What are the Waves Modeling when Referring to the Atomic Orbitals It is taught that the orbital shapes derive from wave functions with different numbers of nodes. For example, the "s" orbital comes from a wave that has one node. But what are the waves modeling? A light wave for example is showing a relationship between an electric field and magnetic field. What relationship does an atomic orbital wave show? In other words, if you were to label the axes, what would the labels be?

Thanks!

In a one-dimensional system (I choose this to keep things as simple as possible) the state of the system is described by the wave function $\psi = \psi(x, t)$, which depends on the coordinate $x$ of the particle at time $t$. Wave functions in general are complex functions of real variables. The symbol $\psi^{∗}(x, t)$ denotes the complex conjugate of $\psi(x, t)$. The quantity $P(x, t) = \psi^{∗}(x, t) \psi(x, t) \mathrm{d}x = | \psi^{∗}(x, t) |^{2} \mathrm{d}x$ gives the probability that at time $t$ the $x$ coordinate of the particle lies in the small interval $[x, x + \mathrm{d}x]$. The probability of the particle being in the interval $(a, b)$ on the $x$ axis is given by $\int_{a}^{b} | \psi^{∗}(x, t) |^{2} \mathrm{d}x$. The probabilistic interpretation of the wave function was proposed by Max Born. By analogy with the formula: mass = density $\times$ volume, the quantity $\psi^{∗}(x, t) \psi(x, t)$ is called the probability density that a particle at time $t$ has position $x$.

Edit: Atomic orbitals

For atoms you have a system that is spherically symmetric. Thus, it is most conveniently expressed in spherical coordinates instead of cartesian ones. The wave function solving the time-independent Schroedinger equation is then expressed in terms of the radial distance $r$, the polar angle $\theta$ and the azimuthal angle $\varphi$.
This wave function $\Psi(r, \theta, \varphi)$ can be split into a radial part $R_{n,\ell}(r)$ and an angular part $Y_{\ell,m} (\theta, \varphi )$, so that $\Psi(r, \theta, \varphi) = R_{n,\ell}(r) Y_{\ell,m} (\theta, \varphi )$, where $n$, $\ell$ and $m$ are the Principal, Azimuthal and Magnetic quantum number, respectively, and the functions $R_{n,\ell}(r)$ are the solutions of the radial Schroedinger equation

\begin{equation}
\bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} }  - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R_{n,\ell}(r) = 0
\end{equation}

with the nuclear charge $Z$ and the mass of an electron $m_{\mathrm{e}}$. Those radial wave functions are the part of the atomic wave functions $\Psi$ that determines their "extent" and their radial nodes, i.e. the nodes that grow with the Principal quantum number $n$: for example a $1\mathrm{s}$ orbital has no radial nodes (it is a simple sphere without any sign-change in the wavefunction) while the $2\mathrm{s}$ orbital has 1 radial node (it is a smaller sphere wrapped in a bigger spherical shell whereby the inner sphere has a different sign than the outer spherical shell such that you get one nodal plane between inner and outer part). The angular wave functions $Y_{\ell,m} (\theta, \varphi )$ are the so called Spherical Harmonics. They determine the angular momentum of the atomic wave function $\Psi$ and its characteristic shape, i.e. $\mathrm{s}$ orbitals are spheres while $\mathrm{p}$ orbitals are dumbbell-shaped (side note: the Spherical Harmonics are complex functions which are rather hard to visualize; the shapes that every chemist is used to comes from the usage of the so called real Spherical Harmonics, which are just linear combinations of the complex-valued Spherical Harmonics).


Do Electrons Really 'Spin'? With regard to the 'Electron Spin Number', lots of websites mention that electrons don't really spin and that the electron spin number has nothing to do with any physical spinning.

However, my chemistry teacher is quite adamant about electrons spinning about their axes in clockwise and counter-clockwise directions.

So my question stands, "Do electrons really spin?"

Could someone cite a reliable source to read up on this (Apart from Wikipedia, thanks)

It depends on what you mean by "spin".

If you mean "have intrinsic internal angular momentum, independent of its trajectory through space", then yes, electrons spin, and that's what the quantum number is measuring.

Though if by "spin" you mean "undergoes rotation" ("there's a little billiard ball, and if I were to put a mark on it and watch it, the mark would be rotationally displaced around the center in a periodic fashion"), then no, electrons don't spin. They're point particles(*): there's no billiard ball to mark, and even if you were somehow to mark a "side" of a point particle (you can't), a point particle can't undergo internal rotational displacement - there's no internal to displace! Likewise, the "frequency" of the spin is also meaningless - attempt to find out how many revolutions per minute an electron "spins" and your calculator will spit garbage at you. It's not something that has any physically realizable meaning.

So the spin on an electron is a very real thing and not just a convenient bookkeeping label. You can do experiments where you couple the spin angular momentum of an electron to "macro-scale" (or at least non-sub-atomic) angular momentum, and you find that the total sum of angular momentum - both "macro-scale" and electron spin - is conserved. Electron "spin" is as real and functional as a gyroscope's spin is, but not in the same way. The electron has "spin" (angular momentum) without actually rotating.

Bizarre, but that's Quantum Mechanics for you.

*) I'm assuming the Standard Model of Quantum Physics here. If you start to get into String Theory things get all sorts of complicated, but there's no universally accepted model of String Theory, so I'm just going to ignore it exists. (As most Chemists do.)

Also note that the Standard Model doesn't even attempt to predict where this intrinsic internal angular momentum comes from. It's just a property of electrons, taken as a given. Asking "why do electrons have spin?" is like asking "why are electrons charged?": They just are.


Which element has larger atomic radius -- beryllium or xenon? I say that xenon radius is smaller comparing to beryllium, but my friend says otherwise.

What I think is that xenon is down in periodic table comparing to beryllium (3 periods down) and it's far toward the right (16 groups ahead). So, the nuclear pull dominate (16 by 3 which is very big) and therefore xenon's radius is smaller than beryllium's.

My friends said that xenon is down in periodic table compared to beryllium and hence it's bigger and the nuclear pull effect has only little effect.

So who is right ?

You cannot really effectively compare the atomic radii of xenon and beryllium empirically; the atoms are too far apart. Once you remove yourself from direct neighbours, it is generally not possible to estimate relative sizes of atoms.

The closest hint you can get are calculated atomic radii on Wikipedia’s data page. These show that xenon is slightly smaller ($\pu{108 pm}$ to $\pu{112pm}$) but the difference is smaller than the expected error. So, for all intents and purposes they have the same size.


Can polyatomic ions (CO₃, PO₄, SO₄, NO₃) be considered conjugated systems? From my perspective these resonance structures allow these specific polyatomic ions to act as donor-acceptor molecules. Many donor-acceptor molecules also tend to be conjugated systems because they have chains of alternating conjugated π orbitals.

So does this imply that these ions (due to their resonance structures) can act as though they had a conjugated system or am I making too big of an assumption?

Curious on others perspectives/if I'm totally of my rocker.

"Conjugated" implies a 1,3-shift to move an electron or hole.  It is a remnant of LCAO modeling that is obviously wrong but fantastically useful short of the Woodward-Hoffmann rules.  I'm not sure inorganikers would like the name as such. MO modeling is accurate but unwieldly.  "Delocalized" or "resonance hybrid" is good, certainly for inorganic systems that may have (virtual) d-orbital participation.

Consider nitrate.  Inorganikers would say $\ce{N^{5+}}$, $\ce{^{-}O-N(=O)2}$, with five bonds to the nitrogen.  The negative charge 1,3-shifts around all three oxygens.  Organikers would see it as $\ce{N^{3+}}$, $\ce{[^{-}O{}-]_2N^{+}=O}%edit$, with four bonds to the nitrogen.  1,3 shifts, etc.  Is nitrate ever a bidentate ligand?  YES!  But it is just another resonance structure.

Inorg. Chem. 35(24) 6964 (1996)
http://pubs.acs.org/doi/abs/10.1021/ic960587b


Reaction of vinylacetylene (but-1-en-3-yne) with HCl Vinylacetylene contains both a double bond and a triple bond. We know that alkenes are more reactive to electrophillic addition reactions, so I thought that the $\ce{H+}$ from $\ce{HCl}$ would get abstracted via the double bond.
However, the final answer says the product is 2-chloroprene, which is possible only if the triple bond abstracts it.

My teacher said that it was because of the carbocation stability and the abstraction of $\ce{H+}$ by triple bond forms a stabler carbocation and leads to conjugation and hence is favoured.

But then Wikipedia says that a 1,4 addition takes place and an allene rearrangement takes place.This means that the triple bond is the one that reacts first , but in a 1,4 fashion to form an allene which then rearranges. I was unable to find a mechanism for this allene rearrangement, why is this 1,4 addition favoured, and why it takes place. Also, what is the correct explanation: the first one or the second one?

Beyond a shadow of a doubt, the second explanation, involving an allene rearrangement is the correct one. As you said, alkenes are more reactive towards electrophilic addition and hence the addition of HCl happens first on the double bond to give 3-chloro-but-1-yne. What follows next is an intramolecular rearrangement, through a cyclic five-membered transition state, which is catalysed by CuCl with NH4Cl. It encompasses the abstraction of a hydrogen from the C4, the migration of the chloride from C3 to C2, following the formation of a double bond at C3 and C4. The product is referred to as chloroprene, which gives neoprene on polymerisation, a useful polymer. However the other explanation offered by your facilitator is used by many as a trick to remember the final product, (including, me!). Hope that helps!


Is silicon oxide=rock? Are common rock (I.e. Rock from mountain, river, bricks, or even concrete) actually silicon oxides? Or it is just one of compositions?

You ask:

  Are common rocks (e.g. rock from mountains or rivers, bricks, or even concrete) actually silicon oxides? Or it is just one of compositions?

Rocks can contain silicon oxides, but that's not part of what the definition of a "rock" implies. From an introductory course in geology at the University of Texas at Austin, a rock is defined as:

  Any coherent, naturally occurring substance generally composed of minerals.

That's all there is to it. Whether or not the minerals include silicon oxides has nothing to do with a rock being called a rock.


How to make Sodium dichromate I've been trying to make Sodium dichromate $\ce{Na2Cr2O7}$ by following these steps:

Dissolve an inox spoon in $\ce{H2SO4}$
Filter the solution
Dissolve the obtained sludge in $\ce{NaClO}$
Heat the solution until slurries appear
Filter the solution

The solution I obtained was yellowish (light yellow), I tried to add some acid to see the orange color of the (hypothetical) dichromate but nothing happened!

I added some ammonia before acidifying the solution, nothing as well.

1st: Is this method the right one to obtain chromate/dichromate?
2nd: If yes, why didn't the orange color appear?

Note that the chromate ion $\ce{CrO4^{2-}}$ has chromium in the +6 oxidation state like dichromate $\ce{Cr2O7^{2-}}$. Chromate is yellow, dichromate is orange.

From the Wikipedia article on sodium chromate: sodium chromate is formed by reaction of dichromate with hydroxide:

$$\ce{Na2Cr2O7 + 2NaOH -> 2Na2CrO4 + H2O}$$

You may be making sodium dichromate, but the strongly basic conditions are converting it to sodium chromate.


Measure the concentration of lactose in dairy products I would like to measure the concentration of lactose in various dairy products : milk, cheese, yoghourt...

How is it done in labs ?
How can I do that using cheap equipments ?

Thanks

If you are looking for economical methods then you have the option of titration and spectrophotometry. See this paper by Folin THE DETERMINATION OF LACTOSE J. Biol. Chem. 1918, 33, 521-524. The titration is based on the reducing ability of lactose.

Another reliable titrimetric method can be seen in this paper by Hites, Determination of Lactose in Milk Products J. Anal. Chem. 1949, 21(8), 993-995.


Biochemical conversion of ethanol to acetic acid I have questions about the involvement of the electron transport chain (ETC) in the biochemical conversion of ethanol to acetic acid. In which of the reaction steps is the ETC employed?

The principal conversion of ethanol to acetaldehyde takes place in the cytosol of liver cells, no ETC involved directly. This is mediated by the enzyme alcohol dehydrogenase (ADH). However, the NADH produced as a consequence would indeed deliver its electrons to the ETS, an electron transport chain.

The acetaldehyde is then metabolized to acetate in the mitochondria. Again, the involvement of the ETS is in accepting electrons from the NADH that is also produced in this step.

Here's the details: https://pubs.niaaa.nih.gov/publications/arh294/245-255.htm


Disposal of Silver Halides dissolved in Ammonium Hydroxide I was reading the Lab Manual for Chemistry prepared by the NCERT . While going through the the test for Chloride ions (pg 88) I came across the following Hazard Warning :

The solution obtained after
dissolving AgCl/AgBr/AgI
precipitate in ammonium
hydroxide should be
acidified with 2M HNO3
and should be discarded
quickly to avoid serious
explosion.

How does it cause a serious explosion?

In the test for Cl- the chloride ions are converted into AgCl by the following reaction,
Cl- + AgNO3
→
NO3- + AgCl
which is then dissolved in ammonium hydroxide by the formation of diammine silver chloride by the following reaction:
AgCl + 2NH4OH
→
[Ag(NH3)2]Cl + 2H2O
On standing, the diammine silver complex subsequently breaks down to Ag3N.
The standard free energy of Ag3N is about +315 kJ/mol . Hence it is unstable and decomposes explosively to metallic silver and nitrogen gas .
2Ag3N → 6Ag + N2
So the diammine silver complex has to be destroyed after use so that it does not break down into Ag3N.
One way of destroying the complex is by adding nitric acid.
[Ag(NH3)2]Cl + 3HNO3
→
AgNO3 + 2NH4NO3 + H2O
So in all reactions involving the formation of diammine silver complex, the complex must be destroyed using acids after the purpose.
The complex is formed when AgCl, AgBr(sparingly soluble)and AgI(the complex will be formed only in trace amounts) are dissolved in NH4OH . The comlex is also formed in the well-known Tollen's test. Tollen's reagent is [Ag(NH3)2]OH. Ammoniacal Silver Nitrate is also used in Fontana–Masson Stain and also in silver mirroring(such as inside insulated vacuum flask).
The Wikipedia page on Silver Nitride
A reported accident


What's the difference between PBE and B3LYP methods? I can't find an answer to that question. I was told that in B3LYP, more variables implemented in the method are empirical, but I can't find anywhere if it's true, and I'm sure it's not the only difference.
I'm new to calculations, and it seems so hard and confusing, so please help me.

PBE

The PBE functional${}^{[1]}$ belongs to the class of generalized gradient approximation (GGA) functionals for the exchange-correlation energy $E_{\mathrm{xc}}$. Considering that the dependence $E_{\mathrm{xc}}[\rho]$ may be non-local, i.e. $E_{\mathrm{xc}}$ may depend on the density $\rho$ at a given point (locality), but also on $\rho$ nearby (non-locality) the assumption made by the local spin density approximation${}^{[2,3]}$ (LSDA) that $E_{\mathrm{xc}}[\rho]$ is strictly local leaves a lot room for improvement. But still LSDA is a good starting point since it is a rather simple efficient model that gives good results. The GGA functionals improve on LSDA in a way that's pretty much in the spirit of a Taylor series with LSDA as a starting point: When we are at a point, what happens further off depends not only on $\rho$ at that point, but also the gradient of $\rho$ at the point, etc. (much as in a Taylor series: if you want to go further you may need not only the gradient, but also the Laplacian, etc.). So, these functionals are typically constructed by adding gradient corrections to the LSDA functionals. But that is not a trivial thing to do and doing it "naively" leads to wrong results. Therefore, some GGA functionals are parametrized by fitting experimental data. But there are also some GGA functionals that achieve the inclusion of gradient correction without introducing experimentally fitted parameters which makes them valid for a wide range of systems. One example of such a parameter-free GGA functional is the one developped by Perdew, Burke and Ernzerhof (PBE). It is known for its general applicability and gives rather accurate results for a wide range of systems.

Side note: GGA functionals are frequently termed non-local
functionals in the literature. This is a somewhat misleading and actually sloppy terminology that should be avoided and since my above statements might be misread in that direction I want to make one thing perfectly clear: All GGA functionals are perfectly
local in the mathematical sense: the value of the functional at a point $\vec{r}$ depends only on information about the density $\rho(\vec{r})$, its gradient $\nabla\rho(\vec{r})$, and possibly other information at this very point and is absolutely independent of properties of $\rho(\vec{r}^{\, \prime})$  at points $\vec{r}^{\, \prime} \neq \vec{r}$. Calling these functionals 'non-local' is only motivated by the fact that these functionals go beyond the 'local' density approximation and of course the observation that knowledge of
the gradients is the first step towards accounting for the inhomogeneity of the real density.

B3LYP

The B3LYP approach${}^{[4,5,6]}$ belongs to the hybrid approximations for the exchange–hybrid correlation functional. The approximation is famous, because it gives very good results and, therefore, is extremely popular. The distinguishing feature of such hybrid approximations is that they mix in a certain amount of the exact Hartree–Fock exchange energy into the exchange and correlation obtained from other functionals. There is actually some justification for that: At the lower limit ($\lambda = 0$, where $\lambda$ is the coupling constant) of the so called coupling constant integration the exchange-correlation hole is equal to the exact exchange hole. This observation led Becke${}^{[7, 8]}$ to conclude that a fraction of exact exchange should be mixed with GGA exchange and correlation. The simplest such hybrid functional is

\begin{align}
E^{\mathrm{hyb}}_{\mathrm{xc}} = a E_{\mathrm{x}}^{\mathrm{exact}} + (1 − a) E^{\mathrm{GGA}}_{\mathrm{x}} + E^{\mathrm{GGA}}_{\mathrm{c}}
\end{align}

where the constant $a$ can be fitted empirically or estimated theoretically ${}^{[9, 10, 11]}$
as $a \approx \frac{1}{4}$ for molecules.
So far so good, but it smells like a witch's brew for the B3LYP exchange–correlation potential $E_{\mathrm{xc}}$: take the exchange–correlation energy from the LSDA
method, add a pinch (20%) of the difference between the Hartree–Fock exchange energy
$E^{\mathrm{KS}}_{\mathrm{x}}$ (well, in fact, this is Kohn–Sham exchange energy, because the Slater determinant wave function, used to calculate it, is the Kohn–Sham determinant, not the Hartree–Fock one) and the LSDA $E^{\mathrm{LSDA}}_{\mathrm{x}}$. Then, mix well 72% of Becke exchange potential${}^{[12]}$ $E^{\mathrm{B88}}_{\mathrm{x}}$ which includes the 1988 correction, then strew in 81% of the Lee–Yang–Parr correlation potential${}^{[13]}$ $E^{\mathrm{LYP}}_{\mathrm{c}}$. You will like this homeopathic magic potion most if you conclude by putting in 19% of the Vosko–Wilk–Nusair potential$^{[14]}$ $E^{\mathrm{VWN}}_{\mathrm{c}}$:

\begin{align}
  E^{\mathrm{B3LYP}}_{\mathrm{xc}} = E^{\mathrm{LSDA}}_{\mathrm{xc}} + 0.2 (E^{\mathrm{KS}}_{\mathrm{x}} - E^{\mathrm{LSDA}}_{\mathrm{x}}) + 0.72 E^{\mathrm{B88}}_{\mathrm{x}} + 0.81 E^{\mathrm{LYP}}_{\mathrm{c}} + 0.19 E^{\mathrm{VWN}}_{\mathrm{c}}
\end{align}

So, you can see B3LYP contains lots of empirical parameters but that's fine as long as you only want the results and don't want to know why this is working :)

References

${}^{1}$ J. P. Perdew, K. Burke, M. Ernzerhof, "Generalized Gradient Approximation
Made Simple", Phys. Rev. Lett. 1996, 77, 3865; J. P. Perdew, K. Burke,
M. Ernzerhof, "Erratum to Generalized Gradient Approximation Made Simple", Phys. Rev. Lett. 1997, 78, 1396 (E). (Link)

${}^{2}$ W. Kohn, L. J. Sham, Phys. Rev. 1965, 140, A1133. (Link)

${}^{3}$ U. von Barth, L. Hedin, "A local exchange-correlation potential for the spin polarized case", J. Phys. C: Solid State Phys. 1972, 5, 1629–1642. (Link)

${}^{4}$ A. D. Becke, "Density-functional thermochemistry. III. The role of exact exchange", J. Chem. Phys. 1993, 98, 5648-5652. (Link)

${}^{5}$ P. J. Stephens, F. J. Devlin, C. F. Chabalowski, M. J. Frisch, "Ab Initio Calculation of Vibrational Absorption and Circular Dichroism Spectra Using Density Functional Force Fields", J. Phys. Chem. 1994, 98, 11623–11627. (Link)

${}^{6}$ K. Kim, K. D. Jordan, "Comparison of Density Functional and MP2 Calculations on the Water Monomer and Dimer", J. Phys. Chem. 1994, 98, 10089–10094. (Link)

${}^{7}$ A. D. Becke, "A new mixing of Hartree–Fock and local density‐functional theories ", J. Chem. Phys., 1993, 98, 1372. (Link)

${}^{8}$ A. D. Becke, "Density‐functional thermochemistry. III. The role of exact exchange", J. Chem. Phys., 1993, 98, 5648. (Link)

${}^{9}$ J. P. Perdew, M. Ernzerhof, and K. Burke, "Rationale for mixing exact exchange with density functional approximations", J. Chem. Phys., 1996 105, 9982. (Link)

${}^{10}$ C. Adamo and V. Barone, "Toward reliable density functional methods without adjustable parameters: The PBE0 model", J. Chem. Phys., 1999 110, 6158. (Link)

${}^{11}$ M. Ernzerhof and G. E. Scuseria, "Assessment of the Perdew–Burke–Ernzerhof exchange-correlation functional", J. Chem. Phys., 1999, 110, 5029. (Link)

${}^{12}$ A. D. Becke. "Density-functional exchange-energy approximation with correct asymptotic behaviour", Phys. Rev. A 1988, 38, 3098-3100. (Link)

${}^{13}$ C. Lee, W. Yang, R. G. Parr, "Development of the Colle-Salvetti correlation-energy formula into a functional of the electron density", Phys. Rev. B, 1988, 37, 785-789. (Link)

$^{14}$ S. Vosko, L. Wilk, M. Nusair, "Accurate spin-dependent electron-liquid correlation energies for local spin density calculations: a critical analysis", Can. J. Phys. 1980, 58, 1200-1211. (Link)


How can I reduce Crystallization temperature of seed oil to be used in winter as biodiesel? How can I reduce Crystallisation temperature of seed oil to be used in winter as biodiesel?
The glycerol has been removed from the oil, and white spirit is used to reduce its viscosity.

Waxing and pour point are typical specs for fuel. How cold is your winter? If it is very cold, and your tank is above ground, there is little to be done. Insulate the tank (metal-backed rock wool) and heat it.  Use a more unsaturated feedstock with no saturated components. That lowers its pour point and removes waxing.  Did you dry your fuel?  Unsaturated fuels oxidize and deposit varnish.  "Biofuel" is a good substrate for microbial growth.

Enviro-whinerism, sustainability, social consciousness, are Luddite lies.  They assume thermodynamics is amenable to legislation and good feelings, and weather is always mild and cooperative.  None of that obtains in the real world.

http://en.wikipedia.org/wiki/Pour_point
http://www.infineum.com/Pages/pourpoint.aspx
http://www.aftonchemical.com/Solutions/LubricantComponents/Pages/PourPointDepressants.aspx
http://www.globalp.com/products/product.cfm?productID=32
Chemicals


Enthalpy of reaction between ammonia and oxygen I have to ask this here because I asked literally everybody I know and I still disagree with them.

  Using the information in the table below, calculate the heat of released from the following reaction in $\pu{kJ}$ per $\pu{g}$ of $\ce{NO}.$ (Mw. of $\ce{NO}$ is $\pu{30.01 g/mol})$

  $$\ce{4 NH3(g) + 5 O2(g) -> 6 H2O(g) + 4 NO(g)}$$

  $$
\begin{array}{lrrr}
\hline
  & \ce{NH3} & \ce{H2O} & \ce{NO} \\
\hline
H_\mathrm{f}^\circ~(\pu{kJ/mol}) & -46.1 & -241.8 & +90.3 \\
\hline
\end{array}
$$

  a) $-13.29$
  b) $-8.372$
  c) $-7.541$
  d) $-30.16$
  e) $-5.678$

I'm convinced the answer is c.

The enthalpy of this reaction for 4 moles of NO is -905.2, to find the enthalpy for 1 mole of NO we have to divide by 4 to get to -226.3.

Then, to find it per gram of NO we have to divide by 30.01, this gives us 7.541.

Am I right ? If not, what did I do that was wrong? Where's my mistake?

Please don't put this as HW help, I asked multiple websites and went through my working 5-6 times but I still couldn't point out where I went wrong.

The heat of reaction of a mixture is defined as the extent of reaction, $\xi$ times the enthaply of reaction, $\Delta H_{rxn}$,
$$Q_{rxn} =\xi\Delta H_{rxn} = \sum^n_{i=1}\nu_iH_{f,i}^{o} =\sum^{n,products}_{i=1}\|\nu_i\|H_{f,i}^{o} - \sum^{m,reactants}_{i=1}\|\nu_i\|H_{f,i}^{o} $$
For a reaction system,
$$\ce{4NH_3(g) + 5O_2(g) -> 6H_2O(g) + 4NO(g)}$$
In the case of the example you've provided, you are assuming this reaction is taking place at standard conditions. If we assume for a stiociometic amount of reactants, then the extend of reaction, $\xi$, should be equal to 1. Therefore,
$$Q_{rxn} = 1*\Delta H_{rxn} = 6\Delta H^o_{f,H_2O} + 4\Delta H^o_{f,NO} - 5\Delta H^o_{f,O_2} - 4\Delta H^o_{f,NH_3}$$
Since, $\Delta H^o_{f,O_2} = 0 \pu{kJ//mol}$, the equation reduces to,
$$\Delta H_{rxn} = 6\pu{mol}(-241.8 \pu{kJ//mol}) + 4\pu{mol}(90.3 \pu{kJ//mol}) - 4\pu{mol}(-46.1 \pu{kJ//mol}) $$
$$\Delta H_{rxn}= -905.2\pu{kJ}$$
Now, per gram of $\ce{NO}$, assuming the basis of 4 mol $\ce{NO}$ from the reaction equation,
$$\frac{Q_{rxn}}{m_{NO}} = \frac{\Delta H_{rxn}}{m_{NO}} = \frac{-905.2\pu{kJ}}{4_{mol, NO}(30.01\pu{g//mol})}=-7.541\pu{kJ//g_{,NO}}$$


Which species dissociate completely in ionic equations? I am tasked with writing a molecular equation, an overall ionic equation, and a net ionic equation for the reaction between an aqueous solution of phosphoric acid, $\ce{H3PO4{(aq)}}$ and an aqueous solution of sodium hydroxide. The products are sodium phosphate and water.

I have already figured out the balanced molecular equation of the following:

$$\ce{H3PO4 {(aq)} + 3 NaOH {(aq)} -> Na3PO4 {(aq)} + 3 H2O {(l)}}$$

However, I am uncertain how to write the proper ionic equations. I know that the ions are separated, but how do you format this properly when writing with multiple molecules. If this is based on the disassociation of the chemicals, how do you determine which ions disassociate?

I have arrived at:

$$\ce{3H+ {(aq)} + PO4^3- {(aq)} + 3 Na+ {(aq)} + 3 OH- -> 3 Na+ {(aq)} + 3 H2O
{(l)} + PO4^3- {(aq)}}$$

However, I still do not know why this is the best answer. Why does the $\ce{NaOH}$ show separated in the equation while the $\ce{PO4}$ does not? How would you know which ones separate by simply looking at the problem?

The second equation you’ve given appears to be the right overall ionic equation. Phosphate is a weak acid, $\ce{NaOH}$ is a strong base. We expect NaOH to dissociate at any pH, where the speciation of phosphate ($\ce{H3PO4}$, $\ce{H2PO4-}$, $\ce{HPO4^2-}$, and $\ce{PO4^3-}$) depends on pH.

A “net ionic equation” doesn’t include spectator ions. Spectator ions exist on both sides of the arrow and can be “canceled like” (think as in algebra). For any acid base reaction, it’s really easy to remember!

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

Lastly don’t forget the state: $\ce{OH {(aq)}-}$. I was taught to right them as subscripts, but I’ve seen then as regular text too: $\ce{OH- (aq)}$.


Why can't electrons be transferred using hydrogen ions? Recognizing how oxidation and reduction occurs is essential for analyzing and understanding biochemical reactions. Which of the following methods would NOT be employed in transferring electrons in biochemical reactions?
a) As hydrogen ions (correct)
b) Through direct combination with oxygen
c) Directly as electrons
d)As hydride ions

My problem is that, it doesn't say why hydrogen ions cannot transfer electrons, also I'm not sure what "hydrogen ions" exactly mean. What if you have a hydrogen ion with 2 or 3 electrons? Why can't that be used to transfer electrons?

Firstly, $\ce{H+}$ would be called a hydrogen ion, and $\ce{H-}$ would be called a hydride ion. It would take a lot of energy to get three electrons onto a hydrogen atom, and considering that we are talking about a biological system, this just won't happen.

$\ce{H+}$ would not be employed in the transfer of electrons because it is a lone proton, and has no electrons to give. Whatever electron the hydrogen atom originally had was left behind, and the proton now exists on its own. This, I believe, should explain your homework question.

To answer your question, $\ce{H-}$ ions are common reducing agents in non-biological organic reactions, in the forms of substances like $\ce{NaBH4}$ and $\ce{LiAlH4}$, but these are extremely unstable in aqueous solution.

Instead, cells uses NAD+ to facilitate the transfer of electrons from biomolecules, like those used in cellular respiration, in the form of $\ce{H-}$. During the transfer, one electron from the $\ce{H-}$ is transferred to a nitrogen, and the other electron along with the proton forms a bond to a carbon directly across the ring from the nitrogen, converting NAD+ to NADH, a reaction that is readily reversible.


Why does bromine increase the viscosity of olive oil? Does bromine bond to the carbon in unsaturated fatty acids and break the double bonds between oxygen and carbon? Does this increase the viscosity of olive oil?
Could someone give an elaborated explanation?

This question arose from a lab experiment. Viscosity goes up because less double bondings makes the fatty acids more “packable”. But why would bromine want to break the bonds?

Olive oil is a triglyceride, and the main fatty acid is oleic acid below.

Now the bromine does not add to the double bond between oxygen and carbon; instead the bromine reacts with carbon-carbon double bond and add itself to it. If you use e.g. coconut oil which has minimal unsaturated (meaning having carbon-carbon double bond) fatty acids (and assume you don't shine light onto it, heating it or otherwise forcing the reaction to start) there should theoretically be no reaction.

  But why would bromine want to break the bonds?

Simply put, bromine loves electron. The carbon-carbon double bond happens to be rich in electrons, so the bromine binds to it. While the carbon-oxygen double bond also has electrons, oxygen loves electron more than bromine and tends to keep the electron for itself, so bromine cannot add there.

  Does this increase the viscosity of olive oil?

The individual molecules that make up olive oil is held together by weak forces, that arise from van der Waals' forces, which is attraction due to small, unevenness in electron distribution (which leads to very small positive and negative charges on the molecules that subsequently attract other molecules).
This is overly simplified concept though so it's not entirely accurate.

Bromine atom is very large (hence bromine, which molecules consists of only two bromine atom, exists as a liquid in room temperature). The more the electron, the more likely there are some unevenness in electron distribution, and hence stronger attraction between molecules. Also, bromine loves electron so it tends to pull electron closer to itself, creating a small but rather permanent positive and negative charge on the molecule. This also strengthen the force between brominated oil molecules, hence the viscosity goes up.

  Viscosity goes up because less double bondings makes the fatty acids more “packable”

This is rather about melting point of the fat / oil, as when freezing, the molecules in liquid need to "pack" into a solid structure. In liquid however, molecules tends not wanting to maintain the "packing", or ordered structure. Adding bromine does increase the melting point but that's another story.


Does intramolecular hydrogen bonding occur in dicarboxylic acids? Consider ethanedioic acid:

Could it form intramolecular hydrogen bonds (between the C=O and HO)?

No, this would not happen the way your second drawing depicts because the hydrogens do not orient themselves that way. Intramolecular hydrogen bonding would be no more feasible in ethandioic acid than any other monocarboxylic acid. See picture below:

Source: http://commons.wikimedia.org/wiki/File:Oxalic-acid-3D-balls-B.png


Why does lactulose turn dark? Why does the standard 10gm/15ml lactulose solution turn dark when exposed to heat above 30 degrees Celsius?

This is probably a simple question. It is funny that the warnings/important information on the drug label of lactulose state that this effect is harmless, but then go on to warn not to use if it becomes too dark. How do I know if it is "too dark"? This is obviously very subjective and simply want to know why it becomes dark or "too dark" so it is scientifically understood why it is safe (or unsafe) to use when it is dark.

When milk is heat-processed, some of the lactose will isomerize to lactulose, a disaccharide.  Lactulose

can decompose by the same routes that most sugars decompose.  These are complicated processes with many products and pathways, so sugar decomposition is rarely, if ever, fully understood.

It can hydrolyze to its constituent monosaccharides fructose and galactose.
Lactulose or its released monosaccharides equilibrate with their ring-opened isomers and the alpha-hydroxyl can air oxidize to produce an alpha-dicarbonyl.  These alpha-dicarbonyls are prone to further oxidation to generate formic acid.  Over time, as the solution becomes more acidic from the formic acid build-up, further oxidations and eliminations (dehydration) can take place to produce a mixture of colored products.
In addition to degrading to smaller molecules, sugars (and the dehydration and oxidation products mentioned above) can also oligomerize to produce a variety of viscous, colored products.  Caramelization (see the "Chemistry" section in the link) is a term applied to this chemical polymerization and coloration process.

If you would like to read more about lactulose and its decomposition, take a look at this reference, especially section 7.2.1 starting on page 236.


Rate determining step for parallel reaction? I have always read that the rate determining step is the slowest step in a reaction (source: Wikipedia).

  In chemical kinetics, the overall rate of a reaction is often approximately determined by the slowest step, known as the rate-determining step (RDS) or rate-limiting step.

However on this website (source), it says that the rate determining step for a parallel reaction is the fastest step. I was quite confused about this statement.

For a parallel reaction $\ce{B <- A -> C}$, where $\ce{A -> B}$ has rate constant $k_1$ and $\ce{A -> C}$ has rate constant $k_2$, the equations (from solving the first-order linear differential equations for A, B, and C, give: $A = A_oe^{-(k_1+k_2)t}$,  $B = \frac{k_1*A_o}{k_1+k_2}*(1-e^{-(k_1+k_2)*t})$, and $C = \frac{k_2*A_o}{k_1+k_2}*(1-e^{-(k_1+k_2)*t})$.

But I do not know how to proceed from there.

Let's suppose that $\ce{A -> C}$ is the fast step in the parallel reaction scheme.  This implies that $k_2 \gg k_1$.  Thus, $k_1 + k_2 \approx k_2$, it is therefore (approximately) true to write the rate of consumption of $\ce{A}$ as $A_0 e^{-k_2t}$.  This is what is meant by the fast reaction "controlling" rate for parallel reactions.


When to run a blank using ICP-MS I am currently writing a paper for my Advanced Analytical Chemistry course, and the topic I am writing on is the analysis of Arsenic in drinking water. The EPA's literature describes how to perform this method but does not describe how many samples should be taken or when and where blanks should be used. The three blanks required for analysis are the rinse blank, calibration blank, and the laboratory reagent blank. My question is, if we assume we are analyzing twenty water samples using ICP-MS, when should each type of blank be run?

This terminology is specific to EPA and it is not universal, because there are so many legal issues associated with any environmental analysis. They have to be extremely careful at each and every step (it may cost someone millions because of a wrong analytical result). Hence the reason for various "blanks." The language of those methods is official and the steps are highly detailed. Hence these robust methods can be defended legally using the terminologies used in the methods.

While preparing a calibration curve by ICP or any other spectrochemical technique, there can be only one blank, which sets the baseline reading. One would run this blank and 5-6 standards. So which blank is this? By EPA nomenclature, this is the calibration blank, which is defined by them as "Calibration Blank – A blank solution containing all of the reagents in the same concentration as those used in the analytical sample preparation. This blank is not subject to the preparation method." Pretty much standard definition.

So what are the other plenty of blank solutions doing? They are used for quality control after the calibration to check the accuracy of the instrument, method, operator etc. One can test them right after performing the calibration process. See how specific the directions are:

a) Laboratory Reagent Blank (LRB) – An aliquot of ASTM Type I water that is treated exactly as a sample including exposure to all labware, equipment, solvents, reagents and internal standards that are used with other samples. The LRB is used to determine if method analytes or other interferences are present in the laboratory environment, the reagents or apparatus.

b) The rinse blank, as the name indicates for rinsing the lines. You should discuss how complex the sample introduction system is in ICPs. >90% sample is wasted and a small fraction goes to the plasma.

EPA defines it as well. The rinse blank should flush the system between solution changes for blanks, standards, and samples. Allow sufficient rinse time (~ 1 min) to remove traces of the previous sample. Solutions should aspirate for at least 30 seconds prior to the acquisition of data to establish equilibrium

Reference: this detailed report with terminologies

(ICP-MS) with - EPA: https://www3.epa.gov/ttnamti1/files/ambient/pb/EQL-0512-202.pdf


Why does bromine add to a double bond instead of a triple bond if both are present? According to my textbook, the addition of bromine to pent-1-en-4-yne at low temperature yields 4,5-dibromopent-1-yne:

Why does bromination of the double bond occur selectively?

I'm going to take "more reactive" as in the alkene undergoes bromination at a faster rate than the alkyne, which seems to imply that the activation energy of the transition state is greater.

A Kinetic Perspective

The formation of the bromonium ion involves a cyclic transition state where the the Br-Br bond is attacked by the alkene/alkyne pi-bond, and the pi-bond donates to the sigma* antibonding orbital to form a sigma C-Br bond. In doing so, both carbons in the alkene/alkyne will have their p-orbitals hybridized with the sp/sp2 orbital to form the sp2/sp3 orbital respectively. However, the sp2 bonding orbital will have more ring strain, as it naturally exists with bond angles of 120 degrees away from other sp2 orbitals, while the sp3 orbital is somewhat less strained with bond angles ~109.5 degrees apart. Thus the higher energy transition state will make the alkyne less reactive, and bromination occurs on the alkene first.

In addition, (as was pointed out here : https://www.masterorganicchemistry.com/2013/05/29/alkyne-addition-pathways-the-3-membered-ring-pathway/) the p orbitals on Cl conjugate to the pi-bond on the alkene transition state, resulting in a 4-electron anti-aromatic system! This further increases the instability of the transition state and results in a slower reaction.

TL;DR: There's more ring strain on the alkyne's transition state because the carbons are sp2 hybridized, and its transition state involves the formation of an anti-aromatic ring, so the reaction occurs slower than bromination on the alkene.


Do surfactants exist in a gas phase? I am wondering if there are surfactants that exist in a gas state that can be used to reduce the surface tension of a liquid.

I know its a broad question, so for example, are there any gases that could reduce the surface tension of water when it is exposed to it?

I am wondering if there are surfactants that exist in a gas state that can be used to reduce the surface tension of a liquid.

The most direct answer is "no".  Gases do not have surface tension and gas-phase molecules do not chemically affect the surface tension of liquids.

  I know its a broad question, so for example, are there any gases that could reduce the surface tension of water when it is exposed to it?

Gases could, however, dissolve in the liquid in question.  Then, they aren't really gases, but solutes in the liquid.  And solutes can affect surface tension.  You can see this earlier question on volatile surfactants for some examples of water-soluble volatile surfactants for some examples.  A paper entitled Surfactant in the gaseous phase was brought up by MAR in the comments and is another good example.


Setting up a water aspirator pump for vacuum distillation I have a question about the functioning of a water aspirator in a vacuum distillation setup.

I have read about how the water aspirator is hooked up to the water source and the vacuum is in turn connected to the vacuum adapter, and that there should be a splash guard or tubing on one end of the aspirator as well. I'm trying to figure out what to make of the highly suggested use of a water trap. Supposedly the water trap needs to be connected to the vacuum as well as the system.

What outlet is there left to connect to the aspirator, already being connected to the water source and apparatus, as well as the splash guard underneath? Also, if a water trap should be connected to the system, where should it be connected, is a flask with additional adapters required?

Since you want the distillate, use the following setup.

This way if the water pressure drops then the water is sucked into the trap flask, not the distillate.


Additives to allow wax to melt in the microwave? I'm working with microcrystalline wax that I would like to be able to microwave to heat up. I've found that wax doesn't seem to microwave, with the most common answer I've come across for why being that it doesn't contain any water.

Adding water doesn't seem like a valid solution in this instance, since even if I could add it to a wax mixture, it would eventually dry out anyway. Therefore, I'm looking for an additive that, once mixed in with the melted wax, will allow me to microwave it after that point. The mix ratio isn't all that important as long as it's kept below 30 percent or so, although this would depend on the additive.

In searching for a potential solution, I came across a concept called the "dielectric constant". Unless I'm misunderstanding, the higher this number the greater the capacity for heating in a microwave. Looking up numbers, this seems to make sense as the DK of paraffin (the closest material I could find to microcrystalline) is around 2, whereas water is around 80.

A potential material I found was titanium dioxide. I'm getting different numbers, but it seems to have a DK somewhere in the 100-300 range. Before going out and buying some, is it feasible that I could use this as a filler for my wax that would allow it to be microwaved?

If not, or if I'm on a completely wrong track with the dielectric constant concept, are there are other suggestions on how I could solve this problem?

Microwave heating works best on polar molecules, such as water and alcohols, as it then causes the molecules to rotate or "jiggle" violently. Waxes are nonpolar, so heat poorly, but if your application can use fatty acids in the mixture, such as palmitic or stearic acid, you might try mixing some into the microcrystalline wax. The dielectric constant is not significant in your application.

However, the longer the fatty acid molecule, the less effective it is in absorbing microwave radiation and converting it to heat, and on the other hand, the shorter molecules have strong odors and do not mix well with wax... do not use butyric acid for that reason. The suggestion of using an absorbent container, by @Li Zhi, seems to be more effective.


Understanding transition state theory Let's say we have standard chemical reaction given by
$$A-B + C \leftrightharpoons [A\cdot\cdot B\cdot\cdot C]^{\dagger} \rightarrow A +B-C$$
I want to estimate the pre-exponential factor of this reaction using Transition State Theory.
From Fundamentals of Chemical Reaction Engineering by Davis and Davis, I see that
$$\text{rate} = \nu _i C_{\dagger} = \nu _i \mathcal{K} C_{AB} C_{C}$$
Where $\mathcal{K} = C_{\dagger}/(C_{AB}C_C)$, and $C_i$ is the concentration of species $i$.
Then we redefine $\mathcal{K}$ as
$$\mathcal{K}_C = \frac{C_{\dagger}}{C_{AB}C_C} = \frac{Q'_{\dagger}}{Q'_{AB}Q'_C}$$ where $Q'_i$ is the partition function per unit volume.
We redefine energies to lowest ground state, so
$$\text{rate} = \nu _i C_{\dagger} = \nu _i \mathcal{K}e^{-E/k_BT}$$
We know that $Q = Q_tQ_rQ_{\nu}Q_{el}$, where $t$ is for translational, $r$ is for rotational, $\nu$ is for vibrational, $el$ is for electronic partition function.
We also know that $Q'_{t} = \frac{(2\pi mk_B T)^{3/2}}{h^3}, Q_{el} \approx 1$.
Everything good so far.
My question lies with the rotational and vibrational partition function.
$$Q_r = \frac{8\pi}{h^2 \sigma} (2\pi k_B T)^{3/2} (I_1I_2...)$$ where $\sigma$ is the rotational symmetry number. My question is, what is the rotational symmetry number? How do I find it? How do I find moment of inertia for molecules? More importantly, does a single atom like $C$ even have a rotational partition function?  How do I calculate $Q_{v}$ for $A-B$? and $C$? The formula says
$$Q_v = \prod_{s=1}^n (1-e^{-hv/k_BT})^{-1}$$
The end result says
$$r = \frac{k_B T}{h} \cdot \frac{Q'_{\dagger}}{Q'_{A}Q'_{B}} e^{-E/k_BT}C_{AB}C_C$$
The final question would be, how do I simply something like this? With the $\frac{k_B T}{h}$, what exactly got simplified?
Edit: I figured out I can get moments of inertia from Calculated moments of inertia for molecules are listed in the NIST Computational Chemistry Comparison and Benchmark Database: http://cccbdb.nist.gov/).

So rotational spectroscopy is something one can spend a lot of time talking about so I'll try to keep this brief. Although this isn't a complete answer, it was too long for a comment, but hopefully this sheds some light on the topic.
For rotational transitions, one of the selection rules tells us that in order for EMR to induce a transition in the rotational energy levels, a permanent dipole moment must exist in the molecule (i.e. a single atom won't be observed in rotational spectra, nor would a molecule such as $\ce{H_2}$).
The rotational symmetry number is a variable that is used to account for the amount of rotations one can perform on a molecule that would make it indistinguishable from the starting point (i.e. a 90 degree rotational of $\ce{PCl_5}$ along the principal axis can be performed three times before returning each atom to its original position in space, however, each rotation produces an identical picture of the molecule each time).
Moment of inertia can be calculated from the rotational constant, however, in practice, one would usually calculate the moment of inertia with computational software and then use that to approximate the rotational constants which are employed to predict the energies and allowed transitions for the rotations of a molecule.
This is a link to an appendix with the equations used for calculating the moment of inertia in molecules based on the geometry (e.g. symmetric rotor, linear)
Here is an article that further discusses symmetry numbers and partition functions.
This one talks about rotational partition functions of diatomic gases.
This article talks about transition moment integrals and selection rules.


Inductive effect comparison of benzylic and allyl group I have to compare the acidity of the above two carboxylic acids. I wish to know which one has a stronger -I effect, the benzylic group or allylic group?

Phenyl has a tendency to withdraw electrons.
see the link for the reasoning:
Inductive effect of phenyl ring
Also the allylic group would be donating via inductive effect, thus making the carboxylate ion unstable( after acid donates H+ ions)
But an sp2 carbon is more electronegative than an sp3 carbon, hence it will donate less.(more electronegative element tries to retain electrons)
So 1 has an electron withdrawing group,2 has an electron donating group which is not as good at donating as 3.This gives us the option  A

An important point to note here is that electron withdrawing groups increase H+ concentration. Think why?

Edit: Now that the Question has been edited, an edit for the solution is required.

Benzylic group has -I inductive effect as in the link. For the second compound in the question, with the allylic group, since less number of sp2 carbon are present(2 instead of 6). So we can infer that -I effect will be less. Infact it should show +I effect, especially since there is an sp2 Oxygen attached 2 carbons away, which will pull electrons as it is highly electronegative.


Single Displacement Reactions: A + AX =? This seems like a stupid question, but how would I find the product of a single displacement reaction like the following?

$$\ce{Pb + Pb(NO3)2 -> \ ?}$$

Normally, a single displacement reaction follows the path of $\ce{A + BX -> AX + B}$. However, what will the equation look like when A and B are the same element?

Nothing. All you would have is just a mixture of white powder and lead metal.

In general, lead (II) nitrate is synthesized by:

$$\ce{PbO (s) + 2 HNO3 (aq) -> Pb(NO3)2 (aq) + H2O (l)}$$

Lead (II) nitrate is soluble in water.

Alternatively, you could do:

$$\ce{PbCO3(s) + 2HNO3(aq) -> Pb(NO3)2(aq) + H2O(l) + CO2(g)} $$

Single displacement reactions are a kind of redox reaction. For single displacement reactions to occur:

1.) A and B must either be different metals OR

$$ \ce{2AgNO3(aq) + Zn(s) -> 2Ag(s) + Zn(NO3)2(aq)}$$

2.) A and B must be a halogen

$$\ce{Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)}$$

This is necessary as for the redox reaction to occur, you need two differing metals on the reactivity series in order for the single replacement to work, or alternatively, use compounds that contain halogens.

Even so, in combining two compounds with metals or halogens, if the element is less reactive than the element in the compound, the single replacement reaction will not proceed.

Some examples are:

$$\ce{Ag (s) + Cu(NO3)2 (aq) -> } \text{No reaction}$$

$$ \ce{I2 + 2KBr ->} \text{No reaction} $$


Order of reactivity of substrates towards Wurtz reaction In a book I refer, it is said that:

Both Ionic [carbanion] mechanism as well as free radical mechanism are widely accepted for Wurtz reaction.

When I read this statement, I was thinking, which of these mechanisms would be able to explain the order of reactivity of different substrates towards Wurtz reaction. My conclusion was that the rate of the reaction should follow the below mentioned order:

$1^\circ > 2^\circ > 3^\circ$ if it takes place through carbanion mechanism; and
$3^\circ > 2^\circ > 1^\circ$ if it takes place through free radical mechanism

where $1^\circ$ refers to primary alkyl halide, $2^\circ$ refers to secondary alkyl halide, and $3^\circ$ refers to tertiary alkyl halide.
So, my doubt is what is the correct consolidated order of reactivity of alkyl halides towards Wurtz reaction?
I know my question could be a possible duplicate of this.
However, I did not get a satisfactory explanation over there and hence have asked it as a new question.

The Wurtz reaction is an organic reaction, in which two molecules of an alkyl halide (e.g., $\ce{R-Cl}$) are coupled in the presence of $\ce{Na, Mg}$, or other reactive metals  to form an alkane (original classical Wurtz reaction used $\ce{Na}$ metal). The mechanism begins with a single electron transfer (SET) from sodium metal to the alkyl halide, which dissociates to form an alkyl radical ($\ce{R^.}$) in a similar fashion to the formation of a Grignard reagent and sodium halide salt (e.g., $\ce{NaCl}$):

$$\ce{R-X + Na^\circ -> R^. + NaX} \; \text{ where } \ce{X = Cl, Br,} \text{ or } \ce{I}$$

A second molecule of sodium undergoes another SET to the alkyl radical to form a nucleophilic carbanion (e.g., $\ce{R-}$):

$$\ce{R^. + Na^\circ -> R^- + Na+} $$

This alkylmetal intermediate can be isolated (e.g., alkyllithium) and has been isolated in a several cases. The carbanion then attacks another molecule of alkyl halide in a nucleophilic substitution reaction ($\mathrm{S_N2}$) to form the final coupled product (e.g., $\ce{R-R}$) and another molecule of sodium halide salt:

$$\ce{R^- + R-X -> R-R + X-} $$

Thus, the mechanism is believed to be undergoing in both free radical and ionic pathways in particular steps. The depicted diagram of the mechanism is illustrated in below:

In a review of the Wurtz reaction (Ref.1) revealed that the reaction is severely limited in scope as demonstrated by the following observations:

In the synthesis of noncyclic systems (dimerization or cross-coupling) reasonable yields require the use of primary alkyl halides, with iodides giving the best results. Secondary halides give very poor results;
Cross-coupling reactions (using a mixture of two different alkyl halides) give approximately statistical ratios of cross-coupled : dimeric products, due to the high reactivity of the organosodium intermediates involved. Thus, in the synthesis of 3-methyluntriacontane (4 ($\ce{R-R'}$); see above diagram), coupling of the iodide (1; $\ce{R-I}$) with 1-iodooctadecane (2; $\ce{R'-I}$) under standard Wurtz conditions gave a mixture of 3 ($\ce{R-R}$) from dimerization of 1, 4 ($\ce{R-R'}$) from the desired cross-coupling, and 5 ($\ce{R'-R'}$) from dimerization of 2, in the ratio $21:50:29$.

However, This classical reaction is of value in the preparation of cyclic systems, notably [2.2]phanes and bicyclobutanes (e.g., bicyclo[1.1.0]butane), as well as in the preparation of symmetric dimers. The Wurtz reaction provides one of the most successful approaches to these class of compounds, by ring-closure reactions. The reaction may proceed either intra- or inter-molecularly, with intramolecular couplings giving better yields (Ref.1):

The modest yields of [2.2]phane products are due to competing oligomerization/cyclization reactions, which is demonstrated by the product distribution obtained on treatment of the dibromide (6) with sodium/tetraphenylethane (TPE) in THF at $\pu{-80 ^\circ C}$ (See above diagram). A homologous series of products has been obtained under these conditions, the approximate yields of the metacyclophanes being obtained were given in the scheme.

In conclusion, the Wurtz reaction is seldom used because of side reactions and it is intolerant of a range of functional groups. It has limited use to the synthesis of symmetric alkanes and cycloalkanes. If two dissimilar alkyl halides are taken as reactants in cross coupling reactions, the resulting product is a mixture of alkanes that is often difficult to separate. In the case of (1,3), (1,4), (1,5), (1,6) dihalides (when used as a reactant), the reaction leads to formation of cyclic products instead of intended dimerization. The Wurtz reaction fails in case of tertiary halides. In addition, since the reaction involves free radical species in one part of the mechanism, a side reaction occurs to produce an alkene (specially with 1,2-dihalides). This side reaction becomes more significant when the alkyl halides are bulky at the halogen-attached carbon atom.

There is an organometalic compound in the mechanism at one stage, which undergoes a $\mathrm{S_N2}$ replacement of halide ion of available second $\ce{R-X}$ molecule. Whether these types of organometalic compounds are ionic (cabanionic nature of $\ce{R-}$) has been discussed and reviewed using quantum chemical calculations recently (Ref.2).

Thus, it is safe to say that the reaction is better for primary ($1^\circ$) halides, regardless of whether the mechanism is free radical oriented or carboanion oriented. Even more than 150 years after Charles Adolphe Wurtz introduced the reaction in 1855, the mechanism of the reaction is still up in air, regardless of involvement of both in the mechanism.

References:

David C. Billington, "Chapter 2.1: Coupling Reactions Between $\mathrm{sp^3}$ Carbon Centers (https://doi.org/10.1016/B978-0-08-052349-1.00068-8)," In Comprehensive Organic Synthesis, Volume 3: Carbon–Carbon $\sigma$-Bond Formation; Barry M. Trost, Ian Fleming, Editors-in-Chief; Elsevier Science Ltd.: New York, NY, 1991, pp. 413-434 (ISBN: 978-0-08-052349-1).
Christoph Lambert, Paul von Ragué Schleyer, “Are Polar Organometallic Compounds “Carbanions”? The Gegenion Effect on Structure and Energies of Alkali‐Metal Compounds,” Angew. Chem., Intl. Ed. 1994, 33(11), 1129-1140  (https://doi.org/10.1002/anie.199411291).


Correct equation for Ionic Conductivity (λ) in Solutions? We haven't started on Electrochemistry at school yet, but I did manage to find some time to read up on the topic. One thing I've noticed from when I started, is that different books and sites use different terms for the same symbols (For example: One book says Molar conductivity is represented by $\Lambda$, but another says it's Molar Conductance, and another book makes use of both Molar Conductance and Molar Conductivity, albeit inconsistently).
Now, my school textbook, as well as another book, Physical Chemistry by Wallwork and Grant employ the following equation for ionic conductivity:
$$\lambda=F\mu$$
Where $F$ is the Faraday constant and $\mu$ is the ionic mobility of the ion under consideration.
But I saw another equation on Wikipedia

$$\lambda=zF\mu$$
Where $z$ is the ionic charge of the ion being considered.

So my question is, which of these equations are right?

Short answer: $\lambda = z \mu F$ is correct. The link  provided by @Loong confirms this (page 73 of the document). To add to that I have provided you with a short derivation.

Prerequisite reading (so to speak) for definitions of a few terms/notation would be my answers to this question, and this question

Let me establish an important distinction (or definition) which can (possibly) be a source of confusion. The two relevant equations from my previous answer  to the first question are:

  $$\Lambda _{m}=\Lambda _{m}^{0}-K{\sqrt  {c}}$$

and,

  $$ \Lambda_{m}^{0}= \nu_{+}\lambda_{+}^{0}+\nu _{-}\lambda _{-}^{0}$$

Before I begin my discussion, I would like to note that "$\Lambda_m$" (uppercase ) is the molar conductivity of the solution, and "$\lambda$" is the molar conductivity of the ion (ionic conductivity). The second equation has a "special name", that I did not use in my previous answer-"Kohlrausch's law of independent migration of ions".

In general for dilute solutions of strong electrolytes, molar conductivity can be decomposed into contributions from the different ions in solution (this is the spirit behind the previously quoted equation):

$$ \Lambda_m^0 = \Sigma \nu_i \lambda^0_i$$

Now, let's attempt to relate the ionic conductivity to the ionic mobility (it clearly makes sense to do so). So I am assuming we have a dilute solution of completely disassociated strong electrolyte with molar concentration $c$. Each formula unit of this electrolyte gives rise to $\nu_+$ cations of charge $z_+e$ and $\nu_-$ anions of charge $z_-e$.

The concentration of each type of ion is therefore, $\nu c$ ($\nu$ could either mean $\nu_+$, or $\nu_-$ to avoid unnecessary notational complexity).

the number density of each type of ions is:

$$\text{No. density} = \nu c N_\mathrm{A}$$

Now picture a small window of area $A$ in your solution. The number of ions that can pass through said window in a time interval $\Delta t$ is equal to the number of ions within the distance $s \Delta t$ (where $s$ is the drift speed of the ions)

$s \Delta t A$, happens to define a volume element and the number of ions in this interval is given by: $ V \times \text{no. density}$, i.e

$$ s \Delta t A \nu c N_\mathrm{A}$$

Thus, the flux of ions is

$$J_\mathrm{(ions)} = \frac{s \Delta t A \nu c N_\mathrm{A}}{ \Delta t A} = s\nu c N_\mathrm{A}$$

remember flux always has dimensions of "something" per area per time

Flux of charge is $$J_\mathrm{(charge)} = ze s\nu c N_A = z s \nu c F $$

again, I am omitting the +/- signs for the sake of clarity. Also, note $F := e \times N_\mathrm{A}$

Now, drift speed is $s = \mu E$, so the flux of charge (henceforth, simply J) is

$$J = z \mu \nu c F E$$

The current is charge flux times area, so

$$I = (z \mu \nu c F E)A$$

Additionally, $E = \frac{\Delta \phi}{l}$ (i.e the potential gradient)

$$I = (z \mu \nu c FA) \frac{\Delta \phi}{l} \tag{I}$$

Using Ohm's Law (i.e $\Delta \phi = IR$}, one obtains

$$I = \frac{\Delta \phi}{R} = G \Delta \phi = \frac{\kappa A \Delta \phi}{l} \tag{II}$$

Comparing the (I) and (II) one is lead to $$ \kappa = z \mu \nu cF$$

This is the ionic conductivity. We need to divide by concentration of ions to the get the molar quantity. The concentration of ions however is not $c$, but rather $\nu c$ and our final result is

$$\lambda = z \mu F$$


Are compounds containing active methylene groups positive for the haloform test? Are compounds containing active methylene groups (e.g. 1,3-diketones) positive for the haloform test? Support your answer with a suitable explanation.

I was stuck on a certain question wherein a diketone was formed and I had to tell whether or not it shows the haloform test. It is clearly mentioned in my class notes that active methylene compounds don't show the haloform test; but in this problem the solution says that the diketone shows positive haloform just because it has an active methylene group. Please help!

Here's a possible mechanism that I came up with. Don't know if its correct. Sorry that it's handwritten; I don't know how to type chemistry.

As orthocresol already commented, apart from some nitpicks, the general direction is correct.

The only issue really is the addition of bromine, which does not occur via a [2+2]-cyclo-addition (2nd from below in scheme).

Ordinarily you would expect addition the addition of bromine to a π-bond to occur first via a polarisation complex and then via a bromonium ion (bottom of scheme).[1] In the case of the enolate, the π-bond is already readily polarised, and sufficiently asymmetric that a direct addition will occur.[2] The following depicts the HOMO -1 calculated at the DF-BP86/def2-SVP level of theory, the HOMO is an (delocalised) in-plane lone pair of the carbonyl-oxygen atoms in the C2v conformation (1st row). In the Cs conformation the π-bond is the HOMO, it is also the low-energy conformer (2nd row). (On mobile the rows might differ.)

Please note that the following scheme is just a proposed mechanism, and as such probably a reasonable simplification; the actual one might depend on more factors like solvent, temperature, pressure, etc., and might have more, or even less steps, other intermediates (ion pairs), hydrogen abstraction might be via water/base bridges, etc. For educational purposes I think this is sufficient though.

See for example Aromaticity-Bromonium Ion, Mechanism for bromoetherification of olefinic alcohol, Why does cyclopropane react with bromine?, What is trans-addition? and more...
Thanks to orthocresol pointing it out in the comments.

  Enolate reaction with bromine (as opposed to ordinary alkene) is traditionally taught to not go via a bromonium ion (img), presumably because the electronic character of the two doubly bonded carbons is very different. But I have not consulted any literature, this is just based off undergrad teaching + textbooks, so...

Since I was unlucky to find any specific literature on the topic (I did not search very thoroughly), I have tried optimising a bromonium analogue at the DF-BP86/def2-SVP level of theory. I have been unable to obtain a stationary point (no local minimum and no transition state). Which is not unexpected given the HOMO/ HOMO -1 is strongly polarised.


Comparing basicity of imidazole and 2-imidazoline I recently sat for an exam in which I was asked to compare the basicity of the above compounds. I thought imidazole would be more basic because it is aromatic but according to the answer key 2-imidazoline is more basic. The reason provided is that the lone pair on imidazole is a part of aromatic system and hence it's not basic.

Shouldn't the other lone pair which is on nitrogen double-bonded to carbon be basic? After all it's not involved in resonance due to being out of plane of the pi-system. What is the real reason behind higher basicity of 2-imidazoline (if it is really more basic)?

I was unable to find experimental $\mathrm{p}K_\mathrm{a}$ values of imidazoline. However, the tools one can use to calculate and thereby predict $\mathrm{p}K_\mathrm{a}$ values are pretty good nowadays as can be seen comparing the predicted and experimental values for imidazole.

$$\textbf{Table 1: }\text{predicted and experimental }\mathrm{p}K_\mathrm{a}\text{ values of imidazole and imidazoline}$$
$$\begin{array}{lcccc}\hline
\text{compound} & & \mathrm{p}K_\mathrm{a,1} & \mathrm{p}K_\mathrm{a,2} & \text{ref}\\ \hline
\text{imidazole} & \text{(e)} & \phantom{0}7.13 & 12.70 & [1]\\
& \text{(p)} & \phantom{0}6.97 & 13.40 & [2]\\
\text{imidazoline} & \text{(p)} & 10.17 & \text{N/A} & [2]\\ \hline\end{array}$$

Note: $\mathrm{p}K_\mathrm{a,1}$ refers to the $\mathrm{p}K_\mathrm{a}$ values of the protonated heterocycle. The higher this value, the higher the basicity of the compound in question.

Obviously, the answer set’s statement that imidazoline is more basic is correct; the question remains why that is the case.

Both compounds obviously have the same protonated form, namely it is the imino nitrogen that bears a proton and thus a formal positive charge. Both of these protonated forms are $C_\mathrm{2v}$ symmetric. At first sight, the electronics of the two are pretty much equal. Yet obviously, imidazoline’s nitrogen must experience a greater electron density for some reason.

The key difference is the size of the π system which spans five atoms in imidazole but only three in imidazoline. At first sight, it seems like imidazole should be more basic since the positive charge can be delocalised onto five atoms compared to only three in imidazoline. However, this explanation fails to acknowledge that all π electrons are delocalised across more atoms in imidazole’s case.

With the π system being larger and spanning more atoms, we expect less electron density to be concentrated on each of the two nitrogens. On the other hand, the much smaller π system of imidazoline means that a greater electron density can be found on both nitrogen atoms at any time.

Further effects, such as the greater electronegativity of $\mathrm{sp^2}$ carbon atoms or the $+I$ effect of $\ce{C-H}$ bonds should be neglegible.

References:

[1]: The Evans $\mathrm{p}K_\mathrm{a}$ table

[2]: Calculated by ChemAxon’s Chemicalize service.


Are all corrosives either acids or bases? Are all corrosives either acid or a base? From what I could remember from school, strong acids and bases can be corrosive, but can substances be corrosive from reasons other than that? I tried my best to check for the answer with my limited knowledge.

If it is not just acids and bases what is it then, how can we categorize them, if possible in simple terms?

Wikipedia lists many corrosive compounds.  While a lot of them function as strong acids or bases, there are also these (not an exhaustive list):

Strong oxidizers including concentrated hydrogen peroxide
Fluorides (they say "fluoride ion", so salts as well as the acid are meant)
Organic compounds that can act as alkylating agents such as methyl sulfate

The moral of the story: Hazards can't be pigeonholed into specific types of chemicals.  Best to find and read the safety data sheet with any chemical.


Can alloys or ceramics have higher density than elemental solids? Are there any examples of alloys or ceramics that have higher density than their densest elemental constituent?  If so, do they have unusual or difficult synthesis processes?  If not, is it provable that this is not possible, or demonstrably unlikely.

Since homogeneous metals are essentially crystals in their solid phase it seems plausible that, in theory, there are elements that could fit within the crystal structure without altering it to increase their density.

This would be particularly interesting for metals that are used in applications primarily for their density: e.g., lead, tungsten, and uranium.

Yes, for both alloys and ceramics it is possible have a density that is higher than their densest elemental constituent. This is because every lattice structure of the denser elemental constituent contains interstitial void spaces, or holes, typically tetrahedral holes (surrounded by 4 atoms) or octahedral holes (surrounded by 6 atoms). These holes exist despite the crystal having an optimal packing structure, thus a maximum density for the pure solid.

The densest elemental solids are typically made up of larger atoms toward the bottom of the periodic table, that have larger atomic radii due to electron-electron repulsion, but which have have much higher atomic mass due to their nuclei. This produces larger interstitial voids that can fit smaller atoms without significantly disturbing the lattice constants of the larger, denser component. An increase in mass without concomitant increase in volume will produce a denser composite.

Specific examples are difficult to find, as densities are not always reported. One common material which is denser than the corresponding "densest elemental constituent" is steel. Steel (maximum density: 8.05 g·cm⁻³) is an alloy of iron (density 7.874 g·cm⁻³) and carbon (diamond density: 3.5 g·cm⁻³), although occasionally some chromium (density: 7.19 g·cm⁻³) and nickel (density: 7.81 g·cm⁻³) may also be included.

Another possible, although less clear-cut example is the alloy heavymet, which is used in ATLAS. It is composed of nickel (9.9 atom %), copper (4.0 atom %), and tungsten (86.1 atom %). Heavymet has a density of 19.3 g·cm⁻³, which despite the less dense diluents, may be slightly larger than tungsten's density which is 19.25 g·cm⁻³.


How useful is the assumption that water doesn't auto-ionize in the following? Now in the first row of the iCe-table, $[\ce{H3O+}]$ is said to be 0. But water autoionizes to some extent, so it shouldn't really be 0. I accept that, because the pH of this solution is 2.5, the autoionization effect is negligible but are there instances where this can't be ignored (e.g. where the pH is close to 7?)? Academic, industrial (or hobby) examples would all be fine.

First, while it is useful to know approximate methods of solving aqueous solution equilibrium problems, you have to realize that in a professional context, these methods are not used other than for back-of-the-envelope calculations. There exist software that perform that kind of calculations without introducing approximations, and they are widely used. So, to answer the question in your title: such approximations are useful for simple calculations, and for those only.

Now, let’s try to evaluate the effect of neglecting auto-ionization of water on your problem.
You can write the full system of equations describing the aqueous equilibrium:

conservation of mass: $[\ce{HA}] + [\ce{A-}]=c_\ce{A}$, where $c_\ce{A}$ is the analytical concentration in chlorobenzoic acid (0.1 mol/L).
electroneutrality: $[\ce{H+}]=[\ce{A-}]+[\ce{HO-}]$
water autoionization equilibrium: $[\ce{H+}][\ce{HO-}]=K_\text{e}$
acid-base equilibrium: $\displaystyle K_\text{a}=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$

You can solve this system of equations, as you have four equations and four unknowns ($K_\text{a}$, $[\ce{A-}]$, $[\ce{HA}]$ and $[\ce{HO-}]$). The answer gives $\mathrm{p}K_\text{a} = 9.19$ (given the precision given in experimental measurements).

You don’t say what figure you obtain, but I can take the same system above and simulate “neglecting auto-ionization” by setting $K_\text{e}=0$. Doing this, I obtain $\mathrm{p}K_\text{a} = 9.18$, which means the error involved in your approximation is 0.01 on the $\mathrm{p}K_\text{a}$. So, this is indeed a good approximation.


What is the difference between an angular and linear skeletal formula? When drawing a skeletal formula, what is the difference between an angular version and a linear version?

I was asked to draw the Z isomer of Resveratrol:

For which I drew:

However the markscheme states that:

  skeletal structure must be correct and angular not linear

I haven't come across the difference between the two before and can't find anything on google to suggest one.  Their drawing of the correct answer is equivalent to mine, however I am concerned that I may have drawn the linear version as my benzene rings are in a line and theirs are not. What am I missing here?

Their version:

Those are $sp^2$ carbons and should have a triangular planar 120° geometry. The angles you've drawn are clearly not 120° .

I've not exactly come across the terminology before, but I believe this is the distinctions:

In a "linear" formula, you disregard the actual bond angles and just try to keep everything in a line.
In an "angular" formula, you draw everything according to the bond angles.

Not sure of this, though.


Does SN2 favor strong nucleophile and SN1 favor weak nucleophile? Does E2 favor a strong base, and E1 favor a weak base? I understand if you have a protic solvent, it will stabilize the strong base (to form weak acid) or the strong nucleophile. The protons would react with them. To have an SN2 or an E2, so you need an aprotic solvent. I also know that in general, a tertiary substrate will favor SN1/E1 reactions, because a tertiary carbocation is more stable. A primary substrate will SN2/E2 reactions, because of less stereo-hinderance.

But why does SN2 favor strong nucleophile and SN1 favor weak nucleophile from the first place? Besides, does E2 favor a strong base, and E1 favor a weak base? I also don't get the reason.

That's not the right view.

The $S_{N}2$ mechanism does not favor strong nucleophiles. That statement doesn't make sense at a high level. Mechanisms can't favor anything.

The key here is to look at what controls reactivity in $S_{N}2$ and $S_{N}1$ reactions.

The former is concerted where the bond breaking and bond forming take place at the same time. Therefore, the strength of the nucleophile is important as is the propensity of the leaving group to leave.

In the latter mechanism, there are two steps, bond breaking and bond forming, where bond breaking is the slower and more difficult step. Since the nucleophile is not involved in that first step, the strength of the nucleophile simply doesn't matter.

Similar considerations apply when looking at elimination reactions.


Representing woody biomass concentration for kinetic reactions Woody biomass is composed of cellulose, hemicellulose, and lignin components. The following diagram provides representative formulas for each component:

Biomass can be represented by it's density, ultimate analysis, and composition. For example, beech wood can be represented by its density, $\rho = 700~\mathrm{kg/m^3}$, its ultimate analysis by weight percent such as $\ce{C}=49.05\%$, $\ce{H}=5.83\%$, $\ce{O}=45\%$, $\ce{N}=0.12\%$, and its composition by weight percent such as $\text{cellulose}=48\%$, $\text{hemicellulose}=28\%$, and $\text{lignin}=24\%$.

The following diagram represents a kinetic scheme for cellulose, CELL, which can be represented by the formula $\ce{C6H10O5}$:

The reaction rates for each pathway are first-order Arrhenius equations where $A$ is the pre-factor in $\mathrm{s^{-1}}$, $E$ is the activation energy in $\mathrm{kcal/kmol}$, $T$ is the temperature, and $\mathcal{R}$ is the universal gas constant:

$$
K = A\cdot \exp{\left(\frac{-E}{\mathcal{R}T}\right)}
$$

To apply the cellulose kinetic reaction to beech wood, I used the density of the wood, $700~\mathrm{kg/m^3}$ to calculate the initial concentration of cellulose as 48% of the wood density, $336~\mathrm{kg/m^3}$.  Using the initial mass concentration of cellulose, the reaction rates at a certain temperature can be applied to that concentration. This approach will give the mass concentration for each group (G1, G4, LVG). Notice that each group represents various chemical species on a mole basis.

Since the individual chemical species are represented on a mole basis, it seems like there are two approaches to solve this problem:

Mass Basis Approach

start with an initial mass concentration of the biomass
divide that mass conc. into cellulose, hemicellulose, and lignin components based on the weight % composition
apply the reaction rates on a mass basis for each component to determine the mass concentration of each group
finally use the mass concentration of the group to calculate the individual chemical species (as mass or moles?) in that group

Mole Basis Approach

represent the biomass as moles of C, H, O, N using the ultimate analysis
divide the biomass into cellulose, hemicellulose, and lignin components
apply the reaction rates on a mole basis to determine the chemical species
atomic balances for C, H, O, N should be checked at each step

How do I attempt to solve this problem on a mole basis?
Or is it best to treat it initially on a mass basis then convert to moles to solve for the individual chemical species?

First step is finding the molar masses of each compound. Since cellulose, hemicellulose and lignin are collections of similar molecules, I picked the following as a crude estimate:

cellulose: $\ce{C6H10O5}$ with $M_c = 162~\mathrm{g/mol}$
hemicellulose (L-Arabinose): $\ce{C5H10O5}$ with $M_h = 150~\mathrm{g/mol}$
lignin: $\ce{C31H34O11}$ with $M_l = 582~\mathrm{g/mol}$

Using a density of $700~\mathrm{g/dm^3}$ you will find the relative masses of the three compounds to be:

$0.48 \cdot 700~\mathrm{g/dm^3} = 336~\mathrm{g/dm^3}$ (cellulose)
$0.28 \cdot 700~\mathrm{g/dm^3} = 196~\mathrm{g/dm^3}$ (hemicellulose)
$0.24 \cdot 700~\mathrm{g/dm^3} = 168~\mathrm{g/dm^3}$ (lignin)

With $n = m/M$ we now find the number of moles of each compound in the sample ($1~\mathrm{dm^3}$):

$2.074~\mathrm{mol/dm^3}$ cellulose
$1.307~\mathrm{mol/dm^3}$ hemicellulose
$0.289~\mathrm{mol/dm^3}$ lignin

Keep in mind that this is a very rough estimate. You'll probably get a better result by taking the average of all species of hemicellulose, etc. in your sample and cross-checking it to make sure it fits the ratios of the elements (weight percent).


When was oil discovered to be of biological origin? Is there an article, book, some account of the discovery that crude oil came from decayed biological material?

If you look at a gas or liquid chromatogram of crude oil, you see a remarkable preponderance of (unbranched) even-numbered carbon compounds (re acetyl CoA homologation).  You also see traces of polycyclics that are net chiral - terpenoids, terpenes, pristane, phytane, cholestane, chlorins, and porphyrins.  Fischer-Tropsch gives a very different distribution and no net chirality.

Mikhailo V. Lomonosov, "Slovo o reshdenii metallov ot tryaseniya zemli"
Proceedings of the Imperial Academy of Sciences, St. Petersburg, 1757.

(Mikhail Tsvet was born in 1872)


Is that correct to say that osmosis happens with water only? Is that correct to say that osmosis happens with water only (based on the definition below) or it can happen with other liquids as well?

I've read the following definition in a book called "BRS - physiology" (6th edition, p.5):

  "Osmosis is the flow of water across a semipermeable membrane from a
  solution with low solute concentration to a solution with high solute
  concentration."

No, it is not correct because it happens with other liquids (solvents) also.

Taking Wikipedia's definition:

  Osmosis is the spontaneous net movement of solvent molecules through a
  semi-permeable membrane into a region of higher solute concentration,
  in the direction that tends to equalize the solute concentrations on
  the two sides.

Only as extra information: osmosis occurs until chemical potentials of every substance on each side of the membrane are equal. As the substance cannot move through the membrane, the liquid (solvent) do.

According to the definition, the book isn't correct, because it says with water only, but take into account that in biology the solvent is always water. So, as an internal definition of biology (or physiology) maybe is not wrong.


Understanding elementary rate laws from a probability stand point I am learning about chemical kinetics and dynamics and as I understand for a general chemical reaction
$$\ce{aA + bB -> cC + dD} $$ whose reaction rate, r, can be described by an elementary rate law can be written as follows: $$r= k[\ce{A]^a[\ce{B}]^b} $$

What I do not seem to conceptually understand is if you look at a chemical reaction from a collision theory standpoint, would it not be true that as one reactant that is in a higher stoichiometric ratio relative to the other, the probabilty that the reactants collide (assuming any collision gives a reaction) would be inversely proportional to the stoichiometry of the reactant. For example in the following:
$$\ce {A + 2B -> C} $$

You basically need 2 mols of B for every mol of A for a successful reaction to produce C, so if the probability that A successfully collides with a mol of B is $\propto$ to $1/z$, then the probability that A successfully collides with TWO mols of B  $\propto 1/z^2$ which is lower, so why does a reaction rate, which is $\propto$ to the probability of a collision, with an elementary rate law $\propto$ to a concentration raised to an exponent?

Let's divide the reaction volume into V little boxes and randomly throw ${\rm N}_A$ molecules of A into it and ${\rm N}_B$ molecules of B.  The probability $p_A$ that a given box has a molecule of A in it is ${\rm N}_A$/V, in the limit that $V >> {\rm N}_A$, i.e. an ideal dilute solution, and likewise the probability $p_B$ that it has a molecule of B in it is ${\rm N}_B/V$.  The probability it has two molecules of B in it is ${p_B}^2$, again in the limit that $V >> {\rm N}_B$ and ${\rm N}_B >> 1$.

Consequently, the probability all three atoms are close to one another is $({\rm N}_A/V)({\rm N}_B/V)^2$, which is proportional to the terms in your rate law.  (This isn't a derivation, by the way, but only a rationalization I hope is appealing to intuition.  I've swept all kinds of detail and qualification under the rug.)

Where your intuition may be leading you astray is in the assumption that squaring the molarity of B makes the RHS of your rate law bigger, and that doesn't match up with your intuition that it should be smaller if it involves a less likely event (a trimolecular collision).  But that is not the case.  Squaring numbers only makes them bigger if they are greater than 1, below 1 it makes them smaller.  And of course the prefactor will be different, too.  What the squaring really does is change the dependence of the rate law on changes in B concentration. It says the rate law is much more sensitive to changes in B concentration than A concentration.  And, given that you need two B molecules to collide with each A molecules, that should be intuitively compelling.


How to apply Le Chatelier's principle? The following reaction is at equilibrium $$\ce{4HCl(aq) + O2(g) <=>2H2O(l) + 2Cl2(g)}\quad \Delta H>0$$Which one of the following changes will result in an increase in the concentration of $\ce{Cl2(g)}$

  adding some $\ce{H2O}$
  decreasing the temperature
  adding a catalyst
  decreasing the pressure
  removing some $\ce{HCl}$

Decreasing the pressure (4) is the answer but I don't understand why. I though decreasing pressure shifts the equilibrium to the side with the larger number of moles which is the left?

Remember that $\ce{HCl}$ (aq) and $\ce{H2O}$(l) are liquids while the oxygen and chlorine are gases . As the number of moles of gases on the left  hand side (one mole of oxygen) is smaller than that of the right hand side (two moles of chlorine), decreasing the pressure makes the equilibrium position move to the right hand side, therefore the concentration of chlorine increases.


Degradation of anode in electrolysis In my electrolysis experiment, I used a 9 volt battery, 1:100 mL baking soda:distilled water, and stainless steel electrodes. I noticed that the anode degraded faster than the cathode. Out of curiosity, is there a phenomena to explain this?

Gosh! You would think that electrolyzing NaHCO3 with stainless electrodes would be a clean reaction to produce H2 and O2.

Perhaps OH. is powerful enough to attach to some atoms at the surface and make it more insulating. So, significant degradation could occur at the positive electrode because of oxidation.

Perhaps lower voltage and currents would be more gentle with the electrodes. And perhaps a more resistant grade of stainless would show less degradation (316 is more resistant than 304, and knife blades are typically 400 series, harder but less corrosion-resistant).

At the cathode, hydrogen would be produced in a nascent (atomic) state, which can reduce the oxide coating on stainless steel, making it active. This activity would not result in corrosion while the electrolysis is going on - it is being cathodically protected. This electrode would be a good conductor even as its protective oxide coat is reduced, but the anode will have its oxide coat thickened till it becomes somewhat insulated - or degraded.

A similar effect can be seen with aluminum.


Hydrogen atom energy level The energy level of a hydrogen atom in its ground state is $-13.6\ \mathrm{eV}$.
How is this value calculated and how can we calculate the same for different orbits?

From the Bohr theory:
$$\large E=-\frac{Z^2e^4m}{8\epsilon_0^2n^2h^2}=-13.6\times\frac{Z^2}{n^2}\ \mathrm{eV}$$
where:

$$\begin{array}{c|c|}
& Z&\text{atomic number}&-\\
& e&\text{charge of electron}&1.60 \times 10^{-19}\ \mathrm{C}\\
& m&\text{mass of electron}&9.10 \times 10^{-31}\ \mathrm{kg}\\
& \epsilon_0&\text{electric permittivity}& 8.85 \times 10^{−12}\ \mathrm{F/m}\\
& n&\text{principal quantum number}&-\\
& h&\text{Planck's constant}&6.62 \times 10^{-34}\ \mathrm{m^2\ kg/s}\\
\end{array}$$

You can put values of $Z$ and $n$ for any shell of a hydrogenic species like $\ce{H}$, $\ce{He+}$, $\ce{Li^2+}$, $\ce{Be^3+}$, $\ce{B^4+}$…


Why does the oxidizing strength of the chlorates decrease with increasing oxidation number? Here is a table of the standard reduction potentials of the chlorates from wikipedia:

Intuitively, one would expect that the more positive the oxidation state, the more oxidizing that compound would be. However, as we see, the standard reduction potentials become more negative as the oxidation state of the $\rm Cl$ atom becomes more positive. So, why does the oxidizing strength of the chlorates decrease with increasing oxidation number?

The chlorate system is very complex. I doubt that the open-circuit potentials can be actually measured in solution, but rather are calculated from heats of formation. If the term "oxidizing strength" is used to compare systems, it should be rigorously defined; in this case, it seems that "oxidizing strength" is defined as "open-circuit voltage". But this could be a poor definition - even woefully poor.

In particular, the comparison of the open-circuit voltages does not include the potential current derivable from the reaction of a mole of each oxychloride. The output energy of the electrolytic cell (if you could just test each reaction, one at a time) would be the output voltage times the number of electrons transferred. By that standard, perchlorate will consume 7 times as many hydrogens as hypochlorite, and thus give 7 times as much current, with only a slightly lower voltage.

A comparison of the heats of reaction (just for the acidic equations) proves to be much more dramatic:

2 moles hypochlorous acid (total heat of formation = -56 kcal (CRC Handbook) produces   2 moles of H2O (total heat of formation = -136 kcal)

2 moles chlorous acid (-28 kcal) --> 4 moles H2O (-272 kcal)

2 moles chloric acid (-46 kcal) --> 6 moles H2O (-362 kcal)

and 2 moles perchloric acid (-62 kcal) --> 8 moles H2O (-544 kcal).

The heats of formation of the chlorine acids are similar, but the heat of the reactions is much higher for the more highly oxidized acids.

A mental picture using water: imagine 4 dams each 200 feet high and 200 feet wide. The lake behind the first is filled with water 163 feet deep and is 2 miles long. The second lake is 164 feet deep and 6 miles long. The third is only 147 feet deep, but 10 miles long. The fourth is 142 feet deep but 14 miles long. A pressure gauge at the dam would indicate the height (similar to the open circuit potentials above), but if the dam is used to provide power, the fourth dam/lake will outshine the other three.

Assuming that all the electrons don't react simultaneously, the open circuit potentials are a measure of the activation energy of the first step of the overall redox reaction: in order to get zero current, you just have to stop the first step. This activation energy will be sensitive to many factors, and we occasionally oversimplify, ignoring the succeeding steps by using non-quantitative words like "strength". The total oxidation strength will be a combination of all these steps, and the open circuit potential may not reflect this combination.


Why does benzyne dimerise and not trimerise? Why does benzyne prefer to undergo dimerisation to stabilise itself, even though the product formed (biphenylene) is anti-aromatic?

Why does it not undergo trimerisation through the following pathway, resulting in an aromatic and presumably more stable compound (triphenylene)?

As per Wikipedia,

  Biphenylene is quite stable both chemically and thermally, and behaves in many ways like a traditional polycyclic aromatic hydrocarbon. However, both the spectral and chemical properties show the influence of the central [4n] ring, leading to considerable interest in the system in terms of its degree of lessened aromaticity.

This implies that biphenylene is rather stable, despite its anti-aromaticity.

As for triphenylene, I'm not sure, and it doesn't say so anywhere, but perhaps the free energy of formation of triphenylene is much more, as it requires more molecules to be put together, leading to the reaction being non-spontaneous. However, benzyne does trimerize as well, forming triphenylene.

So benzyne can actually undergo both dimerization and trimerization, forming two different, but stable products.

As requested, a more primary source is available here. This site cites other sources, whose links I have provided in the quotation.

These lines say more or less the same thing:

  Several recent publications (2 3 4 5 6) concluded that biphenylene possesses signiﬁcant strain energy of the order of 50–60 kcal/mol. These numbers are at least confusing since molecules that possess such tremendous strain can hardly exist. In the meantime, biphenylene is a stable molecule that has been synthesized a long time ago while its chemical properties show typical aromatic behaviour.


Dilution of whisky – the molecular perspective Not a chemist. I'm trying to understand this abstract. In particular:

  We found that guaiacol is preferentially associated with ethanol, and, therefore, primarily found at the liquid-air interface in mixtures that contain up to 45 vol-% of ethanol.

So guaiacol is amphipathic which means that is a largish molecule in which one end is hydrophilic and the other hydrophobic (or lilophilic). At least that's my understanding from Wikipedia. So, how did they arrive at the "therefore primarily found at the liquid-air interface" part? At the surface the molecule attaches one end to the alcohol and one end to the air?

Sorry for any naive interpretation of chemistry on my part.

Edit:
Just run across this article discussing the matter.

What the paper says is that in that range of concentrations, that the excess of alcohol at the air-liquid interface is enough to extract some of the guiacol from the bulk solution. At higher bulk concentrations of ethanol, there's no particular benefit for the guiacol to separate to the interface (there's enough ethanol in the bulk to keep the guiacol satisfied, so to speak). I forget whether the paper determined if there was a concentration (proof) below which the guiacol did not segregate, but if there was, it would be because the slight increase in alcohol concentration at the interface wasn't attractive enough. In other words the guiacol only is preferentially concentrated at the interface when the difference between the concentrations of alcohol at the interface to the concentration in the bulk makes it worth-while for it. (Worthwhile here means the difference in energy is large enough). Here's an analogy (My family would now roll their eyes, my analogies are lousy). A little kid is deciding whether to get up from his video game and go trick-or-treating. There's no candy in his house, and maybe he'll get a half dozen or a dozen pieces if he hits 20 or 30 houses/doors. Nah, he says, not worth it. But what if he's in a rich neighborhood? If he hits 10 doors, he's sure to get at least 2 dozen, maybe 3 or 4 dozen pieces of candy. Well, if his home hasn't much candy then he'll probably go, but if his home is chock-full of candy (rich neighborhood, remember?) then he may decide it's not worth it either. The choice he makes depends not just on the difference in the absolute amount of candy he will end up with, but also depends on how much he values an extra couple of pieces. It's non-linear, or as Liberals might say "context sensitive". (added: just to be clear, going trick-or-treating is analogous to concentrating at the air-liquid interface, not going is analogous to staying in the bulk phase.)


Any other Colligative properties? Are there any other colligative properties other than lowering of vapor pressure, osmotic pressure, boiling point elevation, and freezing point depression? I was wondering whether surface tension of an aqueous solution is colligative or not, but I mostly think it should be. Can anyone help?

Edit: I am pretty aware of what a colligative property is. It's a property's of the solution that doesn't depend on the nature of solute, rather only concentration of the solute, and the nature of solvent.

Firstly I would like to expand a bit on what a colligative property is, in order to bring more understanding and also be able to apply this to a named physical property.
Colligative properties depend mainly on the number of particles in a solution.
The values of the colligative properties are approximately the same for equal
concentrations of different constituents in solution regardless of the species or chemical nature of the constituents. Examples are osmotic pressure, vapor pressure lowering, freezing point depression, and boiling point elevation (as you have noted).
Furthermore, physical properties of substances are not just limited to the colligative nature. In fact a number of properties have been defined:

additive (depend on the total contribution of the atoms in the molecule or on the sum of the properties of the constituents in a solution e.g molecular weight).

constitutive (depend on the arrangement and partly on the number and kind of atoms within a molecule e.g interfacial characteristics).

other thermodynamic physical properties are known (extensive, intensive)

Having gained this background, it is now evident that surface tension (a force that pulls the molecules of the interface together) is dependant on the arrangement and also on the nature of atoms or molecules , therefore is more of a constitutive property rather than a colligative property.
I have included a screenshot to better visualise the phenomenon:

Hope this helps.