IUPAC Recommendations for Concentration Units (ppb, ppm) In a recent interesting post, (Is 1 ppb equal to 1 μg/kg?)
it was pointed out that IUPAC advises to abandon parts per million, parts per billion quantities and instead it suggests to employ micromole/ mole for ppm and nanomole/mole for ppb. An example is given on page 98 of the Green Book
"The volume fraction of helium is 20 ppm".

May be I am missing a point or I see a problem here with the moles in the denominator of these suggested replacement definitions.

a) If we analyze Zr in a rock and find 5 mg Zr in a 20 gram rock, how would one apply the mole definition because one cannot define the moles of a rock. We have to stick to the classical definition:

ppm =10^6 * (mass of analyte / mass of sample).

b) How can one express ppm of Ar in air, and how should we define the moles of air? If we wish to generalize this case, what if our solvent system consists of several components.

The ambiguity I feel in micromole/mole definition is the mole in the denominator. Are we referring to the moles of the solvent or  moles of solvent plus solute.

If this replacement is analogous to molal concentration, where the denominator is clearly definied to be kg of solvent not the kg of solvent plus kg of solute, then IUPAC should clearly define it.

There are not this kind of problems with the IUPAC recommendation. The only point in it is to specify the parts of  what we refer to, nothing more.
In such a sense it seems clearer than expression as molality or molarity.

Take an amount of atmosphere at given conditions. The concentration of Ar or whatever component can surely be expressed in terms of mole number in a million or whatever number of moles that amount contains.

As for the rock analysis case in the question, note that IUPAC umol/mol is an example, it doesn't mean it should be preferred to mg/kg (or ul/l etc.). The measured amount remains the most convenient both practically and mathematically.

I add that following recommendations would - perhaps - also help clear another practical con of unspecified ppm and percentage concentration values.  Often people forget that 1 l --> 1 kg of solution is acceptable only with very dilute solutions and when the solvent is water and go around telling and even publishing % concentration in various solvent so that the reader must assume a wrong procedure in order to follow up. Explicitly saying mg/kg instead of ppm should evidence that when mg starts to be hundred, than one has to switch concentration units or being more careful when reporting.


Is there a connection between UV-degradation and the photoelectric effect Recently I've been learning about the photoelectric effect in my physics classes.

I only have a rudimentary understanding of chemistry, but I've been wondering what the connection is between the photoelectric effect and UV degredation.

For the sake of simplicity let's focus on the degredation of polypropylene.

My hypothesis is, that the UV light frees some electrons from the material. These will either weaken or break the bond between two molecules or atoms in the molecule itself.
I would ultimately except the molecule chains to be broken into smaller pieces due to this.

Is that a reasonable assumption or does the process work completely different?

Yes and no.

In a metal, the photoelectric effect is when a UV photon has enough energy to push an electron from it's current energy level to at least the $n=\infty$ level. Extra energy goes into electron kinetic energy.

For UV degradation, we are promoting elections to higher energy orbitals in an analogous way. This produces an energized state that is highly reactive and may undergo radical reactions that ultimately lead to the formation of different species.

The difference is that we assume the electron is essentially removed from the atom in the photoelectric effect. Elections are weakly enough bound in a metal for a UV photon to achieve this. But in general, for UV degradation, the UV photon can't actually remove the electron from the molecule. But the photon can energize electrons that lead to degradation reactions.


Relation between equilibrium constant and entropy change Like the Van't Hoff equation, which relates change in enthalpy to equilibrium constant, is there a similar equation for the relation between change in entropy and equilibrium constant?

Consider the following specific case,

  For a reaction taking place in a container in equilibrium with its surroundings, the effect of
  temperature on its equilibrium constant K in terms of change in entropy is described by

  [A] With increase in temperature, the value of K for exothermic reaction decreases
  because the entropy change of the system is positive

  [B] With increase in temperature, the value of K for endothermic reaction increases
  because unfavourable change in entropy of the surroundings decreases

  [C] With increase in temperature, the value of K for endothermic reaction increases
  because the entropy change of the system is negative

  [D] With increase in temperature, the value of K for exothermic reaction decreases
  because favourable change in entropy of the surroundings decreases

  Answer: (B) and (D)

My attempt:

Since equilibrium with surrounding is given, the reaction must be reversible.

This implies, $∆S=0, ∆S_s=∆S_{surr}$.

Also, at equilibrium $∆G=0$.

This implies, ∆$H=T∆S_s=-T∆S_{surr}$

(where the s subscript denotes system, and surr denotes surrounding)

Writing the van't Hoff equation in differential form, I got

$= d(lnK)=∆H(dT/RT^2)$

$=d(lnK)=-∆S_{surr}(dT/RT)$

Now, aren't options A and C true too?

As on integrating, we can see directly the variation of $K$ with $∆S$.

What is the correct way to solve this?

Edit: Where ever there is a big space, it implies ∆.
Example,     '       ' H=T '       '  ,

Means, ∆H=T∆S

Like the Van't Hoff equation, which relates change in enthalpy to equilibrium constant, is there a similar equation for the relation between change in entropy and equilibrium constant?

A relation between $K$ and $\Delta S^\circ$ can be obtained as follows:

$$\begin{align}T\log K &= -\frac{\Delta G^\circ}{R} \\
\left(\frac{\partial(T\log K)}{\partial T} \right)_p &= -\frac{1}{R}\left(\frac{\partial \Delta G^\circ}{\partial T}\right)_p= \frac{\Delta S^\circ}{R}\end{align}$$

As regards the multiple choice question, the best way to answer it is to understand that $\Delta G$ represents a sum of the change in the entropy of the system and of the surroundings, scaled by $T$:

$$\begin{align}\Delta G &= \Delta H - T\Delta S \\&= -T\Delta S_{surr} - T\Delta S_{sys} \\&= -T(\Delta S_{surr} + \Delta S_{sys}) \end{align}$$

Exchange of heat at constant pressure (equal to the enthalpy) alters the entropy of the surroundings, so when discussion turns to the heat exchanged at constant pressure, you don't need to consider the entropy of the system, only that of the surroundings (at least for this type of problem - this alone eliminates a and c as possible choices). Next you need to remember that endothermic means the surroundings lose heat to the system, which lowers the entropy of the surroundings. The opposite holds for an exothermic process (surroundings gain entropy).

To summarize: an endothermic reaction lowers the entropy of the surroundings. The decrease in the entropy of the surroundings decreases if you increase T. The result is therefore a net increase in the total entropy. Finally, since $$\Delta S ^\circ_{\text{total}} = -\frac{\Delta G^\circ}{T}=R \log K_{\text{eq}}$$
increasing the total entropy results in a higher equilibrium constant.


Need help to calculate results of an electrochemistry cell potential lab work So, the lab work is this. I made an AgCl electrode of a silver wire by coating it with AgCl. This was done by electrolysis by immersing the wire in HCl and using platinum as a counter electrode and applying current to the system. After that, the AgCl electrode was immersed into HCl dilutions of different molalities and the cell potential was measured against a hydrogen gas electrode (normal hydrogen electrode NHE).
The measurements were done at temperature of 25 °C. The results of the measurements are as follows:
\begin{array}{llc} \hline
c\,(\pu{mol/kg}) &\sqrt{c}\, (\sqrt{\pu{mol/kg}}) & \mathrm{E\,(V)} \\ \hline
0.001 & 0.032 & 0.6322 \\
0.002 & 0.045 & 0.6815 \\
0.004 & 0.063 & 0.6497 \\
0.006 & 0.077 & 0.6280 \\
0.008 & 0.089 & 0.6153 \\
0.010 & 0.100 & 0.6049 \\
0.020 & 0.141 & 0.5700 \\
0.050 & 0.224 & 0.5030 \\ \hline
\end{array}
Because the cell potential can be calculated by (as the pressure gradient of the hydrogen electrode was 0):
$$ E(m)= E^0 + \frac{2ART}{F} \sqrt{\frac{m_{\text{HCl}}}{m^0}} $$
By plotting the table values we get (excluding the first data point as an outlier):

We can see from the equation that the $E^0$ would be about 0,708 V. However, literature references say the standard potential of the AgCl electrode is about 0,222 V. Any idea what I am doing wrong here?
Then I also should plot the HCl mean activity coefficient values calculated from the formula:
$$
E(m) = E^0 - \frac{2RT}{F} \ln \left[ \gamma_{\text{HCl}} \left( \frac{m_{\text{HCl}}}{1\ \text{mol/kg}} \right) \right] \\
\Leftrightarrow \ \gamma_{\text{HCl}} = e^{ \frac{(E^0-E)F}{2RT} } / \frac{m_{\text{HCl}}}{1\ \text{mol/kg}}
$$
Thus I get values:
\begin{array}{cc} \hline
m & \gamma_\ce{HCl} \\
\pu{mol/kg} \\ \hline
0.001 & 4371.98 \\
0.002 & 837.44 \\
0.004 & 777.51 \\
0.006 & 790.73 \\
0.008 & 759.33 \\
0.01 & 743.74 \\
0.02 & 733.46 \\
0.05 & 1080.77 \\ \hline
\end{array}
But literature references (CRC Handbook of Chemistry and Physics - 97th Edition, p. 5-101) gives greatly different values:
\begin{array}{cc} \hline
m & \gamma_\ce{HCl} \\
\pu{mol/kg} \\ \hline
0.001 & 0.965 \\
0.002 & 0.952 \\
0.004 & 0.929 \\
0.01 & 0.905 \\
0.02 & 0.876 \\
0.05 & 0.832 \\ \hline
\end{array}
Any help would be greatly appreciated.

The silver-silver chloride is a half cell. When the silver wire coated with AgCl is dipped into HCl solutions with no Ag$^+$ cations, as confirmed by the OP, it is not a half cell with AgCl in equilibrium with silver and silver chloride, but a half cell that is a silver anode corroding in acid solution (with oxygen from air?).
After some time (hours?), with stirring of the solution, enough AgCl may dissolve, still leaving enough on the silver wire, dropping the 0.5-0.7 V to wind up with the desired 0.222 V. Perhaps the HCl solutions need to be saturated with AgCl and deoxygenated before the immersion of the wire electrode.


Why does ligand exchange take place? Why does ligand exchange happen?
Are there specific factors that determine when a ligand can replace another ligand?
When ligand exchange takes place, does that mean that the "previous" ligand is not bonded to the complex ion at all, or both ligands are bonded?

Ligand exchange is in equilibrium, so it is driven by thermodynamics
There are many reasons one ligand would replace another, but it often has to do with Lewis basicity, HSAB (see trans effect), or the thermodynamics of a subsequent reaction.
Ligand Exchange is a broad term for a set of mechanisms that include various types of ligand association and dissociation, and these all occur for different kinetic reasons. I'll try to provide a brief overview, but the answers to all of these questions are varied depending on specific situations.

In the covalent model, both $\ce{1e-}$ "X" ligands and $\ce{2e-}$ "L" ligands are called "ligands", so it's important to point out that we're only talking about L ligands here.

Generally speaking, there are three types of ligand exchange mechanisms: association, interchange, and dissociation, and each mechanism is further broken into the associative and dissociative subclasses, depending on the exact character of the transition state. If the higher energy TS‡ of a ligand association has the geometric character of the dissociation, it is called a "dissociative association" (the Ad mechanism), for instance, but this is somewhat rare. The equilibrium expressions of the Dissociation, Interchange, and Association mechanisms are as follows, respectively:(1)

$$\begin{align}\ce{ML5X &<=>[$k_1$][$k_{-1}$] ML5 + X}\\
\ce{ML5 + Y &->[$k_2$] ML5Y}\\
\ce{ML5X + Y &->[$k_1$]ML5Y + X}\\
\ce{ML5X + Y &<=>[$k_1$][$k_{-1}$] ML5XY}\\
\ce{ML5XY &->[$k_2$] ML5Y + X}\end{align}$$

The exact nature of a ligand exchange mechanism is something that requires computations, and physical chemistry to determine, and you should look at (1) to read more about this. Really, the mechanisms are part of a continuum:

$$\mathrm{D_d \cdots I_d \cdots I \cdots I_a \cdots A_a}$$

Once the ligand is fully displaced, it is no longer part of the inner sphere of coordination, so the "'previous ligand" is no longer bound to the complex, but there exist many ligand exchange mechanisms that involve transient intermediate in which both the leaving and the incoming ligand are bound to the central atom. All associative mechanisms will have such an intermediate, while dissociative and interchange mechanisms will not. Another important example is the formation of a preassociation complex, in which "rapid equilibrium occurs between the incoming ligand and the 6-coordinate reactant to form an ion pair" (1) (or a neutral preassociation complex with strong dipole-dipole interactions, depending on oxidation states). In the case of a preassociation complex, the outgoing ligand is no longer in the inner sphere of coordination, but it is still "bound" to the complex, albeit through weaker noncovalent interactions. Such a reaction has equilibrium like this:

$$\begin{align}\ce{ML5X + Y &<=>[$k_1$][$k_{-1}$] ML5X . Y}\\
\ce{ML5X . Y &->[$k_2$] ML5Y + X}\end{align}$$

Ligand exchange kinetic and thermodynamic arguments also frequently invoke the chelate effect and the trans effect.

Another important concept is that the driving force for individual ligand dissociations and associations involved in catalytic cycles often implicates the driving force of the overall cycle. In catalytic reactions, the rate of ligand dissociation or association is not dependent not on the thermodynamics/kinetics of the individual reaction, but instead on the step in the cycle with the highest activation energy. This step is known as the  turnover limiting step (2). In a catalytic cycle, all equilibria have the same rate. "On a circular track, on average the same number of trains must pass each point per unit time." (2) For example; in this catalytic alkene hydrogenation, the turnover limiting step is the dissociation of the phosphine from Wilkinson's catalyst (3), so this is the step that determines the turnover frequency (TOF). The rate of association of the ethylene ligand later on is driven by the cycle's overall TOF, so such a ligand association would be indirectly driven by a different step in the cycle. If the final step of the cycle is irreversible, the substrate release might also be a thermodynamic driving force.

$\hskip1.4in$(3)

These are just some of the many possible answers to your questions.

(1) G. L. Miessler, P. J. Fischer, D. A. Tarr. Inorganic Chemistry (5th Ed.), Pearson, 2014, p. 442-444

(2) R. H. Crabtree, The Organometallic Chemistry of the Transition Metals (3rd Ed.), John Wiley & Sons, 2001 p. 225

(3) J. Luo, A. G. Oliver, J. S. McIndoe, Dalton Trans. 2013, 42, 11312-11318


Ionic Character, on which factors do it depend I was watching a lecture of chemistry teacher. He gave the following question:

Q Arrange in terms of increasing ionic character: LiF,NaF,KF,RbF,CsF?

My try: I tried two approaches:
Approach 1: As F is common to all compounds, therefore its metal that affects the ionic character. As we move down the group as in this case, the shielding effect increases and hence it becomes more easier to abstract an electron from the outermost shell or in other words, Ionization energy decreases as we move down the group, therefore CsF>RbF>KF>NaF>LiF should be the order.

Approach 2: Li, Na, K, Rb, Cs have same electrovalency. As Cs has more larger size, the positive charge intensity on the atom decreases and thus by formula of electrostatic force F=-kq1q2/r^2, the electrostatic force of attraction decreases and thus lattice energy decreases and therefore order is:
LiF>NaF>KF>RbF>CsF.

According to me lattice energy should determine ionic character as lattice energy determines ionic bond strength I think and therefore approach 2 should be more legitimate but when I checked the answer, the answer by Approach 1 was correct. Why is it so? Which of the approach is right and which is wrong and why? Is there any other approach?

When you are considering lattice energy as the parameter for judging the ionic character of compounds, you are assuming two things.

The constituting elements of this alkali metal halides are purely ionic,and thus the Lattice energy (U) is the released energy when those two isolated gaseous ions forms the crystal lattice. But, in practice the constituting ions are never purely ionic (especially smaller alkali metal halide such as $\ce{LiF} $ or $\ce{NaF}$.)
The ions constituting the crystal lattice are hard non-elastic spheres. They hardly (in general) obey this postulate. This postulate is again not totally valid for smaller alkali metal halides due to distortion of their electron cloud due to large polarization.

So, it is seen that if you just theoretically say that the electrostatic force as simply Coulomb's law and thus lattice energy decreases, it doesn't conclude anything as it is seen that the Lattice energy is not a proper parameter for judging ionic character.

The correct approach in judging the covalent character present in an ionic compound is to follow Fajans' rules. According to this rule, $\phi _{cation}$ is inversely proportional to $r_{cation}$ and directly proportional to $Z_{cation}$ Here, $Z_+$ for all the cation is same. But radius of $\ce{Li^+}$ is smallest and ionic radius gradually increases from $\ce{Li^+}%$ to $\ce{Cs^+}$. So, $$\ce{$\phi$_{Li^+} > $\phi$_{Na^+}  > $\phi$_{K^+}  > $\phi$_{Rb^+}  > $\phi$_{Cs^+} }$$ Here  $\phi$ is the measure of polarising power of cation. As, the value of $\phi$ increases, the polarising power increase and thus the anion's electron cloud gets attracted towards cation nucleus which tends to form a kind of overlap between the orbitals of cation and anion, which increase covalency and decreases ionic character.

So, according to Fajans' rules, increasing ionic character (i.e. more covalency to less covalency) is, $$\ce{LiF < NaF < KF < RbF < CsF}$$


Is NaCl paramagnetic? My teacher told yes that salt is paramagnetic and it is attracted by magnet slightly. But when I tried that at home it did not work.

Table salt (NaCl) is diamagnetic. For paramagnetism, one needs free unpaired spins.
The NaCl lattice is built up from closed shell cations and anion, there is no such unpaired spin pure NaCl.

Also: experimentally it is difficult to measure paramagnetism, and you generally cannot just try it with a magnet and observe it without some tricks.


Conditions for imine formation Firstly, note that this is not a duplicate of Acid in Imine Formation. Clayden, Warren, & Greeves (2012) states explicitly on p. 231 that imine formation requires acid catalysis, and that the reaction takes place optimally under slightly acidic conditions, close to neutrality. This makes sense as we need to form the $\ce {-OH2^+}$ allowing water to leave. However, if we look at this list of literature, we would observe that most of the imine formation reactions do not necessarily require acidic conditions and some even take place under solely basic conditions. How do we explain this?

Reference

Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

Formation of imines is usually a nucleophilic subsitution of oxigen with nitrogen. For this to happen, the oxigen usually must be bound beforehand as carbon-oxigen bond is quite strong. The most straightforward way is to use acidic catalysis.

Most reactions that you referenced use acidic catalysis, though sometimes it isn't immediately apparent because they use solid catalyst such as molecular sieves a.k.a. H-form zeolites. A few use Lewis acids, such as $\ce{Ti(OEt)4}$. They still count as acids.

The cases when no acid is used are not pure nucleophilic substitution. The reference you provided features various oxidative and other catalytic reactions, proceeding through different mechanism, that do not need acid, and a few reactions where reagent itself bears acidic function, like tosyl isocyanate, that is a strong electrophile and can bind with oxygen on its own.

TL;DR: the cases when acid apparently isn't used usually include it in non-obvious way. Other cases proceed through wild varying different mechanisms, usually involving oxidation.


Why does hydride acidity increase across period and down group in periodic table? Why exactly does hydride acidity increase across period and down group in periodic table? What is the explanation with respect to electrons? I can't figure this out because for 1st period etc H is an anion while for the right and middle of the periodic table H is a cation in compounds formed. I feel confused. Does this have something to do with electronegativity of element that forms hydride?

Yes, but that is somewhat circular reasoning, as electronegativity is largely just putting a numerical value to the trends in properties you are observing.  A better answer is to compare the effective nuclear charge holding on to the valence electrons as you go across or down the Table.  Going across the table, each electron gets added to the same shell, which means it shields the other electrons from the added proton poorly.  Consequently, Z* (the positive charge effectively felt by the outermost electrons) increases, and these electrons are more tightly held.  At some point the atom holds onto not only its own valence electrons, but those of the H atom when the H atom leaves, and so the H atom leaves as H+.  Conversely, at the far left of the table, the Group 1A atoms hold on so loosely to their lone electron that the H atom is able to leave with the metal's electrons, as H-.

Going down the table, you add a whole electron shell between the nucleus and outermost electrons with each period.  That is counterbalanced a bit by the fact that the shells become more closely spaced, but nevertheless for main group elements the extra shielding of the extra shells usually matters more, and so effective nuclear charge falls going down a group.

Finally, bear in mind the "acidity" of a hydride is the tendency of the compound to break up with the H atom leaving as H+.  So the greater the Z* of the partner, the more likely the H atom leaves without its electron.


What is the best hydrogen bond acceptor in 2-amino-N-(3-hydroxyphenyl)acetamide? I was preparing for my first exam in organic chemistry and came across this question.

  Draw a water molecule that is hydrogen bonded to the best anticipated hydrogen bond acceptor.

  2-amino-N-(3-hydroxyphenyl)acetamide, PubChem CID: 20238693

I was thinking that it would be one of the oxygens, because of its greater electronegativity compared to nitrogen. However, I'm having a little trouble distinguishing which oxygen.

I'm leaning towards the alcohol group. My reasoning is that there is less steric hindrance preventing solvation, but I am not sure.

As a general rule of thumb, a nitrogen is usually a better hydrogen bond acceptor than an oxygen. There is a brief overview on Wikipedia with a couple of values and where they were derived from.

This makes sense intuitively, too. Ammonia is a stronger base than water as it is a better proton acceptor. An amine is a stronger base than an alcohol.

In your example, 2-amino-N-(3-hydroxyphenyl)acetamide, there are three functional groups (or four if you want to count the aromatic ring, too): a primary amine, a secondary amide, and a phenolic alcohol.

The phenolic alcohol is rather acidic and not such a good hydrogen bond acceptor, as some of the lone pairs are in conjugation with the aromatic ring.
The amide is also a delocalised system, which makes the nitrogen pretty unavailable for hydrogen bonding. For this functional group, if it were separate, the π-orbital would be the HOMO, and this is where the protonation/ proton-accepting will happen; due to the delocalisation this is less so the case. An amide usually is not a good hydrogen-bond acceptor.
The amine is primary, which means there is much space around it and it can directly interact with the solvent. This is a good hydrogen-bond acceptor.

Therefore neither oxygen will be the best hydrogen-bond acceptor.


Does Langmuir model predict evaporation rates? Slide #16 of this presentation on amalgams says that "Langmuir's equation" provides a specific "theoretical maximum evaporation rate" for liquid Hg as a function of temperature and surface area.

The only common equation bearing Langmuir's name is his adsorption model.  I can vaguely imagine that model being applied in reverse to characterize evaporation rates, but I can't see or find how in general, much less in the specific case asserted in the paper.  Can someone provide a reference or example of how that (or any other model) provides a "theoretical maximum evaporation rate" of a particular liquid?

That most likely refers to the Hertz-Knudsen equation, also known as Knudsen-Langmuir equation. It is the most simplified model of evaporation in which no discussion is given into the diffusivity and the chemical potential gradient (let alone other driving forces like gravity and pressure gradient): one just assumes the leaving mass flux is dictated by the system's thermal energy. This model then suggests a direct relation between evaporation rate and vapor pressure, temperature and molar mass.


Effect of pressure on adsorption of a gas I am currently studying about the basic terms of surface chemistry, and am facing some conceptual doubt regarding the following:

What effect does pressure have on the adsorption of a gas (adsorbate) on a adsorbent? I don't understand the fact that the adsorption of a gas increases very rapidly as the pressure is increased from small values at low temperature.

I have the idea that increasing the pressure brings the molecules of the gas close together so that they have more intermolecular forces of attraction, but how do I correlate this with adsorption?

Short Answer

The following equilibrium roughly represents what goes on during physisorption of a gases onto a a surface.
$$\ce{Gas + Surface <=>[adsorption][desorption] Gas--Surface}  $$

The effect of pressure can be understood on the basis of Le Chatelier's principle, which states that if a stress (such as a change in pressure) is applied to a system in equilibrium, the equilibrium will shift so as to tend to counteract the effect of the constraint.

In this case, we are increasing the pressure, and the system would respond by wanting to reduce the number of gas phase molecules, and in this case it can do so by favouring the forward reaction wherein the gas phase molecules are immobilised on the surface.

Adsorption is usually described through isotherms, that is, the amount of adsorbate on the adsorbent as a function of its pressure (if gas) or concentration (if liquid) at constant temperature. Several isotherm models exist to describe this phenomenon.

Bonus Round: Additional Commentary

One of the earliest mathematical fits for experimental data was the Freundlich adsorption isotherm. Experimentally it was determined that extent of adsorption varies directly with pressure and then it directly varies with pressure raised to the power 1/n until saturation pressure $P_s$ is reached. Beyond that point rate of adsorption saturates even after applying higher pressure. Thus Freundlich adsorption isotherm failed at higher pressure.

$$\frac{x}{m} = Kp^{\frac{1}{n}}$$

where $x$ is the quantity of adsorbate adsorbed in moles, $m$ is the mass of the adsorbent (this is basically a normalisation to allow for comparison between different materials), $P$ is the pressure of adsorbate and $K$ and $n$ are empirical constants at the given temperature.

The function is not adequate at very high pressure because in reality, as I have already mentioned, because $\frac{x}{m}$ has an asymptotic maximum as pressure increases without bound. As the temperature increases, the constants $k$ and $n$ change to reflect the empirical observation that the quantity adsorbed rises more slowly and higher pressures are required to saturate the surface.

Irving Langmuir was the first to derive a semi-empirical model with a kinetic basis and was derived based on statistical thermodynamics.

It is quite simple, (and elegant in my opinion), and finds extensive use. it is based on the following four assumptions

The surface containing the adsorbing sites is a perfectly flat plane with no corrugations (assume the surface is homogeneous) .
The adsorbing gas adsorbs into an immobile state.
All sites are equivalent.
Each site can hold at most one molecule of A (mono-layer coverage only).
There are no interactions between adsorbate molecules on adjacent sites.
There are no phase transitions.

Unfortunately,these four assumptions are seldom all true: there are always imperfections on the surface, adsorbed molecules are not necessarily inert, and the mechanism is clearly not the same for the very first molecules to adsorb to a surface as for the last. The fourth condition is the most troublesome, as frequently more molecules will adsorb to the monolayer; this problem is addressed by the BET isotherm for relatively flat (non-microporous) surfaces.

Anyway, the surface coverage is given as follows $$\theta = \frac{K_{eq}p}{1+ K_{eq}p}$$


Is there a light cured resin (UV or another) that I could make in my house with easy to find chemicals? Is there a light cured resin that I could make in my house with easy to find chemicals? It does not matter if the finished product have practical or commercial applications as long that is not too toxic.

I want to use it for SLA 3D printing of prototypes.

There are simple light-curing chemicals, such as the gelatin-dichromate system, that you could make at home. However, the specifications for stereolithography are rather rigid, and it is doubtful that you could synthesize a monomer that doesn't corrode the system, that has the correct viscosity, has long-term stability and is not toxic.

You might do better taking up gum dichromate photography, which is 2D printing.


How can carbon dioxide be converted into carbon and oxygen? How can $\ce{CO2}$ be converted into carbon and oxygen?

$$\ce{CO2 -> C + O2}$$

Alternatively:

$$\ce{CO2 + ? -> C + O2}$$

I'm aware that plants are capable of transforming $\ce{CO2 + H2O}$ to glucose and oxygen via photosynthesis, but I'm interested in chemical or physical means rather than biological.

In my opinion, the catalytic, solar-driven conversion of carbon dioxide to methanol, formic acid, etc. is much more interesting and promising, but since Enrico asked for the conversion of carbon dioxide to carbon itself:

The group around Yutaka Tamaura was/is active in this field. In one of their earlier publications,[1] they heated magnetite ($\ce{Fe3O4}$) at 290 °C for 4 hours in a stream of hydrogen to yield a material which turned out to be stable at room temperature under nitrogen. This material, $\ce{Fe_{3+\delta}O4}$ $(\delta=0.127)$, i.e. the metastable cation-excess magnetite is able to incorporate oxygen in the form of $\ce{O^2-}$.

Under a $\ce{CO2}$ atmosphere, the oxygen-deficient material converted to "ordinary" $\ce{Fe3O4}$ with carbon deposited on the surface.

This remarkable reaction however is not catalytic, but a short recherche showed that the authors have published a tad more in this field. Maybe somebody else finds a a report on a catalytic conversion among their publications.

Tamaura, Y.; Tahata, M. Complete reduction of carbon dioxide to carbon using cation-excess magnetite. Nature 1990, 346 (6281), 255–256. DOI: 10.1038/346255a0.


Alcoholic potassium permanganate for oxidation A question in my textbook:

  Though alkaline and acidic $\ce{KMnO4}$ are both oxidizing agents, in the manufacture of benzoic acid from toluene, we only use alcoholic $\ce{KMnO4}$ as an oxidant. Why?

I am not able to figure out the answer. But I think maybe the acidic and alkaline $\ce{KMnO4}$ are too strong and may oxidize the benzene ring. Or the benzoic acid may react with the alkaline medium. (Though I wonder why it wouldn't react with alcohol to give ester)

I also tried digging around the web but couldn't come up with anything satisfactory.

Practical considerations, mainly that toluene is non-polar, inorganic. Aqueous $\ce{KMnO4}$ wouldn't be miscible so the oxidation can only take place in the interface. By using alcoholic $\ce{KMnO4}$, we can get the $\ce{KMnO4}$ into the organic layer much more easily, thus making the reaction much faster.

To form an ester, you generally need a dehydrating environment. Plus, the oxidation of toluene with $\ce{KMnO4}$ in neutral conditions will produce base which neutralizes the benzoic acid, thus deactivating it making it unable to undergo esterification.


Why is nylon formed using carboxylic acids, while aramids are formed with acyl halides? I was reading about carbon polymers and noticed something interesting about the difference between the reaction used to form nylon and the reaction used to form aramids: in the former, the monomers are an amine and a carboxylic acid, while in the later the monomers are an aromatic amine and an aromatic acyl halide.

Formation of an aramid

Adipic acid, used to create Nylon 66

This struck me as interesting because it seems like the reaction should be roughly the same for the creation of either polymer, with the only difference being the presence of the aromatic ring in the aramid in place of a simple carbon chain. Is the halide required for polymerization in the aramid, but not the nylon, or is there another, perhaps economic, reason for its use? Why are halides used in the creation of aramids, but not in simpler nylons?

Aromatic amines are less nucleophilic than aliphatic amines because the aromatic amines have the non-bonding electron pair "distracted" in the resonance with the aromatic ring. Therefore, they need a more reactive counterpart (acyl halide instead of the acid).


What is pairing energy of electrons? What exactly is pairing energy when electrons are paired in orbitals and what is responsible for it.

Imagine a sky scraper where everyone wants to live on as low floor as possible, but landlord prohibits then to live more than 2 people per floor. These people will occupy lowest floors in pairs and the highest occupied level might have a single person or a pair. If someone moves out of the apartment on the 1st or 2nd floor, then someone from higher floors will move down to the vacant state. Electrons do the same. Floors are energy levels and the evil landlord is the Pauli exclusion principle.

Electrons are fermions. An electron's spin can be either +1/2 or -1/2. Two options. Two fermions (and, hence, electrons) cannot be the same. Electrons are the same if their spins are the same and they occupy the same orbital (Pauli Exclusion Principle). At lower temperatures (below red heat temperatures) electrons occupy the lowest energy level possible.

As a result, each new electron occupies the lowest allowed level. Because their spins can take two values 2 electrons are occupying each level. If you populate each energy level by a single electron you will get a high energy excited state that well emit light and go to lower energy two electrons per level state. If you try to squeeze more than two electrons it will not work because of Pauli principle. So, since they want to stay in lower energy levels and have to obey the exclusion principle we have electron pairs.

Bonus: here is why radicals recombine. For instance, why $\ce{2 H^. -> H2:}$

If an electron is the last one it is occupying the energy level alone.
So, electrons would like to all occupy the lowest level, but only two per level is allowed. In radicals only one electron occupies a level. When radicals A$^.$ and b$^.$ meet each of them sit on half empty orbital attracted to a single nucleus.  It is energetically favorable to share the same orbital (form a pair, which they are OK with) and orient fragments so that they "fly" (have the highest probability) between two nucleus (A and B). Staying between two opposed charged ions is better than staying next to just one charged ion. And electrons are allowed to stay in pairs (Pauli excluded 3+ electrons staying together).


Why doesn't the N-Br bond dissociate quickly in NBS? NBS is used to provide a constant, small amount of $\ce{Br2}$ which then dissociates to give $\ce{Br^.}$ free radical. Why doesn't the $\ce{N-Br}$ bond simply dissociate to give $\ce{Br}$? This dissociation should be just as fast or faster because the free radical formed would be stabilized by delocalization over the two oxygen atoms.

Actually, the sequence of reactions leading to a low but constant concentration of $\ce{Br^.}$ radicals is very simple:

$$\begin{align}\ce{R2N-Br &<<=> R2N^. + Br^.}\tag{1}\\[0.4em]
\ce{Br^. + Br-NR2 &<=> Br2 + R2N^.}\tag{2}\\[0.4em]
\ce{Br2 &<<=> 2 Br^.}\tag{3}\end{align}$$

The $\ce{N-Br}$ bond is very weak and can easily dissociate homolytically to give both bromine radicals and succinimide radicals. Bromine radicals can combine with other NBS molecules to form free bromine and another succinimide radical and bromine molecules can dissociate to give bromine radicals.

All of these steps are reversible and the equilibrium favours the non-radicals side. Whether your reaction needs a constant supply of bromine or a constant supply of radicals doesn’t matter as NBS can supply them.

Only if there is already some bromide in solution can a fourth reaction take place:

$$\ce{R2N-Br + Br- -> R2N- + Br2}\tag{4}$$

This is the only reaction that has a dedicated forward-direction. If bromide is present, bromine molecules can be generated much more rapidly than radicals, so the generation of radicals will have to come from bromine dissociation.

While it is true that the NBS radical is somewhat well stabilised, that does not make it that much easier for the $\ce{N-Br}$ bond to dissociate: radicals are still radicals and two-electron-two-centre bonds are still energetically favoured. The $\mathrm{S_N2}$ displacement is much better since an anion is equally well stabilised but has a  lower general energy.


Why must NaOH be kept dry during storage? Why should the Erlenmeyer flask used to store $\ce{NaOH}$ be dry? Why should it be kept covered at all times?

Without a proper closed container, sodium hydroxide readily attracts water and become a viscous solution (concentrated sodium hydroxide). This solution readily attracts carbon dioxide in air and reacts with it.

$$\ce{CO2(g) + H2O(l)<=>H2CO3(aq)}$$
$$\ce{2OH^{-}(aq) + H2CO3(aq) -> 2H2O(l) + CO3^{2-}(aq)}$$

(The $\ce{OH^{-}}$ in the second equation comes from $\ce{NaOH}$.)

This reaction decrease the amount of $\ce{NaOH}$ and thus the titration result will contains error.

Actually, the first equation can happen even without sodium hydroxide. Carbon dioxide in air is known to spontaneously dissolved into water and become carbonic acid. This is why the distilled/deionized water which in contact with air will have pH of around 6 after some time. With the presence of sodium hydroxide in water, the second equation greatly drives the equilibrium of the first solution to the right. Concentrated sodium hydroxide is used in some experiment as a carbon dioxide absorber.

Another possible problem that might occur is that concentrated sodium hydroxide can react with glass. Althought this is a very slow reaction, it might causes some unexpected error or ruin the glassware.


CaSO4 precipitate in mixture of Na2SO4 and CaCl2 solutions I have been stuck on a chemistry problem for a long time now and if anyone here can help me I would be eternally grateful.

40. Du tillsätter $\pu{100 ml}$ $\pu{0,01 M}$ $\ce{Na2SO4}$-lösning i $\pu{100 ml}$ $\pu{0,02 M}$ $\ce{CaCl2}$-lösning. Det bildas en $\ce{CaSO4}$-fällning. Hur mycket av $\pu{0,01 M}$ $\ce{Na2SO4}$-lösning måste ännu tillsättas för att den bildade $\ce{CaSO4}$-fällningen just och just ska upplösas helt? $K_\mathrm{s}(\ce{CaSO4}) = \pu{2,5e-5 M}$. $(t = \pu{25 °C})$ (2p)
a. $\pu{0,43 l}$
b. $\pu{0,48 l}$
c. $\pu{0,53 l}$
d. $\pu{0,58 l}$

Translation: We add 100 ml of 0.01 M $\ce{Na2SO4}$-solution to 100 ml of 0.02 M $\ce{CaCl2}$-solution. A precipitate of $\ce{CaSO4}$ is formed. What is the (minimal) volume of 0.01 M $\ce{Na2SO4}$-solution that needs to be added to the mix for the $\ce{CaSO4}$ precipitate to be just dissolved completely?
The problem then lists the solubility constant, $K_\mathrm{s}$, for $\ce{CaSO4}$ as $2.5\cdot 10^{-5}$.
The only way I can think of solving this myself is by looking at the added $\ce{Na2SO4}$-solution as just water and using the $K_\mathrm{s}$ value for $\ce{CaSO4}$ to see what concentration of $\ce{CaSO4}$ is possible in just water. Then calculate the amount of moles of precipitate that was actually formed to see how much water I would need to add to the mixture:
$$K_\mathrm{s}(\ce{CaSO4}) = \pu{2.5e-5 M^2} = [\ce{Ca^2+}][\ce{SO4^2-}]$$
$$\therefore [\ce{Ca^2+}] = [\ce{SO4^2-}] = \sqrt{\pu{2.5e-5 M^2}} = \pu{5e-3 M}$$
$$C = \frac{n}{V} \to n = CV$$
Amount of precipitate formed:
$$n(\ce{CaSO4}) = \pu{0.02 M} \cdot \pu{0.1 dm^3} = \pu{0.002 mol}$$
Required total volume of water to dissolve $\pu{0.002 mol}$ of $\ce{CaSO4}$:
$$V = \frac{n}{C} = \frac{\pu{0.002 mol}}{\pu{5e-3 M}} = \pu{0.4 dm^3}$$
This is wrong, but I can't figure out the right way to go about this. The answer surely has to do with how well $\ce{CaSO4}$ is dissolved in a $\ce{Na2SO4}$-solution as opposed to just water.

As you've already noted, there are 3 substances of interest here: $\ce{Ca^{2+}}, \ce{SO4^{2-}}, \ce{H2O}$.

Let $V$ be the additional volume of sodium sulfate solution. Then:

The total volume is $V + 0.2\ \mathrm{L}$.

The total amount of sulfate is $(V + 0.1\ \mathrm{L}) (0.01\ \mathrm{M})$.

The total amount of calcium is $(0.1\ \mathrm{L}) (0.02\ \mathrm{M})$.

We have then the following relationship:

$$\ce{[Ca^{2+}][SO4^{2-}]}=\frac{(0.1\ \mathrm{L}) (0.02\ \mathrm{M})}{V + 0.2\ \mathrm{L}}\frac{(V + 0.1\ \mathrm{L}) (0.01\ \mathrm{M})}{V + 0.2\ \mathrm{L}}\leq K_{\mathrm{sp}} = 2.5\times 10^{-5}$$

Now, it's just algebra to solve for the threshold value of $V$, keeping in mind that there are two solutions for $V$ because this is a quadratic equation. In this case, the correct solution is the positive root:

$$V \approx 0.48\ \mathrm{L}$$

The mistake you made is that the calcium and sulfate ion concentrations are definitely not equal, since the additional volume of solution contains only sulfate and not calcium.


Method of deriving some properties or possible method of synthesis of a specific monomer This is kind of like two questions but since they're related I have put them together.

Consider a monomer under a hopefully anatomically correct name 'Phenyl Carbolithia Divinylene Oxide [$\ce{C11H9LiO4}$] (For ease of understanding.)

Would it be possible for me to find or estimate some basic properties (physical and chemical)? Or the 3D structure of the molecule (As opposed to the one I plotted, which is based on my estimation of dipole with minimal reasoning.)

By separating the molecules into parts:

Lithonate (Lithium Carbonate)
Benzene (Phenyl group)
Ethylene (Vinylene group)
Divinyl Ether

Perhaps now it is possible to make some predictions based on some functional groups? (Such as ether.)

Though theoretical chemistry has shown that such achievement in predicting a physical structure is unlikely, I am wondering if anyone would know specific properties of this molecule and the way they have deduced it in which I haven't consider.

The second part of the question may lead to the solution for the first part, by synthesizing such molecule.

My humble attempt at theorizing the chemical synthesis with my limited knowledge on the subject is based on condensation polymer. Perhaps having the separate parts of molecules mention previously dehydrated to form a chain, the molecule may be created if a similar polymer is formed except to have the carbon-oxygen backbone chain to form a ring instead.

I also have a slight inquiry on whether it is possible for the double bond to exist in this monomer; and whether my nomenclature for this substance is anatomically correct.

I doubt that your "monomer" is very stable. Orthoesters, $\ce{R'-C(OR)3}$, are hydrolyzed under acidic conditions: $\ce{R'-C(OR)3 + 2 H+ -> R'-COOR + 2 ROH}$. The fate of your monomer is:


What is the formula for heat of vaporization under reduced pressure? I haven't taken a chemistry class in several years, and I have honestly been searching the web for the past hour and can't find anything. I know this is something basic we learned back in high school chemistry but maybe I am not searching for the right term...

Is there a formula? I know how to calculate distillation speed based on wattage, and the vaporization constant of the molecule, but cannot find any particular formula that can calculate how that number changes under reduced pressure.

You are looking for the Clausius-Clapeyron equation.  It comes in various forms:

The exact version is

$$\frac{\mathrm dp}{\mathrm dT}=\frac{\Delta H}{\Delta V T}$$

or, since the volume of the gas phase is so much larger than the volume of the condensed phase

$$\frac{\mathrm dp}{\mathrm dT}=\frac{p \Delta H}{R T^2}$$

or even more succinctly, assuming $\Delta H$ remains constant

$$\frac{\mathrm d \ln p}{\mathrm d \frac{1}{T}}=-\frac{\Delta H}{R}$$


Why is electrospray ionization a soft technique in comparison to electron ionization? I know that electrospray ionization (ESI) is a soft ionization technique and that electron ionization (EI) is not, because it causes a great amount of fragments in a fast time due to the fact that a single electron is accelerated up to 70 V. ESI MS can cause fragments, but slower and not as good. But why is this effect less strong than in EI?

Thank you for your help!

The title of the Nobel Lecture of the person who invented electrospray ionization was "Electrospray Wings for Molecular Elephants". By that he meant that this technique is good for very large molecules and it does not break them. The interesting part is nobody fully understands how the big molecules get a charge. There are two models and who can explain it better than John Fenn, the inventor. Have a look at his lecture: https://www.nobelprize.org/prizes/chemistry/2002/fenn/lecture/

Coming to the electron impact ionization, it is a physical contact (don't take it literally) of a very high energy electron beam with the molecule, the energy of this process is sufficient enough to knock out an electron and break other bonds too in the gas phase. This breakage is understood in terms of molecule's cross section. Note that in ESI, no free electrons are used to ionize molecules. The ionization process is a mix of solution phase chemistry (i.e., what ions were present, pH, etc) and the molecule's own properties. Not all can be ionized by ESI. Certain ions will suppress electrospray phenomenon like trifluoroacetic acid.


Elimination mechanism in alkenes Do E2 elimination reactions occur in alkenes? If not, what mechanism is used?

If they do, are they limited by any stereochemical rules as alkanes do (antiperiplanar)?

My question is pernitent to the alkyne reaction mechanism, where both the E and Z alkenes undergo elimination to form the alkyne.

My guess is E2 mechanism is possible for the Z alkene, while the E alkene undergoes E1

No E1elimination. E2 in both cases. The E-isomer has a good overlap; syn elimination. The Z-isomer goes through anti elimination.


Difference between the absolute thermodynamic scale and ITS-90 What is the difference between the absolute thermodynamic temperature scale and the practical temperature scale: ITS-90?

The absolute thermodynamic temperature scale is a scale fixed by absolute zero and (at least for the kelvin, I'm not sure about the Rankine scale) the triple point of water, where absolute zero is defined as 0 K and the triple point of water is defined as 273.16 K. The rest of the scale is just a straight line from zero through 273.16 K off to infinity. However, this scale relies on only two defined points (only one of which can actually be measured, absolute zero being established from thermodynamic relations).

In practical use, for the purposes of calibrating thermometres and the like, this is not a great scale as the two calibration points may be far away from the temperature region one is interested in actually measuring. The usual way to deal with this is to use other reference points outside the actual definition. So that different people do this as consistently as possible, ITS-90 was created as a practical implementation of the absolute scale, setting additional reference points (all phase transitions) to defined values, interpolating between them to get other values. It's technically not a scale in itself, just a standard for calibrating temperatures that are as close to the absolute scale as possible. By making sure everyone uses the same reference points, measurements can be more consistent than if reference points are chosen arbitrarily and possibly assigned different values.

For example, if one wanted to make a thermometre for general lab use, one might pick the triple point of mercury, the triple point of water, and the melting point of gallium from the ITS-90 standard for reference points, knowing that other thermometres in this range conforming to the standard would also be calibrated this way, rather than choosing say the melting and boiling points of water, increasing the likelihood of consistent measurements with other thermometres.


Why is carbon dioxide bent in its first excited state? R. H. Crabtree writes in The Organometallic Chemistry of the Transition Metals (p 145, 6th ed.) that

  $\ce{CO2}$ is linear in the free state but bent [...] in the first excited state...

Why is this so? (The context refers to an electronically excited state.)

This effect is the so-called Renner–Teller effect and is a consequence of a coupling between vibrational and electronic motion and thus a breakdown of the Born–Oppenheimer approximation. You might be familiar with the Jahn–Teller effect which is related to the Renner–Teller effect. In degenerate states of linear molecules with three or more atoms, the total energy of the molecule can be reduced by bending the molecule. Consider a three atomic linear molecule in a degenerate state (such as $\ce{CO2}$ in its excited $\Pi$ state). When the molecule bents, the degeneracy is lifted as it can bend in the $x$ or $y$ direction. Because of symmetry requirements, the potential has to be an even function of the bending coordinate. Ignoring the lifting of the degeneracy we may approximate the zero-order potential with a Taylor expansion as
$$
V^0=ar^2+br^4+\cdots,
$$
where $r$ is the bending coordinate and $a$ and $b$ are constants describing the potential ignoring the vibronic interaction. When we do consider the vibronic interaction, the degenerate potential splits in two, but the splitting must still be an even function of $r$.
$$
\Delta V=\alpha r^2 + \beta r^4 + \cdots
$$
If the interaction is so large that $a<\alpha/2$, the potential will not have its minimum at $r=0$, but will have two minima at $\pm r_e$ and the molecule has a lower energy when it is bent.
See also Herzberg Vol 3, p 26.


Thermite reaction beween zinc and iron(II, III) oxide I have zinc metal and iron oxide in the form of magnetite which I would like to obtain iron metal from. I was wondering if I could use a thermite reaction to create iron and zinc oxide? I would like to know whether such a reaction would occur between these two substances, how violent it is and whether the reaction will sustain itself?

I'd check the Galvanic series.

$\ce{Fe(II)}$ and $\ce{Fe(III)}$ have an higher standard potential than $\ce{Zn}$ and therefore your reaction should take place.

I don't know how reactive it is but I had good experiences with also using $\ce{HCl}$. Use $\ce{Zn + HCl}$ to produce $\ce{H2}$ and heat $\ce{Fe2O3}$ in the presence of $\ce{H2}$ to produce $\ce{Fe}$ and $\ce{H2O}$.


Why lone pair occupies more space than bond pair? I'm not a chemistry expert and merely a high school student. In my book there is a slight introduction to VSEPR theory and therein they mention that lone pair-lone pair repulsions are maximum, followed by lone pair-bond pair and then bond pair-bond pair repulsions.

Now the reasoning they provide to this is that as lone pairs are localized on the central atom, and each bond pair is shared, the lone pair occupies more space and hence the repulsions are greater.

Now I don't understand how localisation of lone pair leads to its large size and secondly what is meant by"size"here, because it's hard to imagine size of a lone pair for me.

It's a request not to mention about advanced theories like molecular orbital theory, because I'm not familiar with aaj that right now.

EDIT
I would say that the other is also in a way similar but I could not find an answer of my question there. That's why I posted it.

Any of your help is appreciated.

Thanks.

The basic premise of this statement is that the repulsion exerted by an electron system depends upon the area it occupies. More the area, more is the repulsion.

As the bond pair is formed from the sharing of electrons, it experiences attraction from not one, but two nucleii. On the other hand, the lone pair is attracted by only one nucleus.

So, the bond pair is more localised while the lone pair is spread out and occupies a larger area.

The 'size' here refers not to the size of electrons but to the size and nature of the orbitals these electrons occupy.


Why do collisions in elementary reactions of higher-orders appear to be more likely? So we are told that a unimolecular elementary reaction has a rate law of $k[\text{A}]$ where a termolecular reaction with three unique reagents, $A$, $B$ and $C$ has a rate law of $k[\text{A}][\text{B}][\text{C}]$. Now, other things being equal, and assuming for the sake of argument that the initial concentrations are all greater than 1M, this means that the termolecular elementary reaction is more likely at the outset. But that doesn't make much sense. There may be some statistical mechanics principle that's lost on me but it seems more likely that A bumps into A than, A, B and C come together all at once. So I don't quite follow this reasoning. Is it the case that $k$ is generally quite smaller in these termolecular elementary reactions?

I second J M’s answer: the rate does not depend only on concentrations, but the rate constant is key. However, I will add two more comments:

First, you cannot compare values of rate constants for reactions of different orders… they don't have the same dimension (unit). So, you cannot say things like “rate constants of third order reactions are typically smaller than those of first-order reactions”.
However, you can say (and it is true) that reactions rates of high-order elementary reactions tend to be small, because of the low probability of the molecules crossing path.


Solutions of several salts What happens when you have a solution of a salt, and then try to dissolve another salt in that solution? I know about the common ion-effect etc, but how do you solve problems like this:

How much AgCl can you solve per L in a 3.8g/L NaCl solution?

To solve a problem like this, we need to know the solubility product constant $K_\mathrm{sp}$ for $\ce{AgCl}$, where at equilibrium,

$$K_\mathrm{sp}=[\ce{Ag+}][\ce{Cl-}]=1.8\times 10^{-10}\ \mathrm{M^2}$$

This relationship stipulates that under equilibrium conditions at all times, the concentrations of $\ce{Ag+}$ and $\ce{Cl-}$ must satisfy this relationship.

If the solution contains just $\ce{AgCl}$, we can determine the solubility. The concentrations of both ions must be equal based on the chemical formula.

$$\begin{array}{lcl}
x&=&[\ce{Ag+}]=[\ce{Cl-}]\\
x^2 &=& 1.8\times 10^{-10} \\
x &=& \sqrt{1.8\times 10^{-10}}=1.342...\times 10^{-5}
 \end{array}$$

We have $1.342\times 10^{-5}\ \mathrm{mol/L}=1.92\times 10^{-3}\ \mathrm{g/L}$ of $\ce{AgCl}$ in soltion.

However, if we already have some $\ce{Cl-}$ in solution, then things are a little different. Let's say we have $3.8\ \mathrm{g/L} = 6.5\times10^{-2}\ \mathrm{mol/L}\ \ce{NaCl}$ in solution, now we have a different concentration of $\ce{Cl-}$. When $\ce{AgCl}$ is added, some will dissolve, and the concentrations of $\ce{Ag+}$ and $\ce{Cl-}$ will both increase by some amount $x$.

We solve these problems using the ICE (Initial, Change, Equilibrium} approach:

$$\begin{array}{c|r|l}
\ & [\ce{Ag+}] &[\ce{Cl-}] \\ \hline
\mathrm{I} & 0 & 6.5\times 10^{-2} \\
\mathrm{C} & +\ x & +\ x \\
\mathrm{E} & x & 6.5\times 10^{-2} + x\\ \hline \end{array} \\
\ \\
K_{\mathrm{sp}}=1.8\times 10^{-10}=x(6.5\times 10^{-2} + x)\\
1.8\times 10^{-10}=x^2 +(6.5\times 10^{-2})x \\
0 = x^2 +(6.5\times 10^{-2})x - 1.8\times 10^{-10}\\
x = 2.77\times 10^{-9}, -0.065$$

One of the values for $x$ is nonsensical. The concentration of $\ce{Ag+}$ cannot be negative. Thus the amount of $\ce{AgCl}$ that dissolved in one liter was $9.77\times 10^{-9}\ \mathrm{mol}=1.38\times 10^{-6}\ \mathrm{g}$.

Note that since $x\ll 6.5\times 10^{-2}$, we could have simplified to $6.5\times 10^{-2}+x \approx 6.5\times 10^{-2}$ and then our maths would not have needed to invoke the quadratic theorem while getting the same answer within significant figures. This approximation is often valid. However, unless you are taking a test, you should always be able to Wolfram Alpha.


Arsenic bound proteins confirmatory test I want to confirm the presence of arsenic in some protein samples, apart from the marsh test can I use the following method?--

Arsenic is present as arsenic (III) ion in the samples. Hydrogen sulfide gas when passed in the sample of arsenic (III) ion gives precipitates which are yellow in color.

The yellow precipitates dissolve on adding concentrated nitric acid. This solution on reaction with ammonium molybdate in presence of concentrated nitric acid again gives yellow precipitate.

This confirms the presence of arsenic (III) ions in the sample.

If not, please suggest another method

P.S I have some info about arsenic bound proteins from-
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3797521/

You may wish to give this suggestion a try, courtesy of the literature on constructing nano-particles.

First, dissolve the protein samples in say ethanol or other flammable solvents.

Next, place it into an alcohol lamp with a wick composed preferably from glass fiber, but one may be able to employ woven cotton or a cotton string.

Lastly, light the lamp and place a bowl over the lamp to capture/condense escaping particles.

My recollection is that the target metal of interest, is slowly draw through the wick and thermally broken down. Still looking for the source, but here is a related article.

This is clearly relates to the final step in the Marsh Test, where the arsenic presence is distinctive.

Note, this proposed procedure, if successful, avoids the conversion of a sample of a compound into deadly AsH3 gas per a nascent hydrogen pathway, which is perhaps not required here, as you already have constructed (via an alcohol lamp), a path to 'burn' the As(III).


electron passing nodes? When one say that the electron goes from 1 orbit , say 2S, to another (higher) orbital, say 5P, it mean that the electron is excited to that orbit by some mean. In doing so it also it will also cross the nodes of 2P etc. But doing so does the electron actually cross the node or what? Cause doing so the, there is probability, although very less, of finding electron in node for a very short time.

When one say that the electron goes from 1 orbit , say 2S, to another
  (higher) orbital, say 5P, it mean that the electron is excited to that
  orbit by some mean.

First off, there are no orbits. Electrons do not circle the nucleus. Rather they are in particular quantum states, or, in other words, they "occupy" particular orbitals. Note that when we say that an electron "occupies" an orbital, we just mean that it is in a particular quantum state. There is no spatial occupation in a classical sense out there: eletrons do not "sit" or "circle" anywhere, they are just in their quanatum states.

Consequently, when we say that electron "goes" from one orbital to another, what we actually mean is that the quantum state of the electrons changes. And again, an electron transition is not a mechanical motion, thus, electrons upon transitions do not "move" through nodes.


Dissolution of protein powder in water I am using this Mocha Coffe.

When I try to dissolve it in a cupful of cold water, it forms a globule of powder and does not dissolve.

But if I add a small amount of water to the powder and mix it, it makes a paste which will dissolve in more water. What is happening here ?

A lot of powders show this behaviour. Try it with flour or 100% cocoa powder, and you get the same result.

What all these powders have in common, is that the grains contain some hydrophobic material.

When you put a bunch of powder in water, the outside layer of powder gets wet and sticks together, but doesn't let the water through, trapping an air bubble inside. Then when you try to stir, the waterproof bubbles just float around, easily avoiding your spoon.

But if you only put a small amount of liquid, the stirring is much better, and you can wet all the grains.

Source: this entry in the Hitchhiker's Guide to the Galaxy (of all places).


Can the titrant and the unknown material be flipped? We know the typical experiment of determining the concentration of $\ce{Fe^2+}$, where we titrate $\ce{MnO4^-}$ which has dark purple color and watch the color of the bottle.

Can we also determine the $\ce{MnO4^-}$ concentration using $\ce{Fe^2+}$ as titrant and how?

You can see that the iron is inside the beaker and we titrate with  $\ce{MnO4^-}$. Can we do it backward, i.e. having $\ce{MnO4^-}$ inside the beaker with purple color and we titrate $\ce{Fe^2+}$.

The titration reaction does not know or care whether one reagent is in the buret and one is in the conical flask (Erlenmeyer flask, it is not a beaker in your figure). And certainly you can standardize/titrate potassium permanganate with ferrous ammonium sulfate solution to determine the concentration of permanganate ion. Ferrous ammonium sulfate hexahydrate is a primary standard in analytical chemistry.

If you are extraordinary careful, you can fill the iron (II) solution in the buret and have permanganate in the conical flask. The end-point will be slightly vague (deep purple to colorless). You really have to be extra-cautious. The question is why bother with this painful eye-stressing procedure? Always take the permanganate in the buret.

A dark solution can be used in the conical flask. For example, in iodometric titrations, we start with a deep brown solution, which fades, to nearly a colorless solution upon titration with thiosulfate. In order to confirm the end-point, you would put a drop of starch before your expected equivalence point. Again, the color change is from deep blue to colorless.


Why are solids and liquids ignored in the equilibrium constant? If a reaction is taking place and is going towards equilibrium, wouldn’t the reactant be consumed to form products and vice versa? If this reactant/product is a solid or liquid, wouldn’t the number of moles when the reaction reaches equilibrium decrease/increase from its initial number of moles? Wouldn’t this change in moles from start of reaction to point of equilibrium be important to the equation, since it’s a ratio of products to reactants at equilibrium?

A solid or liquid that should take part in a solution or gas phase reaction has to dissolve or evaporate before it can react in any meaningful way.

If you add more solid solute to a saturated solution, its concentration does not change, and so doesn't the equillibrium constant of a reaction of the dissolved solute.

(The solute might also react directly out of the solid. Theoretically. But this different reaction needs much more energy than the one from solution, and it will only add a minuscule shift to the total equillibrium, even if your solid has a huge surface. Surely neglectable, most likely not even measurable.)


Synthesis of chloroquine from 4-amino-7-chloroquinoline Can we get chloroquine from 4-amino-7-chloroquinoline (7-chloroquinolin-4-amine) by adding 4-chloro-N,N-diethylpentan-1-amine $(\ce{CH3CHCl(CH2)3N(Et)2})$ ?

My intuition tells me that the nitrogen in the ring is more nucleophilic than the exocyclic amine. A quick-and-dirty electrostatic potential calculation using B88LYP/6-31G(d) on the PM3 geometry shows a large electron density (blue and violet) on the ring nitrogen and lower density (yellow) on the exocyclic amine:

Are there conditions that might work? Yes. Should you try it, if you need that molecule? Yes. Will it be difficult? Yes.


Diagonal line going through groups IIIA-VIA on the periodic table Why is there a diagonal line cutting through some of the nonmetals and metalloids on the periodic table in groups IIIA, IVA, VA, VIA, and VIIA?

Is there any historical background about it?

The red line is the dividing line between metals and non-metals. It is rather arbitrary; as Mendeleev himself wrote: “It is...impossible to draw a strict line of demarcation between metals and nonmetals, there being many intermediate substances.” Most modern periodic tables do not feature it.

The elements with a blue background are the metalloids. Against, it's a rather arbitrary choice, but it corresponds to elements that have some properties of metals. Wikipedia describes them in the following way:

  Physically, metalloids usually have a metallic appearance but they are brittle and only fair conductors of electricity; chemically, they mostly behave as (weak) nonmetals. They can, however, form alloys with metals. Ordinarily, most of the other physical and chemical properties of metalloids are intermediate in nature


Determine products of NH4NO3 hydrolysis What are the products of the following hydrolysis reaction? How do you know?

$$\ce{NH4NO3 + H2O <=> ?}$$

It will never happen. You can not hydrolyze a salt to any strong acid (nitric acid in your case) or base because the reverse reaction is predominant.


What determines whether an aldol reaction will occur again or eliminate? The product of an aldol reaction is a $\beta$-hydroxy ketone. This can go on to eliminate to form an alkene thus:

However, the species on the far left of the above diagram (the $\beta$-hydroxy ketone product of the first aldol reaction) can react through an aldol reaction again given that it has two more available protons at the $\alpha$ position.

How is it possible to tell which reaction pathway will occur? Is this the issue that is referred to when the Aldol reaction is referred to as "hard to control"?

Partially yes. The big problem with aldol reactions in general is that there are a lot of species that can react together to form stable compounds resulting in low yield. There are numerous ways to counter this:

Specific enolate equivalants, like lithium enolates, silyl enol ethers, aza enolates, enamines, etc. These are among the most important.
Lack of proton in $\alpha$-position.
Strong thermodynamic favoring of certain products in combination with reversibility of the reaction.
etc.

As for your example the $\beta$-hydroxy ketone is more sterically hindered than acetone and is therefore not very likely to be the major contributor (although polymerization can still happen of course).

In general you cannot predict whether or not elimination of $\ce{H2O}$ occurs after the aldol reaction, but the likeliness can be increased by forcing certain reaction conditions.


Mesomeric Effect and its Strength While solving a question in organic chemistry, I needed to compare the strength of mesomeric effects of various groups.
I looked it up on Wikipedia, and it goes like this :

+M EFFECT ORDER :

–O− > –NH2 > –NHR > –OR > –NHCOR > –OCOR > –Ph > –F > –Cl > –Br > –I

-M EFFECT ORDER :

–NO2 > –CN > --S(=O)2−OH > –CHO > –C=O > –COOCOR > –COOR > –COOH > –CONH2 > –COO−

What is the way to understand such comparisons ? . Why does the -OH group show stronger mesomeric effect than the -NH2 group ? Also, why is an -OR group weaker than the -NH2 group ? Can somebody please explain ; I was unable to find a reasonable explanation through a google search.

Some background knowledge
You need to be aware of electronegativities, assignement of s and p character, relation of structure to the donating or accepting nature of the substituent, and most importantly, the overlap efficiency of the donor and acceptor positions (roughly estimated by overlap integral , although for the sake of this discussion, most of the mathematical rigor is not required)
I will be comparing most of the order for the +M effect, and provide some hints for the -M effect.
Group 1: from $\ce{O-}$ to $\ce{OCOR}$
This group of substituents consists of two kinds of donor atoms from the 2nd period, namely $\ce{O}$ and $\ce{N}$. As they are from the same period as $\ce{C}$ , therefore, overlap efficiency will be maximum in this case, for 2p-2p overlap.

Recall that p-p overlap> s-p overlap> s-s overlap as per overlap efficiency

Due to this reason, this group of substituents will have better +M effect as compared to the halogens, because as we move down the group 17, overlap efficiency for 2p-3p,2p-4p overlap and so on decreases.
However, a member of period 2 does exist among the halogens, namely $\ce{F}$ .But since it is more electronegative than both $\ce{O}$ and $\ce{N}$, it will again have lesser donating tendency as compared to this group.
Now, let's get down to the specific cases:
The group leader $\ce{O-}$ :
First,let us look at some enthalpy values:
$\ce{ O(g) + e^- -> O^-(g) 1st EA = -142 kJ /mol }$  (1)
$\ce{ O^-(g) + e^- -> O^2-(g) 2nd EA = +844 kJ /mol }$ (2)
As you can clearly see, adding an electron to $\ce{O-}$ ( or adding a second electron to $\ce{O}$) is an endothermic process, which means energy will have to be provided externally for pushing more electron density inside $\ce{O-}$.
That is pretty obvious as well, because you are forcing an electron into an already negative ion. It's not going to go in willingly! The second electron affinity of oxygen is particularly high because the electron is being forced into a small, very electron-dense space.[source]
Conversely, this means that the reverse reaction for (2) will be pretty favorable, in the sense that $\ce{O-}$ will quite readily give away electron density to a suitable acceptor atom,as is evident from the enthalpy values. Hence, $\ce{O-}$ leads the charge in terms of +M effect

It is to be kept in mind that the factor of compatibility of overlap still holds. Donation from $\ce{O-}$ to, say, $\ce{Si}$ will still not be as effective as that to $\ce{C}$

Subgroup 1:The curious case of $\ce{-NH2}$,$\ce{-NHR}$ and $\ce{-OR}$
First of all, since we have got neutral donor atoms in this subset, it stands to reason that simple electronegativity will be the major governing factor here to judge the overall donation tendency. Hence, both the $\ce{N}$ donor groups should be more donating than the $\ce{O}$ donor group.
The slightly peculiar judgement crops up when you are deciding between $\ce{-NH2}$ and $\ce{-NHR}$ . On a preliminary basis, you might say that since the $\ce{R}$ will most probably be exerting a strong +I effect ,and thereby the latter should have had more +M effect than the former.
But here's the catch: the electron density of the $\ce{N-R}$ bond will mostly shift towards N as compared to the $\ce{N-H}$ bond, but the donation to the benzene ring will ultimately take place by the $\ce{N}$ atom. And hence, it becomes important to look at what the donor nitrogens of these two groups have to say.
We are basically going to compare the bond angles of $\ce{H-N-H}$ and $\ce{H-N-R}$ .Due to the sterically bigger $\ce{R}$ group , the latter should be greater than the former. This means that the s-character will go up on the $\ce{N}$ atom of the latter, and this directly means that electronegativity for it will increase.

(Think of the various hybridization schemes, namely $\ce{sp}$ , $\ce{sp^2}$ and $\ce{sp^3}$, the percentage s character in each, the bond angles for each, and the electronegativity trend of $\ce{sp}$>$\ce{sp^2}$>$\ce{sp^3}$ )

Hence, donation power of $\ce{-NHR}$ should be less than $\ce{-NH2}$

A similar argument will also explain why the +M effect of $\ce{-OR}$ is less than $\ce{-OH}$

Subgroup 2: $\ce{-NHCOR}$ and $\ce{-OCOR}$
Both these group are analogous structurally, and that's the key to their relative comparison for our purpose. On a preliminary basis, we see that the number of highly electronegative atoms( that is, $\ce{O}$, $\ce{N}$ and $\ce{F}$) has increased in both these groups as compared to the previous members of this group. $\ce{-NHCOR}$ has an $\ce{O}$ and $\ce{N}$, while $\ce{-OCOR}$ has two oxygens. This immediately tend to make them less donating than the previous members of the group.
Another reason to see why they would be less donating is because of the distribution of partial charges in the resonance hybrid. The donor atom's lone pairs in both the cases are conjugating between a $\ce{C=O}$ group and the benzene ring. Since negative charge is more stable on $\ce{O}$ than on $\ce{C}$, the oxygen of the $\ce{C=O}$ will end up possessing much of the negative charge which would otherwise be available for donation to the ring.
In other words, we can say that the charge will be dispersed over a competing electron sink, namely, the $\ce{C=O}$ .Hence, their mesomeric effects will decrease considerably as compared to other members of the group.
As for their individual comparisons, you can easily judge that by the electronegativities of their individual donor atoms, $\ce{N}$ and $\ce{O}$
The phenyl group
Wikipedia says:

It is generally considered an inductively withdrawing group (-I), because of the higher electronegativity of sp2 carbon atoms, and a resonance donating group (+M), due to the ability of its π system to donate electron density when conjugation is possible.

Drawing from this, one can say that since the phenyl group is donating by using bond pair of electrons(namely, it's $\ce{C=C}$ π bonds) rather than lone pairs(unlike all the previous groups which were using lone pairs of $\ce{N}$ or $\ce{O}$) therefore it will have lesser donation tendency than the preceding members as lone pairs are generally more "freely available" for donation, being attracted by a single nucleus rather than bond pair of electrons which are held by two nuclei.
Still, as it is a donation from a $\ce{C}$ donor atom, it has better +M effect as compared to the halogens.
Comparison between $\ce{-Ph}$ and $\ce{-F}$ is a tough case, as it may seem that fluorine which is using it's lone pairs should have been a better donor than $\ce{-Ph}$. To this I would say that maybe since fluorine is the most electronegative element, it's electronegativity would dominate the donating tendency by the lone pairs(in technical terms, HOMO-LUMO gap would be more in case of F's non-bonded electrons than the $\ce{C=C}$ pi-bonded electrons)
But again, this reasoning would seem unsatisfactory.  Anyway,the mesomeric effect of the $\ce{-Ph}$ group usually holds little chemical interest. It mainly serves as a control, and we are much more interested in the effect of hetero atoms on the ring as compared to the effect of one benzene ring on another(in case of mesomeric effect)
Group 2:The halogens
I have already given enough idea about them above, more details can be found here
So there you have it. A similar idea can be applied for deducing the logic of the -M ordering as well.

Hints for -M effect

$\ce{-NO2}$ has 3 highly electronegative atoms, and an electron sink in the form of $\ce{-N=O}$, so it should lead the pack.

Just like the previous ordering, we are again interested in the characteristics of the acceptor atom here. While for +M effect, it increased with decrease in s-character(and hence electronegativity) on the donor atom, and more freely available electron density

-M effect will increase with increase in s-character(and hence electronegativity) of the acceptor atom. Since in your list, the remaining substituents are $\ce{C}$ acceptors, then the hybridization of $\ce{C}$ needs to be considered.
Accordingly, $\ce{-CN}$ should fall next with an $\ce{sp}$ hybridized $\ce{C}$

Now most of the items I can see on the list are substituents with $\ce{sp^2}$ carbons, having different combinations of $\ce{O}$ and $\ce{N}$ attached to the acceptor atom. Your job would now be to checking the varying extents of cross conjugation happening at the electron sink directly attached to the acceptor atom(and thereby comparing the electron accepting ability of the same) by looking at the electron donating abilities of the attached heteroatoms

You can ask for further clarification if these hints should not prove to be sufficient.


How do we know if this following molecule is a superimposable mirror image or non-superimposable mirror image ? [See Image attached] How do we know if this following molecule is Super Imposable mirror image or Non Super Imposable mirror image ? [See Image attached]

Take 2 round, soft objects like potatoes or apples.
Pin into each 4 pins of 3 different kinds ( 1 + 2 + 1 ),
pointing to the most different directions.
Call the 1st pin bromine, the 2nd one hydrogen, the 3rd one ethyl.
Do the same for the other potato to be the mirror image of the first one.
Try to rotate one of them to look like the other.
If you succeed, they are superimposable.
Now, replace one of the same pins on both objects by another -
4th kind - of pins in such a way that they are still mutual mirror images.
Try to match them the same way you tried previously.
If you succeed, they are superimposable.

The summary:

While there are 2 hydrogen atoms on a carbon atoms - or more generally, there are only 3 kinds of groups on a C atom - the groups/atoms of the same kind are interchangeble and you can rotate the molecule to match the mirror image.
If there are 4 different groups on C atoms, you cannot make such a match.
Said by other way, the former case has a plane of symmetry, the latter case does not.


What is product obtained when B2O3 is reacted with NaOH? Will the product be a metaborate ($\ce{BO2-}$) or a borate ($\ce{BO3-}$) or anything else and why?

As @Ivan said, it may form metaborate, borax or a tetrahydroxoborate complex all depending on reaction conditions(temperature, pressure, medium, stoichiometry etc.).

$$\ce{B2O3 + 2NaOH ->[\Delta] 2NaBO2 + H2O}$$

Boron trioxide react with sodium hydroxide to produce sodium metaborate and water. This reaction takes place at a temperature of 400-550°C.

$$\ce{2B2O3 + 2NaOH → Na2B4O7 + H2O}$$

Boron trioxide react with sodium hydroxide to produce sodium tetraborate and water. Sodium hydroxide - diluted solution. Boron oxide is amorphous.

$$\ce{B2O3 + 2NaOH + 3H2O → 2Na[B(OH)4]}$$

Boron trioxide react with sodium hydroxide and water to produce sodium tetrahydroxoborate(III). Boron oxide is amorphous. Sodium hydroxide - concentrated solution. The reaction proceeds at room temperature.


Why is a cationic trialkylammonium functional group inductively withdrawing in nature? The alkyl groups on the nitrogen ion should donate electrons (alkyl groups are electron donating in nature). So, irrespective of the nitrogen ion being a cation (which is an electrophile), the alkyl groups should compensate the electrophilicity of the nitrogen cation, and the ion as a whole should be electron donating in nature.

Then why is $\ce{R3N+}$ considered to be electron withdrawing?

Nitrogen is a very electronegative element and the fact that nitrogen bears a positive charge in the ammonium ion makes it even more electrophilic(as you pointed out) . Alkyl groups are only weak electron donating groups and can't help much about nitrogen's positive charge.Hence the ammonium cation is inductively withdrawing in nature.


Show that decomposition of AsH3 follows first order kinetics The decomposition of $\ce{AsH3}$ into solid $\ce{As}$ and $\ce{H2}$ is given in terms of pressure of the gas in a closed vessel from time to time.
$$\ce{2AsH3 -> 2As + 3H2} $$

Show that this reaction follows first order kinetics.

My attempt :

So to prove that the reaction follows first order kinetics, I tried finding the relation between pressure in the vessel and the time taken.

I assumed the initial concentration of $\ce{AsH3}$ (which is the pressure of the whole vessel at $t=0$) as '$2a$'. Assuming the dissociation factor to be $\alpha$, at any time $t$, the composition of the vessel can be :
$$\ce{AsH3} \rightarrow (2a - 2\alpha)$$
$$\ce{As} \rightarrow 2\alpha$$
$$\ce{H2} \rightarrow 3\alpha$$
(We are working with gases so the concentration of solid As can be assumed to be constant and have no effect on the pressure of the vessel)

Assuming all gases to be ideal.

So I was able to pull through till here. Assuming temperature to be constant, $a$ can found out:
$$2a = {\frac{P}{RT}} \Rightarrow a = {\frac{733.22} {2 \cdot 0.081 \cdot  T}} = {\frac{4465.4}{T}}$$

I am not able to move forward after this.

In a first order reaction $\ce{A \rightarrow B + C}$ with rate constant k the rate equation is
$$ {\frac{d\ce{[A]}}{dt} = - k\ce{[A]} } $$
which  if integrated to give the concentration vs time where the initial concentration is $\ce{[A_0]}$ produces
 $$\ln\left(\frac{\ce{[A]}}{\ce{[A_0]}}\right)=-kt$$ or $$ \ce{[A]= [A_0]}e^{-kt}$$

Thus a log of the ratio concentrations or pressures (or any measurement proportional to concentration) vs time should be a straight line if the reaction is first order.

In your case you are observing the product which appears as the $\ce{AsH3}$ disappears therefore
$$ {\frac{d\ce{[B]}}{dt} = + k\ce{[A]} } $$

You can substitute for $\ce{[A]}$ and integrate again to get $\ce{[B]}$.

$$\ce{[B]-[B_0]=[A_0]}(1-e^{-kt})$$
However the total pressure is measured which is the sum of $\ce{2[A] + 3[B]}$ and if $\ce{[B_0] =0}$ then

$$\ce{3[B] + 2[A] =3[A_0]}(1-e^{-kt}) + \ce{2[A_0]}e^{-kt}$$
if the measured pressure is proportional to sum of concentrations with $p_0$ as initial pressure then

$$p = p_0(3-e^{-kt})$$

At $t=0$ the initial pressure is $2p_0$ which is the value given in the table. At very long times the pressure becomes constant at $3p_0$ and calculating $3p_0 - p$, rearranging and taking logs produces

$$\ln\left(\frac{3p_0 - p}{p_0}\right) =  -kt $$

which when plotted is the 'normal' equation for a first order equation.


Optimization completed on the basis of negligible forces I am performing relaxed scan in Gaussian (B3LYP/6-31G(d,p)). There are two small organic molecules, one of which is a radical, the other has a benzene ring.

I'm trying to understand the optimization output. Specifically, the line Optimization completed on the basis of negligible forces. I understand that these forces are the maximum and RMS forces above.

Some examples of "negligible forces" from my output file:

        Item               Value
Maximum Force            0.000004
RMS     Force            0.000001
Maximum Displacement     0.019917
RMS     Displacement     0.004747

        Item               Value
Maximum Force            0.000002
RMS     Force            0.000000
Maximum Displacement     0.002231
RMS     Displacement     0.000490

        Item               Value
Maximum Force            0.000002
RMS     Force            0.000001
Maximum Displacement     0.003132
RMS     Displacement     0.000805

However, some small values like

        Item               Value
Maximum Force            0.000002
RMS     Force            0.000001
Maximum Displacement     0.000666
RMS     Displacement     0.000149

are followed only by Optimization completed. So what does "negligible" mean? How small do they need to be to say they are negligible? Or is this connected to at least one NO in convergence criterion section?

Next question, how reliable is optimization when such message pops up? If it's unreliable, what do I do about this? Would altering the initial geometry fix it when in only happens in a couple of scan steps?

According to the Gaussian Frequently Asked Questions:

In general, a stationary point is found when one of two criteria is met for a point in an optimization calculation:

All four values listed in the output are smaller than the indicated thresholds.
The Maximum Force and RMS Force are two orders of magnitude smaller than the thresholds shown, regardless of the values of the displacements.

You indicate three molecules where the second criteria activate - the forces are really, really small. So the optimization gives up.
In your final case, you can see that the displacements are also small, so the first criteria activate.
You then want to know how "reliable" the geometries are when the forces are small, but the displacements are not. That depends a lot on your needs. The optimization quits because such small forces imply a very flat potential energy surface and it may take a long time before "complete" convergence. My guess is that you still have some imaginary frequencies, but that might not matter depending on your needs.
It's worth pointing out that many quantum chemical programs have different criteria for declaring an optimization "finished." (In other words, your calculation in a different program might be declared done because of negligible energy and force/gradient change.)
If you want to push for a true minima, you should take that optimized geometry, run a force / frequency calculation to get an accurate Hessian, and save the result to a checkpoint, then re-start the optimization using the saved Hessian:
%Chk=myfile
# B3LYP/6-31G(d,p) OPT=ReadFC Freq Geom=AllCheck Guess=Read

Of course, that means probably breaking your relaxed scan calculation into a set of separate steps. (Honestly, this is how I usually do relaxed scans anyway - a set of separate jobs with each stepped geometry.)
One other very important point. If you're doing a scan of intermolecular interactions with B3LYP, you really should be using a dispersion correction such as EmpiricalDispersion=GD3BJ. I often see incorrect repulsive potential energy curves from B3LYP without dispersion corrections.


Is it possible to form bubbles of ionized hydrogen and ionized oxygen under the right pressure/temperature conditions during electrolysis of water? When we run a current through water (pure or not), $\ce{O_2}$ forms at the anode while $\ce{H_2}$ forms at the cathode. If I understand correctly, as water is already slightly ionized, $\ce{H}$'s snap together (and so do $\ce{O}$'s) during the electron exchange. Is it possible to apply enough heat or enough pressure to water to prevent $\ce{H}$'s and $\ce{O}$'s from snapping together and so we could thus form bubbles of ionized hydrogen and ionized oxygen?

The electrolysis will not form any ionized oxygen or ionized hydrogen, but instead proceeds through radical species. Some experiments have attempted to show the presence of $H^{\bullet}$ and $OH^{\bullet}$ radicals at the anode / cathode respectively (see P.H. Kasai, D. McLeod J. Am. Chem. Soc (1978)). Provided that the surface concentration is sufficient, these radical species will very quickly recombine to lead to $H_{2}$ and $O_{2}$. The surface of the electrode is very likely to influence the process, as the radicals species are likely to interact with it and not be released in high concentration in the solution.

Other factors are also influential, and the radical species might also react with other constituents of the solution (see for instance T. Mokudai,  K. Nakamura,  T. Kanno,  Y. Niwano, PLOS One (2012) on the formation of hydroxyl radical vs formation of hypochlorous acid through electrolysis).

Heating the solution is likely to increase radical concentration, but the temperature should be extremely high to prevent radical recombination within the gas. As for pressure, my guess would be that a reduced pressure is likely to improve radical isolation by displacement of the solvation equilibrium.


Positively charged gases Theoretically, if you could could have a positively charged gas say O$^+$ enclosed (by a sufficiently inert material) around a conductor which is grounded, what would the reaction be at the conductor if the conductor was made of;

1) Tungsten, 2) Stainless Steel,  3) Copper.

Would the gas gain electrons from the conductor and form O$_2$, react with the conductor to form a coating, react with each other, react with the container or something else?

You totally can't have a gas (or any other state of matter) made of positive ions alone.

That's not what you asked, though. The question was: what if we theoretically could have it. Well, then in a pretty short instant you will stop having it. The said instant would be rather violent, too(*).

To put things into perspective, imagine an indestructible spherical vessel about as big as to contain 1 mole of gas at normal conditions, which implies the radius of 17.5 cm. Now let's magically put 1 mole of $\ce{O+}$ inside. That would give it an electric charge of about 96500 coulombs, which is quite a lot. The electric field outside the sphere would be $$|\vec E\,|={1\over4\pi\varepsilon_0}\cdot{q\over r^2}=2.836\cdot10^{16}\;\rm V/m$$

Do you know how big is this? Wrong! it is bigger than you think. Mere $3\cdot10^6\;\rm V/m$ would cause a breakdown in dry air. Good solid insulators can handle more, but I think we may safely assume that $10^{10}$ is enough to break anything that is made of atoms. Mind you, we have six orders of magnitude on top of that(†). The conductor will be instantly vaporized and thrown out with immense force, enough to start many nuclear reactions and create a bunch of weird elementary particles. What metal it was made of would be the least of your concerns.

So it goes.

(*) Reminds me of a certain what-if episode, but they used a bigger sphere.
(†) Thought we would break the Schwinger limit, but we didn't.


What is the keto form of a hexahydroxybenzene? 159. The keto isomer of the following compound is:

The answer has been given as (c). I cannot seem to understand why is it not possible for the keto form to only have one OH convert into keto form shown below:

The question is badly phrased because it uses the term ‘the keto form’, implying that there be only one. In the case of hexahydroxybenzene, you are correct in that there is a monoketo form, at least two different diketo forms and potentially even more triketo forms.

All of these polyketo forms are created in the same way as the monoketo form except you do the isomerisation multiple times. So one way of arriving at the answer would be to check whether you can get to any of those molecules by tautomerising the hexahydroxy form often enough.

It may, however, be easier to think in the other direction, i.e. which ketones can I tautomerise to give a fully aromatic ring. Obviously, as your monoketo example shows, a ketone needs an α-hydrogen atom, so a) and b) cannot tautomerise at all and d) can tautomerise only twice, not three times. Upon careful examination of c) you should be able to realise that this can tautomerise three times to give hexahydroxybenzene.

Entirely out of scope of the textbook question is whether this tautomerisation is probable. None of the keto forms provides any advantage over the hexa-ol isomer in no solvent. We cannot argue with better hydrogen bonding and we cannot argue with polarity (both would favour the hexa-ol in all environments I can think of). Thus, the question in itself is somewhat academic. Yet it remains helpful to see how to get from polyketo to poly-ol forms and back.


Source for thermodynamic and/or transport properties I'm working on some combustion problems using Cantera and I can't find references for the thermodynamic or transport properties for many of my chemical species. I'm looking for NASA-7 or NASA-9 curve-fits for the thermodynamics, although if there is another equation format that exists, I will make it work.  And I also need properties for transport that Cantera calls "well depth", "diameter", and "rotational relaxation".

Specifically I am looking for: HClO4, HCl (transport information), ClO3, ClO2 (transport information), ClO (transport information), Cl2 (transport information), Cl (transport information), and ClOH.

So I suppose the real question is where can I find thermodynamic and transport properties for gaseous species with chlorine? Finding data on hydrocarbons and air is simple, and I found data on liquid refrigerants but nothing on gaseous chlorine species.

Note: I checked NIST and JANAF and neither had the information. I can find constant values, but running a very detailed reaction mechanism with calorically perfect equation of state and power or Sutherland law transport isn't worth the computational cost.

I have a CHEMKIN files with HCL:

HCL                1   344.7       3.339     0.000     0.000     1.000  !(singh)

344.7 should be $\varepsilon/k$ in K and 3.339 should be $\sigma$ in Angstroms.

! Species: HCL              CAS Number: 7647-01-0
! Name:    Hydrogen Chloride
! Source:  SNLL fit to 1986 JANAF tables, 4/21/89
! H0(298K) =      -22.0600 (Kcal/mole),  S0(298K) =       44.6500 (cal/mole-K)
HCL                42189CL  1H   1          G   300.000  5000.000 1000.00      1
 2.75533500e+00 1.47358100e-03-4.97125400e-07 8.10865800e-11-5.07206300e-15    2
-1.19180600e+04 6.51511600e+00 3.33853400e+00 1.26820700e-03-3.66691700e-06    3
 4.70399200e-09-1.83601100e-12-1.21315100e+04 3.19355500e+00                   4

Also, Poling [1] has some data on HCl, Cl$_2$, ClNO. You could estimate the other properties using group contributions to define critical properties and then corresponding state principle methods (like Chung's [2]) to get your temperature/pressure dependencies. You can then make your own curvefit for your conditions of interest. Not as good as curvefits from experimental data but it should get you in the ballpark...

[1] B. E. Poling, J. M. Prausnitz, and J. P. O’Connell. The Properties of Gases and Liquids, Fifth Edition. McGraw-Hill, 2001.

[2] T. H. Chung, M. Ajlan, L. L. Lee, and K. E. Starling. Generalized multiparameter corresponding state correlation for polyatomic, polar fluid transport properties. 27:671–679, 1988.