scale. Consider the reaction

$$\ce{$r_1$R1 + $r_2$R2 + \ldots + $r_n$R_n <=> $p_1$P1 + $p_2$P2 + \ldots + $p_m$P_m}\tag1.$$

Reaction $(1)$ is elementary if and only if it proceeds in a single step. At some time $t = t_c$ all reactant particles $\ce{R1, R2, \ldots, R_n}$ collide. Quantum mechanically this collision is not too well defined, so view it to be Newtonian in essence. Not to mention: the probability $P$ of the simultaneous collision goes to zero as $n \to \infty$. This is why in actuality most elementary steps involve at most three individual constituents.

We therefore sacrifice some generality and instead consider the easier to handle

$$\ce{R1(g) + R2(g) <=> P1(g) + P2(g)} \tag{1'}.$$

Energies

Collision theory postulates that particles only initiate a transformation when they have enough energy. This is called activation energy and denoted by $E_a$. Maxwell-Boltzmann distribution

$$\frac{\mathrm{d}N_v}{N} = 4\pi\left(\frac{m}{2\pi k_bT}\right)^{3/2}v^2\exp\left(-\frac{mv^2}{2k_bT}\right)\mathrm{d}v\tag2$$

allows us to estimate the fraction of particles that have the required amount of movement. Note we are technically assuming $\mathrm{total\ energy} = \mathrm{kinetic\ energy}$. Similarly relativistic effects are ignored, and no cap is placed on maximum velocity. With these simplifications, the fraction of interesting particles is

$$f = \frac{N^*}{N} = \frac{1}{k_bT}\int_{E_a}^\infty\exp\left(-\frac{E}{k_bT}\right)\mathrm{d}E\tag3$$

which evaluates to

$$f = \exp\left(-\frac{E_a}{k_bT}\right)\tag{3'}.$$

Number of interesting collisions

Say $Z^*$ is the number of collisions that matter. We will look for a function $Z$ such that

$$Z^* = Z\cdot f.\tag4$$

The Maxwell-Boltzmann distribution $(2)$ gives for a single kind of particle (say two $\ce{R1}$'s) that there are

$$Z = 2N^2\left(\sigma_\ce{R1}\right)^2\sqrt{\frac{\pi RT}{M_{\ce{R1}}}}\tag5$$

collisions in $1$ second per $1\ \mathrm{cm^3}$ if $N$ is the number of particles per square centimeter. (This is the traditional starting point unit.) Hence the rate of such a reaction would be

$$v_r = \frac{2Z^*}{N_A} \cdot 10^3\ \ \ \left(\frac{\mathrm{mol}}{\mathrm{s\cdot dm^3}}\tag6\right).$$

Reaction rate $v_r$

If you like, you are now able to substitute $f$ from $(3')$ and $Z$ via $(5)$ into $(4)$. Then insert $(4)$ into $(6)$.  After necessary manipulation, this yields $(7)$

$$v_r = \underbrace{4 \cdot 10^{-3} N_A\left(\sigma_\ce{R1}\right)^2\sqrt{\frac{\pi RT}{M_{\ce{R1}}}}\exp\left(-\frac{E_a}{k_bT}\right)}_k \cdot  \overbrace{\left(\frac{N}{N_A}\cdot 10^3\right)^2}^{c^2}.$$

Coming back to our actual reaction $(1')$ involves more relative quantities. For example, $\sigma_\ce{R1}$ becomes $$\sigma_\text{aver} = 0.5\left(d_\ce{R1} + d_\ce{R2}\right).$$

So it is a bit more difficult. But the result is analogous to $(7)$.

$$v_r = \underbrace{2\sqrt{2} \cdot 10^{-3} N_A\left(\sigma_\text{aver}\right)^2\sqrt{\pi RT\left(\frac{1}{M_{\ce{R1}}} + \frac{1}{M_{\ce{R2}}}\right)}\exp\left(-\frac{E_a}{k_bT}\right)}_k \cdot c_{\ce{R1}} \cdot c_{\ce{R2}}$$

More compactly, if $\ce{R1} = A$ and $\ce{R2} = B$

$$v_r = k[A][B]\tag8$$

which is what we set out to prove.

This derivation assumes that every active collision leads to a reaction. When theoretically computed $k$ were compared to experimental, an extra factor was introduced, $P$. This is called the steric factor, and in classical collision theory $P$ remains empirical in essence.

Collision theory is for elementary steps

As your quote suggests, collision theory is a first theoretical explanation for the proportionality to products of concentrations. It does not generally explain various exponentiation. It does not have to either. The higher (or non-integer or negative) exponents usually derive from the mechanism itself. In other words, the fact that common reactions are not elementary comes into play.

For instance, the transition

$$\ce{2Br- + H2O2 + 2H+ -> Br2 + H2O}$$

is experimentally found to follow

$$v_r = k\ce{[H2O2][H+][Br-]}\tag{a}.$$

Mathematically, we can verify that one possible mechanism is

$$\ce{H+ + H2O2 <=>[K] H2O+-OH,} \tag{fast equilibrium}$$ $$\ce{H2O+-OH + Br- ->[k_2] HOBr + H2O,}\tag{slow}$$ $$\ce{HOBr + H+ + Br- ->[k_3] Br2 + H2O.}\tag{fast}$$

Indeed,

$$K_c = \frac{\ce{[H2O+-OH]}}{\ce{[H+]}\ce{[H2O2]}} \implies \ce{[H2O+-OH]} = K_c\ce{[H+]}\ce{[H2O2]}\tag{b}.$$

Applying the method of stationary concentration gives

$$v\left(\ce{HOBr}\right) = 0 \implies k_2\ce{[H2O+-OH]}\ce{[Br-]} = k_3\ce{[H2O2]}\ce{[H+]}\ce{[Br-]}\tag{c}.$$

The overall rate of the reaction is characterised by the rate of formation of bromine $\ce{Br2}$. So,

$$v_r = v\left(\ce{Br2}\right) = k_3\ce{[H2O2]}\ce{[H+]}\ce{[Br-]} \overset{(c)}{=} k_2\ce{[H2O+-OH]}\ce{[Br-]} \overset{(b)}{=} k_2K_c\ce{[H+][H2O2]}\ce{[Br-]}$$

or more briefly using $k_2K_c = k$

$$v_r = k\ce{[H2O2][H+]}\ce{[Br-]}\tag{d}.$$

I repeat: this only shows that it might be a valid mechanism, not that the reaction actually follows such a pathway. Therefore, while we can use collision theory to derive the law of mass action as a first approximation, the exponents for non-elementary reactions are determined via experiment.


Flow rate of the fuel consisting of ethanol and water burning in air of a given flow rate I hope that someone could give me some hints or tips on how to approach this question, even though these types of questions are usually closed off as homework.

A burner is run by a fuel mixture of $\pu{90 mol.\%}$ ethanol and $\pu{10 mol.\%}$ water. Assume a complete combustion with air ($\pu{79 mol.\%}$ $\ce{N2}$, $\pu{21 mol.\%}$ $\ce{O2}$). The whole process takes place at $\pu{100 kPa}$.
Calculate the inlet flow rate of the fuel given that the flow rate of air is $\pu{100 mol s-1}$ and that the excess of air is $20\%$.

The answer key only gives the answer $\pu{6.48 mol s-1}$ of the fuel without explaining the calculations.
My attempt at the question was:
$$\ce{C2H5OH + H2O + 3O2 -> 2CO2 + 4H2O}$$
Knowing that we have $20\%$ excess in air, we could calculate the theoretical moles needed.
$${\text{fractional excess} = (n_\mathrm{feed} - n_\mathrm{theo})/n_\mathrm{theo}}$$
$$\to n_\mathrm{theo} = \pu{83.3 mol}$$
I then used the molar ratio from the reaction to calculate how much ethanol and water was needed. I just don't see how the answer becomes $\pu{6.48 mol s-1}$. Any help is greatly appreciated!

I think you overcomplicate the problem. There are many variables you don't need and many components that are duds (nitrogen, water, pressure). What important is that we burn ethanol:

$$\ce{C2H5OH + 3 O2 -> 2 CO2 + 3 H2O}$$

so that the ratio between ethanol and oxygen is $1:3$. I have taken chemistry engineering classes long time ago and I don't remember exact notations, so I'd rather denote molar flow rate with an n-dot notation:

$$\dot{n} = \frac{\mathrm{d}n}{\mathrm{d}t}$$

where $n$ – amount; $t$ – time. Since fuel is only 90% alcohol, then

$$\dot{n}(\text{fuel}) = \frac{\dot{n}(\ce{C2H5OH})}{x(\ce{C2H5OH})}$$

where $x$ denotes molar fraction. From the combustion reaction

$$\dot{n}(\ce{C2H5OH}) = \frac{\dot{n}(\ce{O2})}{3}$$

and for the air the following is fulfilled:

$$\dot{n}(\ce{O2}) = x(\ce{O2})\frac{\dot{n}(\text{air})}{1 + α}$$

where $α$ is the partial excess of air. Now we can put everything together and plug in the numbers:

$$
\begin{align}
\dot{n}(\text{fuel}) &= \frac{x(\ce{O2})\dot{n}(\text{air})}{3x(\ce{C2H5OH})(1 + α)}\\
  &= \frac{0.21\cdot\pu{100 mol s-1}}{3\cdot 0.9\cdot (1 + 0.2)}\\
  &= \pu{6.48 mol s-1}
\end{align}
$$

One hell of a burner.


Correct Electron Pushing with HCl and Water My book indicates that the following electron pushing is wrong:

So what would the right electron pushing diagram be?

I have the following in mind. My rationale is:

1) Electrons attack protons, not the other way around. This is just convention right? Is this just a conceit in chemistry; just a standardization? On a related note is this why one person described organic chemistry as:

"The foundation of every reaction in organic chemistry can be found in one statement: Nucleophile Attacks Electrophile."

Rather than electrophile attacks nucleophile?

2) Water is the base here; water is the Lewis Base (nucleophile) so water will attack the hydrogen cation (which is a bare nucleus).
3) HCl is the acid here; H+ is the Lewis Acid (electrophile, so it's going to go toward the lone pairs on the water).

So:

Would this be right?

1) Nucleophiles often have excess electron density while electrophiles are characterized by a lack thereof. The electrons flow in the direction of lower electron density (and not the other way around), because they are "attracted" by the positive partial or formal charge of the electrophile. This is the reason why the first scheme from your book is wrong, because it shows a nucleophile attacking another nucleophile.

2), 3) and the reaction scheme you have drawn are correct.


What happens when pyridine N-oxide reacts with dichlorocarbene? It seems that the combination of sodium methoxide (a base) and chloroform will generate dichlorocarbene, $\ce{CCl2}$, by α-elimination.

What happens after that, though? It is an ortho formylation like in the Reimer–Tiemann reaction? Or does the nucleophilic oxygen simply attack the carbene?

Dichlorocarbene has been shown to deoxygenate pyridine N-oxide producing phosgene as an initial byproduct. In this case, since methoxide is present, dimethyl carbonate is ultimately formed.[1]

The proposed mechanism involves nucleophilic attack on the carbene by the oxygen, followed by an elimination:

(1) Schweizer, E. E.; O'Neill, G. J. A Novel Deoxygenation Method for Pyridine N-Oxide. J. Org. Chem. 1963, 28 (9), 2460–2461. DOI: 10.1021/jo01044a516.


Testing for Iron-Gall Inks I have to bottles of Lamy blue-black fountain pen ink. It turns out that some months ago the composition of the ink was changed by the manufacturer and there is no difference in the package.
Before, this ink used to contain Iron-Gall. Which makes the ink more permanent (old inks contained most of the times Iron-Gall). As Iron-Gall is an oxidizer, it has been removed from the ink because it can damage the pens if not cleaned regularly.

I don't know much about chemistry and I wonder if any of you could tell me a way to know if my ink has Iron-Gall and therefore is safe for long lifespan documents.
I was wondering if litmus paper would do the trick comparing this ink with normal black ink (which does not have Iron-Gall).

A simple and efficient method for the analysis and comparison of inks is thin layer chromatography (TLC).

In order to separate the various carboxylic and phenolic compounds that may be contained in iron gall ink, the mobile phase should be a polar solvent (e.g. ethyl acetate) or solvent mixture and made acidic (e.g. by adding formic acid).

The coloured spots can be compared under normal lighting; other spots may be detected under UV light. The phenolic compounds may be made visible by spraying with $\ce{FeCl3}$ solution.


Ag-Sn-Cu Phase diagram to formulate homogenization temperature / duration I am working on Ag-Sn-Cu alloy phase diagram to figure out the best temperature / duration to homogenize the alloy (annealing). The %age of elements in the alloy are

Ag - 40 %
Cu - 27.8 %
Sn - 32.2 %

(Composition can be varied if necessary but not more than 5 %)

Now the main goal is:

Find out the temperature / time duration at which the ingot of the above said alloy should be annealed to get the maximum Ag3Sn phase in the final ingot. Ag3Sn is the most desired phase with Cu3Sn being the second desired phase. Anyhow the main goal is to make as much Ag3Sn as possible.
The time duration for which the ingot should be placed at the desired temperature. (The total ingot is usually 6–7 kg)
[Supplementary question] Should the temperature be gradually reduced to the desired level of maximum Ag3Sn yield (or gradually increased)?

The phase diagram of Ag-Sn-Cu is as follows:

Much of this answer is based on a document available from NIST at Properties of Ternary Copper-Silver Systems [originally J. Phys. Chem. Ref. Data, 6(3) 621-673 (1977)] - this has an equivalent to the diagram in the question (which is just the surface of the liquidus). Now, since you are talking about balances of solid phases, it is also worth looking at the binary diagrams to discuss those.

For Ag-Sn, one gets:

For Cu-Sn there is:

Now I'll throw out a few observations:
Ag$_{3}$Sn is a line compound, which in the binary system is not stable above 480C. In the NIST report this phase is referenced as $\theta$, but called Ag$_{3}$Sn($\epsilon$), causing great confusion since the $\zeta$ Ag-Sn phase is designated $\epsilon_{2}$, and Cu$_{3}$Sn($\epsilon$) is designated $\epsilon_{1}$. So, watch it...

The Cu$_{3}$Sn line compound converts to the fcc $\gamma$ phase at around 676C. The $\gamma$ phase is sometimes also referred to as Cu$_{3}$Sn (yet more confusion possible). The Cu$_{3}$Sn and Ag$_{3}$Sn line compounds are different crystal structures, and do not form a solid solution across the ternary.

For both Ag-Sn and Cu-Sn, another thing to notice is the various other phases that are in close proximity to the Ag$_{3}$Sn and Cu$_{3}$Sn line compounds. Going into the ternary system, you get questions like how adding Cu into $\zeta$Ag-Sn impacts the free energy.

So, what now? In the NIST report, on page 664 is a pseudo-binary diagram where the Sn wt% is held at 20%, and Cu wt% is varied from 0 to 80 (balance Ag). Personally I hate weight percent, but...  The point that one can eventually pick out is that the Ag$_{3}$Sn phase, designated $\theta$ appears almost nowhere - there is just a little sliver way over near 0 wt% Cu and under 250C. It appears that with the addition of Cu, the $\zeta$Ag-Sn phase rapidly replaces the Ag$_{3}$Sn phase.

So, as far as I can tell, I'm not sure that the kinetics even matter here. You basically can't get much Ag$_{3}$Sn to form within the ternary parameters you have chosen - you will get $\zeta$Ag-Sn instead.


Does benzene structure stand for a single resonance form or the whole molecule? It is known that benzene has two main resonance forms

Often we still draw benzene molecule as one of them

My question is, actually when we draw the lower figure, do we refer to the whole benzene molecule or one of the resonance structure or depend on the context?

The answer is you are referring to neither of them. That is because resonance structures don't actually exist in reality. We only use them to give us a rough idea what the actually molecule and bonds look like. A common way to explain resonance structures is this:

An explorer from a far distant land travels to a new continent and sees a strange animal that he has never seen before, a rhinoceros! When he comes back home, not really sure how to describe this strange animal, he tells everyone that it looked like an unicorn and a dragon mixed together.

So what the explorer did was he tried to explain some weird, not easily imagined object (a rhinoceros) by describing it as a mixture of two mythical, non-existent objects, but which are easily imagined by people.

That is exactly what the resonance structures of benzene (also for any other molecule) are. They don't actually exist, but rather benzene exists as a weird mixture of these resonance structures. The reason why we use resonance structures is that they give us an idea how to imagine what benzene looks like.


How do pharmaceutical manufacturers ensure 100% purity in their medications? What purification measures are taken to ensure no side products are present in the medication? Because hardly any organic reaction is perfectly efficient, I'm guessing that multiple recrystallizations are performed. Also how would recrystallizations even be performed on an industrial scale? How much product is then lost and can it be recovered or is not economically advantageous to do so?

Most registered specifications will contain the main active pharmaceutical ingredient (API) at an assay of 98-102% to allow for hplc variation(the most common analytical technique for pharmaceuticals). The list of known impurities usually follows with their maximum limit ; this is usually derived from toxicology studies. But not always.

An exception is something like Sativex which has an assay of <90% for the main API, but the other ingredients are not detrimental but generally regarded as safe (GRAS); sterols, triglycerides.

The drug product then has the API in  specified range and can comprise most of the medication e.g. paracetamol tablets, or not very much e.g. acyclovir in ointment form or cisatracurium besylate in water for injection. Excipients are added which could be seen to reduce the "purity" of the medication, but they are present to aid bioavailability, taste etc.

Any technique you can think of can be used industrially... at a cost. So while liquid -liquid extraction is cheapest and easiest,  it rarely gives the desired purity. Recrystallisation is a cost effective technique and is frequently used industrially. Put solid in a reactor, pump in solvent, heat, cool. Extract, distil solvent,  solvent switch, heat, cool is also valid.

Yield loss from industrial scale recrystallisation is usually 5% or greater.  I have seen 30% for achiral material. I've seen 55% for a diastereomeric salt crystallisation (so at least 50% is to be expected).

Sometimes one crystallisation is enough; sometimes three are required. It depends on the chemical transformation efficiency and co-crystallising impurities.

As for recovery ; economic viability depends on the cost of the material. So at 20 dollars/kg, probably not. 15000 dollars/kg, probably yes.

Not all impurities are bad as they can be very similar to the API and simply have a different effect e.g. anti-histamine.  Some could genotoxic and are controlled at the ppm level.

Unknown impurities are tightly controlled to, typically, the total being no more than (NMT) 0.5%. Again, there are exceptions.


How to find the number of valence electrons? I need to understand the following:

Considering an element Sulfur - S which has 16 electrons.

How do we calculate the number of valence electrons of S?

Please correct me if I am wrong:2+2+6+2+4=16. So, the valence electron = 4+2=6

Is it that always valence electron is the addition of the last two shells? In that case for Na, sodium, if we take 11 and arrange it as 2+2+6+1, then the valence electron is 1 or 6+1?

Now, considering the fact valence electron is 1, if we go by, up down up down rule, then as it is 1, hence the spin is up?

For 6: is it up(1) down(2) up(3) down(4) up(5) down(6) as it is both up/down hence it would be +-1/2 spin?

For Phosphorous 2,8,5, what would be the valence electron? 5 or 13?

If 5 then up down up down up, will it have an up spin?

Is it that for even numbers like 4,6 the spin will be both up/down i.e. +-1/2?

'p' and 'd'.

There is one type of 's' orbital. It is spherical. So an s orbital can hold 2 electrons only.
There are three types of 'p' orbitals. They are the same shape (like a dumbell) but are oriented in different directions - along the x, y and z axes.  So a 'p' orbital can hold $2$x$3$ = 6 electrons.
There are five types of 'd' orbitals (10 electrons) and seven types of 'f' orbitals (14 electrons).

Fill up the orbitals in the following order (following the red arrows), which is increasing order of energy. Lowest energy fills up first.

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p

So let's take the example of sulfur.
The total number of electrons is 16. We'll distribute them in the following manner.

$1s^2, 2s^2, 2p^6, 3s^2, 3p^4$, where the superscripts of each orbital represent the number of electrons in it. Now, we have a total of 6 electrons in the last (3rd) orbital. Therefore the valence number is 6.

It is not simply the last 2 numbers while writing the electronic configuration.

For Phosphorous, we have:
$1s^2, 2s^2, 2p^6, 3s^2, 3p^3$, which gives us 5 electrons in outermost (3rd) orbit.

For Lithium (3 electrons), we have:
$1s^2, 2s^1$, which gives us a total of 3. Outermost orbit is the 2nd orbit and the total number of electrons in it is just one. Therefore valency of Lithium is one.

However, it is much simpler to do it the way I've shown above as with Sodium. Don't need to bring in all the orbitals

For the spin, just remember that you have to fill up the entire orbital with up spin electrons before you start putting the down spin electrons. Don't just alternate up and down.

Let's talk about the 'p' orbital. Each 'type' of the p-orbital must have up-spin electrons before adding the down-spin. So there will be 3 up spin electrons before you start adding the down-spin ones. Also, you can not move on to the next orbital without completely filling the previous one with an EQUAL number of up and down spins.

For Phosphorous; let's say we've filled the 1s, 2s, and 2p orbitals. Now we add one up spin to the 3s, followed by a down-spin. Which leaves us with 3 more electrons to be added to the valence orbit (3rd). It is a 'p'-orbital, which needs to have 3 up spin electrons before any downspin ones can be accommodated. But we have only 3 remaining! So all will be up-spin electrons in the valence orbit. So we'll have a total spin of +3/2.

Hope I didn't complicate too much, but I've actually tried to make it as short as possible. I've still left out quite a bit.


What happens to the methanol formed by the hydrolysis of methyl 2-hydroxybenzoate (oil of wintergreen) when synthesising salicylic acid? Here are the reaction equations for reference (you may recognise this as a preliminary step for the synthesis of aspirin).

Steps for synthesis of salicylic acid:

1) methyl 2-hydroxybenzoate + sodium hydroxide -> sodium 2-hydroxybenzoate + methanol

2) sodium 2-hydroxybenzoate + hydrochloric acid -> salicylic acid + sodium chloride

Does the methanol play an underlying role in the reaction somewhere? Is the methanol filtered off? I do not know.

Thanks in advance.

The methanol plays no underlying role in the reaction

Without more detail of the reaction procedure I cannot give a complete answer, but my expectation is that after the HCl addition the reaction will be quenched by addition of water to either precipitate the salicylic acid or then be extracted into an immiscible organic solvent. The methanol will stay in the aqueous phase in either case.


How to use C1V1=C2V2 How do the units work? I am trying to do a dilution where I know I have 0.749 g/ mL and I want to use this concentration to calculate the grams of aspirin in each 100 mL solution. I am specifically worried about the units. I want grams out. Is it valid to do the following?
One:
$$C1V1=C2V2$$
$$(0.749 g)(1 mL)=(C2 g)(100mL)$$
$$C2=0.00749 g$$

or must I do this?
Two:
$$C1V1=C2V2$$
$$(0.749 g/mL)(1 mL)=(C2 g/mL)(100mL)$$
$$C2=0.00749 g/mL$$

The difference being the final units. g vs. g/mL

Concentration has units mass over volume; or particles (moles by convention) over volume.

It's fine to use $g/mL$ for concentration in a basic dilution equation.

$C$ is never a mass.

So your second equation has proper units.

If you want mass of aspirin in a given solution with concentration $C$, the mass is given by $m = C\times V$


What is the relation between ionization energy and electropositivity? Electropositivity of an element is defined as a measure of element's ability to donate electrons while Ionization energy or Ionization potential is defined as the energy required to dispel a loosely held electron from the valence shell. These two terms sound closely related to me or they are just measure and energy?

Electropositivity is rarely used by itself. It is much more commonly seen as just the opposite of electronegativity. While electronegative is more or less defined to the point where numeric values can be found for each element in a given system (sometimes excluding noble gases), electropositivity has no such scale I am aware of.

Ionisation energy or potential are clearly defined energies that can be experimentally directly determined and are tabulated across physical chemistry textbooks with little to no variance (the main variance being experimental error).

The two terms are hardly related; neither by definition/physical meaning nor by value trend.


Enzyme Kinetics two enzymes in complex with one substrate Given an enzyme reaction where two enzymes are required to form the complex, for example $\ce{2E +S-> E3S->2E +P}$, how would I calculate complex concentration assuming steady state approximation ($\frac{d\ce{E2S}}{dot}=0$). Does the 2 in front of $\ce{E}$ multiply the rate or is it the exponent when looking at the rates going into the complex?

As this seems like a homework question (and may be closed), as an example consider the reaction  $\ce{H2 + I2 = 2HI}$. The rate, r is $\displaystyle r=\frac{1}{2}\frac{d\mathrm{[HI]}}{dt}=k_1\mathrm{[H_2][I_2]}-k_2\mathrm{[HI]}^2$ where the rate constants $k_1,\;k_2$ are for the forward and reverse reactions respectively. (This assumes that the reaction occurs as written.) As   steady state the intermediate or transient species has a rate of change of zero and you note in your question.

In the general case if the reaction is $aA+bB= dD+eE$ then $\displaystyle -\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=\frac{1}{d}\frac{d[D]}{dt}=-\frac{1}{e}\frac{d[E]}{dt}$


Why is Potassium in French and English not called Kalium? The symbol from the periodic table for Potassium is "K". This fits my native language (dutch), where Potassium is called "Kalium". The same question applies to Sodium (Na), which again in Dutch is called "Natrium". A quick look at Wikipedia showed that Potassium is mainly used as a word in Roman based languages (French, Spanish) and to my surprise in English. What is the history behind this?

(with thanks to @NilayGhosh, who provided the key source for much of this answer)

Short answer: Potassium was called Potassium (and Sodium Sodium) by Humphry Davy, who first isolated both; it was then renamed Kalium in Germanic countries because of a previous naming proposal by Klaproth, who was the first to show that potassium carbonate and sodium carbonate were different.

Until at least the Middle Ages no distinction was drawn between potash (potassium carbonate) and soda (sodium carbonate). Martin Heinrich Klaproth first distinguished them in 1797, and suggested the name 'kali' for the first and 'natron' for the second.

Metallic potassium and sodium were then isolated for the first time in 1807 by Sir Humphry Davy, by electrolysis of potassium hydroxide (KOH) and sodium hydroxide (NaOH) respectively. He named his discoveries Potassium and Sodium in the Bakerian lecture at the Royal Society on the 19th of November the same year:

  Potasium and Sodium are the names by which I have ventured to call the
  two new substances

(source)

(During the lecture he also apparently produced a nugget of metallic potassium and tossed it into a flask of water, "where the lump skittered around the surface of the water before exploding in lavender flames".)

This paper was translated into German by Ludwig Wilhelm Gilbert (supposedly in Annalen der Physik, although I can't find the original paper in any of the volumes from  1809). According to the article on vanderkrogt.net Gilbert replaced Potassium with Kalium and Sodium with Natronium "in agreement with German nomenclature" (i.e. following Klaproth's suggested names). From this point on, generally speaking, English- and French-speaking countries followed Davy's naming system, while the Germanic countries followed that of Klaproth/Gilbert, as can be seen from the sidebar lists here (potassium) and here (sodium).

Regarding the atomic symbols mismatch: the international system of atomic symbols was devised by the Swedish chemist Jacob Berzelius, a rival of Davy's, who followed the naming convention used by Germanic countries. Although actually, when first published (1813, in Thomas Thomson's Annals of Philosophy), Berzelius did follow Davy's names and used the symbols Po and So. Within a year he had switched to K/Na. So since the first publication of Berzelius' system actually followed Davy and used Po/So, the real surprise is that we ended up using K/Na.


Can every chemical compound be melted? Are there any chemical compounds that disintegrate (without going into other chemical reactions; let's say in a vacuum) before reaching a melting temperature?

Some compounds decompose before they reach the melting point. In the literature, you'll see something like "m.p. $40\ ^{\circ}\mathrm{C}$ (dec.)"

Ammonium chloride is an example.


How to determine omitted units in a publication I have found a 1954 paper in J. Am. Chem. Soc. which gives various temperatures in degrees but does not specify units.

A snippet of the introduction is:

  Titanium reacts appreciably with fluorine above 150°…

If I were feeling charitable, I might excuse the "units are for wimps" attitude as reflecting a commonly understood standard for that publication at that time. Perhaps my searching was poor, but I couldn't find an ACS standard or style guide for the 1950s era.

Given the location and time, I would guess that the temperatures are in degrees Fahrenheit, but I can't be sure they're not Celsius. Is there any reasonable way to determine which units the authors were actually using?

I don't have access to the full paper, just the first page at the link below.

Haendler et al., J. Am. Chem. Soc., 1954, 76 (8), pp 2177–2178
DOI 10.1021/ja01637a038

These are degrees Celsius.

To prove that, you could search JACS for the publications of the same year/issue for the temperature-related physical constants such as melting point:

Douglas et al. [1] lists m.p. of $\ce{LiF}$ as $\pu{848 ± 1 ^\circ}$, whereas currently Wikipedia lists $\pu{845 °C}$.
Yankwich et al. [2] lists m.p. of malonic acid as $\pu{140 ^\circ}$ whereas currently Wikipedia lists $\pu{135 to 137 °C}$.

You may find further examples, but even at this point it's fairly obvious what those temperature units are.

References

Douglas, T. B.; Dever, J. L. Lithium Fluoride: Heat Content from 0 to 900°, the Melting Point and Heat of Fusion. J. Am. Chem. Soc. 1954, 76 (19), 4826–4829. https://doi.org/10.1021/ja01648a016.
Yankwich, P. E.; Promislow, A. L. Intramolecular Carbon Isotope Effect in the Decarboxylation of Liquid Malonic Acid Near the Melting Point. J. Am. Chem. Soc. 1954, 76 (18), 4648–4651. https://doi.org/10.1021/ja01647a050.


How long can iodine bottle left open? Can it go bad sooner than expiration date? I have bought iodine about a year ago but the bottle cap got damaged during the transportation and it leaked all over the package, but I have decided to keep it anyway. Unfortunately I did not use it to anything and it was just sitting on my shelf with cracked cap with about a one millimetre sized hole.
I wonder if it could go wrong and now it's useless. As far the expiration date, it's still not expired, but that's probably when the bottle is fully sealed.
So is it still safe to use if needed or I can return it to the drug store so they can safely dispose it?
Content for 100ml: iodopolyvidonum 7,5g with 10% content of Povidone-iodine (M.h. 40 000)
Thank you

A solution of iodine antiseptic is a medical item that is used for improving health. There are two ways to treat your cracked cap.
#1. If you inspected the bottle in the drug store before you bought it, you would not buy this bottle. Almost all over-the-counter medicines have warnings: Do not use if the safety seal is broken. So, out of utmost caution, do not use this iodine, because it might be contaminated (even tho you are the only one who could have contaminated it). Look at it this way: you bought a bottle of iodine solution, you had its potential usefulness for a year to keep your skin abrasions antiseptic (even tho you did not have the need to use it). It can be considered to have served its purpose usefully; get a new bottle and be more careful.
#2. Being the cheapskate that I am, I would examine the bottle: has the liquid level dropped and dried out? Is it gooey so as to be difficult to apply? If not, it might be OK to keep and use. On the other hand, a cracked lid is an invitation to disaster - if the bottle gets knocked, it will leak - probably onto the item you just laid down, which will, of course, be very important and/or delicate. To avoid catastrophes of this sort, dispose of the iodine solution. You will feel more secure. Get a new bottle.
Or, get a tube of antiseptic ointment, which will have less tendency to spill, unless you step on it.


How is dioxygen stabilised by coordination to haemoglobin? I wonder how the binding of $\ce{O2}$ to a molecule of hemoglobin stabilizes it to the extent that I can carried around the body with the risk of it being involved in oxidation reactions minimized. The same applies to hemocyanin.

I am looking for an answer invoking orbital interactions, back-bonding etc.

I think you have a misconception. The binding of oxygen to haem and its release are processes that are entirely equilibrium-controlled by the equilibrium shown in $(1)$.

$$\ce{haem + O2 <=> haem{-}O2}\tag{1}$$

Meaning that wherever high partial oxygen pressures are observed, oxygen preferentially binds to haem but likewise it is released where low partial oxygen pressures are observed. Thereby, Le Chatelier’s principle nicely controls where oxygen is released from the red blood cells’ haem groups — it will always be in those areas where oxygen is required. Haem itself has evolutionally acquired the role of binding oxygen rather well but only to make sure that it actually reaches the distant parts of the body.

Oxygen in its $\ce{O2}$ modification is also not as bad as your initial post seems to assume. Yes it can react with quite a few substances and many of its reactions have a low activation barrier — but the concentrations in which it would be released into the blood stream ‘undesiredly’ are concentrations the body has learnt to cope with over millenia of evolution. So while minor damage may occur, most damage is quickly repaired by dedicated enzymes. Most importantly, any liberated oxygen will always be $\ce{O2}$ and never superoxide or peroxide.

Only while being bound to haem do oxygen’s (and iron’s) electronic properties change. On a microscopic level, the oxygen molecule causes a one-electron oxidation of iron converting itself to superoxide. This also causes a spin-flip in the iron centre which goes from high to low spin. The combined $\ce{\overset{\mathrm{+III}}{Fe}-O2}$ system can finally undergo antiferromagnetic coupling to give an overall observed diamagnetic ground state. Details can be found in this answer. However, while it is superoxide the $\ce{O2}$ moiety will never leave the iron centre — or, more precisely described, as soon as $\ce{O2}$ leaves, the entire process is reversed again. Thus, even though the bound state includes a superoxide ion no superoxide can ever be directly released into the surrounding blood stream.

$$\ce{[\overset{\mathrm{+II}}{Fe}(por)] + ^3O2 <=> [\overset{\mathrm{+III}}{Fe}(O2^{.-})(por)]}\tag{2}$$


Is there an easy way to detect methanol in spirits (which may have been adulterated)? Is there an easy way to detect methanol in spirits, which, you suspect, may have been adulterated?

Well that depens on your possibilities.

a) Destillation / Boiling Point (easy, unsafe)

b) Refractive index (depends on impurities)

c) GC (best, very expensive)

d) 1H NMR


What condenser can be used for simple distillation? I think I need some distilling (boiling) and collection flasks and a 3-way adapter with a thermometer.  I was thinking of using an open vacuum adaptor to join the lower end of the condenser to the receiving flask.  Does this sound ok?  And if so, what kind of condenser should I use?

That sounds good.  Here's a drawing of the set up.  I'd use a water-cooled condenser.


Rate of hydrogenation The rate of hydrogenation is faster for which alkene : cis or trans. I know it depends on steric hindrance on the carbon, but both have steric hindrance. Can someone please help

Are you asking about catalytic hydrogenation or non-catalytic hydrogenation (e.g. diimide)?  For the time being, let's assume that your question is about the catalytic process.  All the catalyst does is lower the activation energy required to reach the transition state (TS) and the TS will look about the same independent of which isomer you start with.  Therefor, ground state stability differences between the two olefins will determine which isomer reacts faster.  Since trans-olefins are generally more stable than their corresponding cis isomers, it will take more energy (the difference in stability between the cis and trans isomers) for the more stable (lower energy) trans isomer to reach the TS, and less energy for the cis isomer to reach the TS.  Consequently cis isomers react somewhat faster than their trans-counterparts during catalytic hydrogenation.  I suspect that the same considerations would apply to non-catalytic processes.

As an aside, remember that in the catalytic process, depending upon catalyst, temperature, solvent, pressure, etc.; many other things can be going on, such as cis-trans isomerization of the starting material, positional isomerization, etc.  So if the reaction conditions are not chosen carefully and 1) if cis-trans isomerization is faster than hydrogenation and 2) let's say the cis isomer reacts 100 times faster than the trans; then both isomers could appear to give the same rate of hydrogenation because all hydrogenation would proceed through the cis isomer.


Use of subscript cc for HCl and H2SO4 Does the subscript $\ce{X}_{cc}$ mean concentrated? I have never seen this notation. How often is it used and what are the general guidelines to denoting concentrated reagents?

I found a very old book that used this notation to describe a plot where $X$ many cubic centimeters of $\ce{HCl}$ are added to a solution.


Is it possible to figure out how many carbon-carbon double bonds are present using just the molecular formula? One of my test review questions reads as follows:

  "The prostaglandin precursor arachidonic acid has the molecular
  formula $\ce{C20H32O2}$. Given that arachidonic acid is an acyclic carboxylic
  acid that contains no carbon-carbon triple bonds, how many
  carbon-carbon double bonds are present?"

My original instinct was to draw out the structure, but having just the molecular formula to go off of and nothing else, it seems like it would take a ton of trial and error to figure out the correct structure. Is there a simpler way to approach a problem like this to figure out the amount of double bonds?

The only other thing that came to mind was to calculate the degree of unsaturation from the formula, which came out to 5. So, I know there can't be more than that. But that could represent both double bonds and/or rings, so. Anything I'm missing?

(The answer is 4, by the way. I just need help figuring out how the process of getting there is laid out.)

Thanks.

Both you and Rapid99 have missed the word "acyclic" in the problem statement.  So you don't have to worry about rings (because they said acyclic).  [Also no triple bonds.]

You also know that it's a carboxylic acid ("COOH group", formula but not structure of carboxylic acid group), so don't worry about something funky with the oxygens.  The formula is then $\ce{C19H31-COOH}$.

If the part of the compound on left was full saturated and sans $\ce{COOH}$ it would be $\ce{C19H_{(2*19+2)}}=\ce{C19H40}$.  Remember any acyclic alkane has formula $\ce{C_nH_{2n+2}}$.

Accounting for the $\ce{COOH}$ group location means the saturated alkane would then be $\ce{C19H39-COOH}$ (take away one $\ce{H}$ "spot" to attach the COOH group).

So we are really 8 hydrogens short: 39-31 (from above) = 8.  Each double bond takes away two hydrogens.  So if there are 8 missing hydrogens, there must be 4 double bonds.  X=8/2.


How can we justify setting the affinity $\sum_i \mu_i \nu_i$ equal to zero at chemical equilibrium? Let $\sum_A \nu_A A \rightarrow \sum_B \nu_B B$ be a general reaction whose progress during time interval $dt$ is measured by $d\zeta$, so the amount of reactants consumed and products generated in mole would be $d N_i=\nu_i d\zeta$, wherein, $\nu_i$ would take negative values for the reactants being consumed.

From on the other hand we already know from the second law of thermodynamics that the reaction will occur in the direction in which $\sum_i \mu_i dN_i \le 0$, wherein, $\mu_i=\frac{\partial G}{\partial N_i}\bigr|_{p,T,N_j\;(j\ne i)}$ are the chemical potentials, so that we would have:
$$\left(\sum_i \mu_i \nu_i\right)\; d\zeta \le 0$$
The equality implies reversibility and should denote the equilibrium (am I right?). At the equilibrium it is clear that the progression of the reaction should vanish and we should have $d\zeta=0$, so that the term $\left(\sum_i \mu_i \nu_i\right)$ should be free to gain any positive, zero or negative finite value. However, this is not what Guggenheim has written in his Thermodynamics book. He has first assumed in a given direction the reaction progresses, so that in that direction $d\zeta>0$, then has canceled out this positive quantity from the inequality and achieved: $\left(\sum_i \mu_i \nu_i\right) \le 0$, and eventually concluded that at the equilibrium the equality holds and we should have $\left(\sum_i \mu_i \nu_i\right) = 0$.

But how is it justified to be true when we already know at the equilibrium $d\zeta>0$ doesn\t hold and we should instead write: $d\zeta=0$?

Thanks for bearing with me, I'm new to chemistry.

Chemical equilibrium is a stable equilibrium, i.e. the system returns to equilibrium to whatever perturbations. This also leads to the condition of equilibrium, as it should  contain this information.

The short answer to your question:

Due to the stability, the system stays/returns to equilibrium at any kind of small perturbation. It also means, $\left(\sum_i \mu_i \nu_i\right)\; d\zeta = 0$ not only in the equilibrium point, but also for any small $d\zeta$. Therefore the $\left(\sum_i \mu_i \nu_i\right)$ itself must be zero.

Somewhat longer answer:

The reaction goes from one direction to the other ($\sum_A \nu_A A \rightarrow \sum_B \nu_B B$)  spontaneously only till it reaches the equilibrium condition. In this direction  $\sum_i \mu_i dN_i \le 0$.

We followed the reaction from a given concentration toward equilibrium. If we follow the same $\sum_A \nu_A A \rightarrow \sum_B \nu_B B$  transformation BEYOND the equilibrium point, $ \sum_i \mu_i dN_i $ must change sign at the equilibrium point, and it must be positive should be valid (since it is not a spontaneous reaction).

Therefore the condition of equilibrium is not $d\zeta = 0 $ , but sign change: $ \sum_i \mu_i dN_i =0$.


charge density as a measure of lattice enthalpy & polarizing power? My official CIE A level textbook, and some mark schemes mention the following:

Ions with the same charge have a lower charge density if their radius is large. This is because the same charge is spread over a larger volume. A lower charge density weakens the electrostatic attraction and thus decreases the lattice enthalpy of an ionic lattice.

From a physics point of view, this can't possibly be right, can it? Modelling the ions as spheres, gauss' law/ shell theorem clearly state that for all points outside a shell/sphere (of uniform charge density), the spheres/shells are equivalent to point particles. Why should the volume of the ion have any affect on the attractive force? And what would, then, be the correct explanation for decreasing lattice enthalpy down the group?
moreover, charge density of an ion is  also said to affect its polarising power. The polarisibing power should depend on the electric field created by the ion which,again, doesn't depend on the size at all. If this too is wrong, why does Be have a greater covalent character than Mg?

Atoms are not static distributions of charge, and the models we use are sometimes heuristic (approximations). It can be useful - if approximate - to describe an atom in terms of a distribution of static charges, and to compare two different atoms by saying they differ by evenly "stretching" the charge distribution in the radial dimension.
Yes, Gauss's law applies (assuming a static charge distribution). Assume we are comparing two ions that differ only in the radial charge distribution (by stretching the distribution). Gauss's law simplifies the comparison, since we need only consider the change in the distance from the ion center to the charged test particle (placed at the surface of the ionic sphere) and not the magnitude of the charge within the sphere (which remains constant; for a uniform charge distribution it can also be assumed to remain centered in the sphere). The Coulombic interaction energy falls as 1/r (the inverse distance between charges) and the electric field as $1/r^2$. The effect of radially scaling the distribution ($r\rightarrow a\cdot r$) is to multiply the energy and the field by a constant ($a^{-1}$ or $a^{-2}$). If you make an ion larger, the distance of a charged particle (outside of the ionic sphere) from the center of the ion increases and both the interaction energy and electric field decrease in magnitude.
Electric polarizability $\alpha$ (here assumed isotropic and written as a scalar - it is generally a tensor) is a measure of the extent to which charges can be separated by an applied electric field, resulting in an induced dipole moment:
$$\vec{\mu} = \alpha \vec{ E }$$
To compare atoms consider the magnitude of the electric field required to induce an equivalent dipole moment. This is determined by the polarizability (or rather its inverse). More polarizable means a weaker field suffices to generate the same dipole moment.
Finally, as explained in the Wikipedia:

Generally, polarizability increases as the volume occupied by electrons increases.[7] In atoms, this occurs because larger atoms have more loosely held electrons in contrast to smaller atoms with tightly bound electrons.[7][8]


Alkane, alkene, alkyne boiling point comparison Which of the following has higher boiling points? Alkanes, alkenes, or alkynes? And why?

Disclaimer: All of this "jazz" will be about reaching a mere rule-of-thumb. You can't just compare whole families of organic compounds with each other. There are more factors to consider than below, mostly based on isomerism notions. However, as most of the A grade exams emphasize on the lighter aliphatic compounds, we can understand each other here. :)

It's all about polarizability.

  Polarizability is the ability for a molecule to be polarized.

When determining (aka comparing) the boiling points of different molecular substances, intermolecular forces known as London Dispersion Forces are at work here. Which means, these are the forces that are overcome when the boiling occurs. (See here for example)

Source

London forces get stronger with an increase in volume, and that's because the polarizability of the molecule increases. (See the answer to this recent question)

Alkanes vs. Alkenes

In their simplest form (where no substitution etc. has occurred) alkanes tend to have very close boiling points to alkenes.

  The boiling point of each alkene is very similar to that of the alkane with the same number of carbon atoms. Ethene, propene and the various butenes are gases at room temperature. All the rest that you are likely to come across are liquids.

  Boiling points of alkenes depends on  more molecular mass (chain length). The more intermolecular mass is added, the higher the boiling point. Intermolecular forces of alkenes gets stronger with increase in the size of the molecules.

  \begin{array}{|c|c|}\hline
\text{Compound} & \text{Boiling point / }^\circ\mathrm{C} \\ \hline
\text{Ethene} & -104 \\ \hline
\text{Propene} & -47 \\ \hline
\textit{trans}\text{-2-Butene} & 0.9 \\ \hline
\textit{cis}\text{-2-Butene} & 3.7 \\ \hline
\textit{trans}\text{-1,2-dichlorobutene} & 155 \\ \hline
\textit{cis}\text{-1,2-dichlorobutene} & 152 \\ \hline
\text{1-Pentene} & 30 \\ \hline
\textit{trans}\text{-2-Pentene} & 36 \\ \hline
\textit{cis}\text{-2-Pentene} & 37 \\ \hline
\text{1-Heptene} & 115 \\ \hline
\text{3-Octene} & 122 \\ \hline
\text{3-Nonene} & 147 \\ \hline
\text{5-Decene} & 170 \\ \hline
\end{array}
  In each case, the alkene has a boiling point which is a small number of degrees lower than the corresponding alkane. The only attractions involved are Van der Waals dispersion forces, and these depend on the shape of the molecule and the number of electrons it contains. Each alkene has 2 fewer electrons than the alkane with the same number of carbons.

Alkanes vs. Alkynes

As explained, since there is a bigger volume to an alkane than its corresponding alkyne (i.e. with the same number of carbons) the alkane should have a higher boiling point. However, there's something else in play here:

Alkynes, have a TRIPLE bond!

I currently can think of two things that happen as a result of this:

London Dispersion Forces are in relation with distance. Usually, this relation is $r^{-6}$. (See here) The triple bond allows two alkynes to get closer. The closer they are, the more the electron densities are polarised, and thus the stronger the forces are.
Electrons in $\pi$ bonds are more polarizable$^{10}$.

These two factors overcome the slight difference of volume here. As a result, you have higher boiling points for alkynes than alkanes, generally.

\begin{array}{|c|c|}\hline
\text{Compound} & \text{Boiling point / }^\circ\mathrm{C} \\ \hline
\text{Ethyne} & -84^{[1]} \\ \hline
\text{Propyne} & -23.2^{[2]} \\ \hline
\text{2-Butyne} & 27^{[3]} \\ \hline
\text{1,4-Dichloro-2-butyne} & 165.5^{[4]} \\ \hline
\text{1-Pentyne} & 40.2^{[5]} \\ \hline
\text{2-Heptyne} & 112\text{–}113^{[6]} \\ \hline
\text{3-Octyne} & 133^{[7]} \\ \hline
\text{3-Nonyne} & 157.1^{[8]} \\ \hline
\text{5-Decyne} & 177\text{–}178^{[9]} \\ \hline
\end{array}
1: http://en.wikipedia.org/wiki/Acetylene
2: http://en.wikipedia.org/wiki/Propyne
3: http://en.wikipedia.org/wiki/2-Butyne
4: http://www.lookchem.com/1-4-Dichloro-2-butyne/
5: http://en.wikipedia.org/wiki/1-Pentyne
6: http://www.chemsynthesis.com/base/chemical-structure-17405.html
7: http://www.chemspider.com/Chemical-Structure.76541.html
8: http://www.thegoodscentscompany.com/data/rw1120961.html
9: http://www.chemsynthesis.com/base/chemical-structure-3310.html
10: https://chemistry.stackexchange.com/a/27531/5026

Conclusion: We can't fully determine the boiling points of the whole class of alkanes, alkenes and alkynes. However, for the lighter hydrocarbons, comparing the boiling points, you get:
$$\text{Alkynes > Alkanes > Alkenes}$$


How to come up with overall equation of a galvanic cell How are overall equations of the galvanic cell written? In a zinc-copper cell, the overall equation is

$$ \ce{Cu^2+(aq) + Zn(s) -> Zn^2+(aq) + Cu(s)} $$

So in order to come up with this equation, do you take the net ionic equation that occurs when $ \ce{Zn} $ is placed into $ \ce{CuSO4} $ solution (the electrolyte for copper half cell)? In other words, do you calculate the overall equation of the galvanic cell by coming up with the net ionic equation of the single displacement reaction that occurs between the anode electrode and the cathode electrolyte?

In this example, it would be

$$ \ce{Zn(s) + Cu^2+(aq) + SO_4^2- (aq) -> Zn^2+(aq) + Cu(s) + SO4^2- (aq)} $$

where $ \ce{Zn(s)} $ would be the anode electrode and $ \ce{CuSO_4} $ would be the cathode electrolyte.

And the $ \ce{SO_4^{2-}} $ will cancel out since it is a spectator ion to give the overall equation of the galvanic cell.

Am I doing this right?

Since $ \ce{Zn} $ never touches any $ \ce{Cu^2+} $ ions in a galvanic cell - the salt bridge/porous plate is preventing this - it would not be helpful to take this chemical equation as a starting point. As the aim of such cells is to separate the reduction and oxidation process, this is where the equations should originate.

Direction of the Reaction

To start with, one should find out which of the 2 electrode materials is more noble, hence will be the oxidizing agent aka cathode material, and will get itself reduced during the process through receiving electrons from the anode over the external circuit. A higher standard electrode potential ($E^\circ$) correlates with higher nobility. Information on different $E^\circ$ values can be found in galvanic series (Wikipedia: Standard electrode potential (data page)). A lookup of $\ce{Cu}$ and $\ce{Zn}$ gives:

$$
\begin{array}{rclc}
\hline
\text{Oxidant} & & \text{Reductant} & E^\circ/\mathrm{V}\\
\hline
\ce{Zn^2+(aq) + 2 e− & <=> & Zn(s)} & −0.7618\\
\ce{Cu^2+(aq) + 2 e− & <=> & Cu(s)} & +0.337\\
\hline
\end{array}
$$

Since $\ce{Cu}$ has the higher $E^\circ$ (is more noble), $\ce{Cu}$ will act as cathode, and $\ce{Zn}$ as anode, to provide the reactions spontaneity.

Setting up the Reaction

Looking at the electrochemical process, first the $ \ce{Zn} $ from the zink electrode gets oxidized according to:

$$ \text{Ox - Anode half equation:}\qquad\ce{Zn(s) -> Zn^2+(aq) + 2 e-} $$

After passing through the external circuit the electrons get to reduce the $ \ce{Cu^2+} $ ions according to:

$$ \text{Red - Cathode half equation:}\qquad\ce{Cu^2+(aq) + 2 e- -> Cu(s)} $$

Which shows, that the anode is going to lose mass through corrosion, while the cathode is going to gain mass through the formation of solid copper on its surface.

After having these two half equation we can sum them up and cancel out the $ \ce{2 e-} $ respectively to yield:

$$ \text{Redox - Total equation:}\qquad\ce{Cu^2+(aq) + Zn(s) -> Cu(s) + Zn^2+(aq)} $$

While these processes happen, the salt bridge/porous plate is allowing the balancing of charges in both half cells.

Note: Of course someone could begin with the reduction, rather than with the oxidation, but it would anyway end in the same total equation, and it's harder then to imagine the electrons wandering through the external circuit.

Edit: As requested in the comments: The reactions with spectator ions:

$$\begin{align}
\text{Ox - Anode:}&& \ce{Zn(s) + SO4^{2-}(aq) &-> Zn^2+(aq) + SO4^{2-}(aq) + 2 e-}\\
\text{Red - Cathode:}&& \ce{Cu^2+(aq) + SO4^{2-}(aq) + 2 e- &-> Cu(s) + SO4^{2-}(aq)}\\
\text{Redox - Cell:}&& \ce{Cu^2+(aq) + Zn(s) + 2SO4^{2-}(aq) &-> Cu(s) + Zn^2+(aq) + 2SO4^{2-}(aq)}
\end{align}$$

Edit 2: Added Direction of the Reaction


Molecule with 13 total electrons Is there a stable molecule with total number of electrons equal to 13? Molecule can be either organic or inorganic

There are quite some molecules that are persistent (= live long enough in solution to allow for a spectroscopical characterization, but cannot be isolated) or even stable (= can be isolated, bottled and/or used as a starting material) that have an odd number n of electrons, however $n \ne 13$.

Among these radicals are the Gomberg radical (1), the Koelsch radical (2), and the TEMPO radical (3).


Speed of ionic reactions? Why are ionic reactions extremely fast,  as compared to other gaseous
  reactions?

The ionic reactions (precipitations, neutralizations etc.) are extremely fast (somewhere about $10^{12}$ times faster) than gas-phase reactions and other ordinary reactions in solutions (inversion, substitution etc). Why is this so?

My thoughts
One possible reason might be the relative freedom of steric orientation in case of simple ionic reactions as compared to organic reactions and complicated gas reactions, but this isn't the case with simple gas reactions or formation and dissociation of ionic complexes, which do require specific steric orientations. The activation energy could be very low for ionic reactions, if this is so, then what is the reason behind it?

Reaction rate is inversely proportional to activation energy for a given process. The examples given have very low barriers and often the diffusion in solution is the rate limiting process.

If you think of reaction of anion with cation, than there should be no barrier at all, because the interaction between oppositely charged species is strongly attractive.


Why do we need sulfuric acid in creating alkyl bromide and not in creating alkyl iodide from alcohol and hydrogen halide? i) $\ce{R-CH_2-OH + HBr   ->[H_2SO_4] R-CH_2-Br + H_2O}$

ii) $\ce{R-CH_2-OH + HI   -> R-CH_2-I + H_2O}$

What role does sulfuric acid play in the first reaction?
Why are we not using it in the second one?

I suspect $\ce{HBr}$ is produced in situ from $\ce{KBr}$ and $\ce{H2SO4}$.

Same process isn't particularly suitable for $\ce{HI}$ generation as it reacts with $\ce{H2SO4}$:

$$\ce{2HI + H2SO4 -> H2SO3 + H2O + I2}$$

I actually seriously doubt second reaction will take place at all. Hydroiodic acid can also act as a reducing agent, converting the resulting alkyl iodides to alkanes. Instead of $\ce{HI}$ is often being used $\ce{I2}$ with red phosphorous, or halide salts (Finkelstein reaction).

As @orthocresol noticed in the comment, the following two reactions are more preferable for alkyl iodides synthesis:

$$\ce{R - OH ->[(C6H5O)3P+ CH3I-] R - I}$$

$$\ce{R - OH ->[ (CH3 - S - CH = N+ (CH3)2) I- , THF] R - I}$$

Also, chapter 2.01.5.4 Alkyl Iodides from Alcohols, p. 16 in Comprehensive Organic Functional Group Transformations II (1) accumulates numerous ways of iodoalkanes synthesis:

  The adduct of triphenylphosphine with elemental iodine ($\ce{Ph3PI2}$)
  is a classic reagent used to convert alcohols cleanly and with
  inversion to iodides. Addition of imidazole is well known to promote
  this conversion with high yields. Furthermore, treatment of alcohols
  with triphenylphosphine and cyanogen iodide affords the corresponding
  iodides in good yields. The reagent $\ce{KI/BF3 * Et2O}$ in dioxane is
  highly selective and effective for the transformation of allylic and
  benzylic alcohols to iodides.

  A mild and effective procedure for directly converting secondary,
  tertiary, and benzylic alcohols into the corresponding iodides
  involves treatment with iodine in refluxing petroleum ether. The
  reaction proceeds with inversion of configuration:

  Alkyl iodides can also be prepared in a single step from the
  corresponding alcohols upon treatment with the standard reagent
  $\ce{P-I2}$. A less well used reagent,
  1,2-bis-(diphenylphoshino)ethane, can be used to prepare iodides from
  primary alcohols in the presence of iodide.

  Cerium(III) chloride, a Lewis acid imparting high regio- and
  chemoselectivity in various chemical transformations, can be used in
  combination with sodium iodide in refluxing acetonitrile to replace a
  hydroxy by an iodo group:

  This method cannot be applied to tertiary alcohols. In this case an
  alkene is derived by dehydration of the alcohol.

  Finally, a classical method for the transformation of alcohols to
  halides is the well-known two- step procedure via a sulfonate ester,
  commonly tosylate (p-toluenesulfonate) or mesylate (methanesulfonate).

  Treatment of the sulfonate esters with $\ce{NaI}$ or $\ce{LiI}$ gives
  iodides in good yields.

(1) Katritzky, A. R.; Ramsden, C.; Taylor, R. J. K. Carbon with One Heteroatom Attached by a Single Bond; Comprehensive Organic Functional Group Transformations II; Elsevier Science Ltd, 2004; Vol. 2. ISBN: 978-0-08-044253-2


Understanding of acids and bases My teacher tried to teach this concept to us in 5 minutes. I'm lost but I've made an effort to understand it all (but I need clarification.)

If we have a compound such as LiOH, what element would be needed to neutralize it? As in, I understand that this element has to be a base but, there are SO MANY bases. Can you maybe teach how to do these using the LiOH example and another example using HCl (aq). Any examples, really. I just need to understand that.

Secondly, we haven't learned solubility tables yet, I can identify an acid (H ...) and bases (... OH) but what if there's a compound like MgCl2, what do I do now? Do I use a table to look up the states or the pH? Is there another approach?

I learned that Alkali Metal + O2 = Oxides but I can't really apply that here, can I?

Literally any help will be appreciated, I'm not cramming rather just don't understand the concept.

As we are discussing acid-base neutralisation reactions they will be in solution (mixed with water).

First consider LiOH when mixed with water:
$$\ce{LiOH -> Li+ + OH-}$$

As we can see $\ce{OH-}$ ions are formed, meaning this solution is a base.  This means we need an acid to neutralise it.  (Acids and bases react to neutralise each other).

Lets say we use $\ce{HCl}$ as our acid.  Consider it mixed with water:
$$\ce{HCl -> H+ + Cl-}$$

When we mix these two solutions together we get:
$$\ce{Li+ + Cl- + H+ + OH- -> LiCl + H2O}$$

Notice how the $\ce{OH-}$ and $\ce{H+}$ react to form $\ce{H2O}$.  The $\ce{LiOH}$ is now neutralised.

This can be written in one equation like this:
$$\ce{LiOH + HCl -> LiCl + H2O}$$

Other acids could also be used to neutralise the $\ce{LiOH}$ such as $\ce{H2SO4}$:
$$\ce{2LiOH + H2SO4 -> Li2SO4 + 2H2O}$$

I am uncertain of your second question.  There is no way of neutralising $\ce{MgCl2}$ as it is a neutral salt (no $\ce{OH-}$ or $\ce{H+}$).  Solubility tables (at this level) also do not affect the pH of your solution.  If you haven't learned about it yet then don't attempt to now, you will likely just confuse yourself for your test tomorrow.

I hope this has been at an appropriate level for you.


What is distorted square pyramidal structure of a compound? I have recently being studying structures of different types of compounds, when I came across the structure of $\ce{[XeF5]+}$ which have a structure distorted square pyramidal. I could not understand what "distorted" square pyramidal is and how it is different from square pyramidal.

Going by a standard argumentation, the $\ce{[XeF5]+}$ cation would be expected to have two different bond types: a classical 2-electron-2-centre bond to the axial fluorine and a pair of two 4-electron-3-centre bonds making up the base of the pyramid with the equatorial fluorines. This structure is shown in figure 1.

Figure 1: Hypothetical perfect structure of $\ce{[XeF5]+}$. Image taken from Wikipedia, where a full list of authors is available.

The actual structure of $\ce{[XeF5]+}$, reported by Leary et al.,[1] is not perfect, though. The axial fluorine atom does not sit at the centre of the pyramid, destroying the perfect $C_\mathrm{4v}$ point group. This is probably an effect of core orbitals on the central xenon atom (I cannot access the paper where I am right now). Because the pyramid no longer has four equal faces, it is distorted.

Reference:
K. Leary, D. H. Templeton, A. Zalkin, N. Bartlett, Inorg. Chem. 1973, 12, 1726. DOI: 10.1021/ic50126a004.


are there any stable four membered chelate rings? With reference to coordination compounds, most chelate rings formed with the metal atom and ligands are 5-member rings (including the central metal atom), such as complexes with Oxalate, ethylenediamine, dimethylglyoxime and glycinate ions as complexing agents.

Are there any compounds which contain stable 4-member chelate rings? Or any 7-member rings as well for that matter? I understand that angle strain makes these complexes hard to form but are there any exceptions to this trend?

Four-member rings can be formed with simple oxy-anion ligands such as nitrate.  An example with iron as the central ion is given here.

Reference

G. R. A Wyllie, O. Q. Munro, C. E. Schulz, W. R. Scheidt,
"Structural and Physical Characterization of (Nitrato)iron(III)
Porphyrinates [Fe(por)(NO3)] — Variable Coordination of Nitrate",
Polyhedron, Oct 10 2007, 26(16):  4664-4672.


drawing epoxides, the 3d shape given the image on the left, when you look at it from the side would it look like the image in the right? since the thicker lines means it's coming out of the page?

From PubChem

So the molecule is sort of boat shaped.


Terminal Alkyne in Suzuki Coupling I've done a very similar reaction to the one below but by TLC I observe only starting materials. It's super clear that nothing else is going on. Conditions: K3PO4, Pd(dppf)Cl2, dioxane/water, heat to 100 deg.C.

What thoughts do you have on why this doesn't work?

I imagine the alkyne chelating to the Pd and potentially 'poisoning' it? I currently lack the words and technical lingo to put this into a formal explanation.
A competing Sonogashira perhaps? Unlikely given the aqueous conditions.
Any other issues?

If TLC is the only analysis we have, it may be difficult to determine exactly what is happening (or not happening). Other things that might be happening include homocoupling (biaryl formation), proteo-deborylation, and debromination. These may be hard to discern by TLC alone.

If there is no reaction of starting materials this might be due to:

Failure to reduce Pd(II) precatalyst to active Pd(0) species: your catalytic cycle never initiates. Consult chapter 1 of Negishi's Handbook of Organopalladium Chemistry for Organic Synthesis. Consider adding a catalytic amount of exogenous reducing agent, changing the solvent to EtOH/PhMe/H2O, increasing the temperature, switching to a Pd(0) catalyst system, or a system that is more easily reduced to Pd(0) like Pd(OAc)2 + xs phosphine ligand.
Stalling in the catalytic cycle: there is insufficient energy to get over the activation barrier in the oxidative addition, transmetallation, or reductive elimination step. Try increasing the temperature.
Catalyst destruction by oxidation: did you degas your system with N2 or Ar?

Alkyne poisoning of the homogeneous Pd catalyst seems unlikely since there are plenty of examples with Sonogashira coupling of terminal alkynes involving dppfPdCl2 that work just fine.


Why do we titrate from a colourless solution and why should we remove unused sodium hydroxide from the burette as soon as possible? In my lab we performed an acid base titration, where NaOH was used to titrate a standard solution, potassium hydrogen phthalate (KHP, or $\ce{K+C6H4(CO2H)(CO2^-)}$), using phenolphthalein indicator.

Why is it preferred to titrate from colourless to a pink solution rather than pink to colourless?

Also, why is it advisable to remove NaOH from the burette as soon as possible after we are done titrating?

Why is it preferred to titrate from colourless to a pink solution rather than pink to colourless?

It is much easier to detect the appearance of a color rather than the disappearance of one.

I really liked the large white ceramic ringstands for phenolphthalein titrations. The bright white background really made it easy to monitor the titration. At first you can just let the base (sodium hydroxide) pour into the solution as your are swirling the solution. You just get a pink cylinder where the base is entering the solution. As you get nearer to the end point then the color will start to go out more into the solution. Finally you do drops at a time, and then a partial drop.

  Also, why is it advisable to remove NaOH from the burette as soon as possible after we are done titrating?

Sodium hydroxide is a strong base and it will etch (dissolve) glass given enough time. If you're using a burette with a glass stopcock the sodium hydroxide can also weld the stopcock joint.

I'll add that you never should pour the unused sodium hydroxide back into the stock solution. Any that you don't use is waste. You want to avoid any possibility of contamination.


Question on ring expansion In the orange box, I think that the blue-carbon should be a carbocation instead. So, option A should be correct.
But, the answer is B. Why is it so?

As orthocresol commented (vide supra), migratory aptitude in cationic reactions has been given as: $\ce{H}=\mathrm{Ph} \gt \ce{(CH3)3C \gt (CH3)2CH \gt CH3CH2 \gt CH3}$ (Wikipedia and Migratory aptitude in pinacol-pinacolone rearrangement). In your question, the secondary carbocation formed when protonated has three groups, which can migrate to give more stable tertiary carbocation. They, with their migratory aptitude, are:  $\ce{-R^1CH2(CH3)CH\!- \gt R^2CH2CH2\!- \gt H3C\!-}$ (where $\ce{R^1}$ and $\ce{R^2}$ are continuation of rest of cyclopentane ring).

Now, if  $\ce{H3C\!-}$ group migrated, it'd give option (D); if  $\ce{R^2CH2CH2\!-}$ group migrated, it'd give option (A); and if  $\ce{R^1CH2(CH3)CH\!-}$ group migrated, it'd give option (B). If nothing migrated you have option (C). Although all 4 products are possible in the reaction mixture ((C) would be very trace amount), asking is the major product, and hence is is the product given but migration of group with highest migratory aptitude. Therefore, it should be the product given in option (B).


Is negative activation energy possible? Is negative activation possible? And, in coupled reactions is there any difference?

Because I saw in the paper - Chemical Engineering Science 1996, 51 (11), 2995–2999 - the following conclusion:

  Negative activation energies of reaction have been observed [...]. This unusual phenomenon can be predicted from classical theories, as a competition between two consequences of a temperature increase: greater thermal energies available to overcome the energy barrier represented by the intrinsic activation energy, and less adsorption to form reactive intermediates represented by the heat of absorption.

Does it simply mean that although the energy to overcome the activation energy barrier is there but the rate that products are formed is lower?

No, it's not possible to have a negative activation energy in a simple reaction such as an isomerisation because there is no possible way to draw to potential energy curves to give a negative activation energy. The best you can do is to be activationless, and is commonly observed in electron transfer reactions (see for example J. Am. Chem. Soc. 2002, 124 (19), 5518–5527).

Experimentally, however, negative activation energies can be observed when the reaction comprises several steps some of which are not identified at the time of doing the experiment. For example if each reaction follows an Arrhenius type rate constant, the first may be $k_a = A_a\exp(-E_a/RT)$ and the second $k_b=A_b\exp(-E_b/RT)$, but because there are, say, equilibria involved in the mechanism, what may be observed without realising it is that the rate is a ratio of the two individual rates:

$$\begin{align}
k_\mathrm{obs} &\propto \frac{k_a}{k_b} \\
&\propto \frac{\exp(-E_a/RT)}{\exp(-E_b/RT)} \\
&= \exp\left(\frac{-(E_a-E_b)}{RT}\right) \\
&= \exp\left(-\frac{E_\mathrm{obs}}{RT}\right)
\end{align}$$

so it is possible for $E_\mathrm{obs} = E_a - E_b$ to be positive or negative.