Equilibrium constant of a reaction whose order of forward and reverse reaction is not same Suppose, a reaction is like,
$$\ce{aA + bB <=> cC + dD}$$
and that A, B, C, D all are gas.
Now it is known that:
$$K_c = \dfrac{[A]^a[B]^b}{[C]^c[D]^d}$$
Now if the forward reaction is second order and the reverse reaction is in first-order somehow then, how will this change.
I read some articles online but it was not actually clear that if all the reactants or products are in the same state like gas or aqueous then how $K_c$ or $K_p$ (for gas) will change depending on the rate of reaction or if they don't depend on that totally.
It will be helpful if one example is provided.

I'll try and answer your question, but a "full" answer would take a book.
Given the reaction:
$$\ce{aA + bB <=> cC + dD}$$
Then assuming an elementary reaction in the gaseous state the concentration equilibrium constant always has products over reactants and will be:
$$K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$$
The above equation relies on two very specific assumptions:

The reaction is an elementary reaction which often isn't true and the actual coefficients must be determined experimentally since the reaction occurs in steps.
That concentrations can be used instead of activities.

Formally using activities explains part of your question immediately. For instance if A, B and C are gases and D is a liquid or solid, then the activity of D would be unity by definition.
Now breaking down the equilibrium expression into two rate equations we have:

$r_\mathrm{f}$ - Forward reaction rate
$r_\mathrm{r}$ - Reverse reaction rate
$\ce{a^*, b^*, c^*}$ and $\ce{d^*}$ = experimentally determined coefficients which may or may not be equal to the stoichiometric coefficients
$k_\mathrm{f}$ and $k_\mathrm{r}$ are constants for the forward and reverse reactions respectively.

Now using concentrations instead of activities:
\begin{align}
r_\mathrm{f} &= k_\mathrm{f} \ce{[A]^{a^*}[B]^{b^*}}\\
r_\mathrm{r} &= k_\mathrm{r} \ce{[C]^{c^*}[D]^{d^*}}
\end{align}
and at equilibrium by definition:
\begin{align}
r_\mathrm{f} &= r_\mathrm{r}\\
k_\mathrm{f} \ce{[A]^{a^*}[B]^{b^*}} &= k_\mathrm{r} \ce{[C]^{c^*}[D]^{d^*}}
\end{align}
so:
$$K_c = \frac{k_\mathrm{f}}{k_\mathrm{r}}= \frac{\ce{[C]^{c^*}[D]^{d^*}}}{\ce{[A]^{a^*}[B]^{b^*}}}$$
To determine the coefficients the experimentalist can manipulate the experiment to simply the kinetic expression. For example let $\ce{[A] \gg [B]}$ then $\ce{[A]^{a^*}}$ is essentially a constant.
Does this answer your question?


Amount of Cl- needed to precipitate AgCl given Ksp and [Ag+] Why is $0.25 \ \mathrm{mL}$ the correct answer in this problem?

I used the Ksp to find the molarity of Chlorine to be $3.47 \times 10^{-2} \ \mathrm{M}$.

Then I used the formula

$$\text{Molarity}_{\text{initial}} \times \text{Volume}_{\text{initial}} = \text{Molarity}_{\text{final}} \times \text{Volume}_{\text{final}}$$

to find the volume of Chlorine, which turned out to be $0.21 \ \mathrm{mL}$  for me. However, my answer does not match the correct answer.

What am I doing wrong?

First, solve for [$\ce{Cl-}$] using the expression for K$_{\mathrm sp}$ and the given molarity for $\ce{Ag+}$:

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

which rearranges to

$$[\ce{Cl-}] = {1.8\times 10^{-10}\over 7.2\times 10^{-5}} = 2.5\times 10^{-6}\, \mathrm{M}$$

Note that the units for K$_{\mathrm sp}$ are actually $\mathrm{M}^{2}$: I have omitted that from the math above but dimensional analysis tells us that the answer for $[\ce{Cl-}]$ is in units of molarity ($\mathrm{M}$).

Because we have 100 $\mathrm{mL}$ of $\ce{Ag+}$ solution, divide the $[\ce{Cl-}]$ molarity by 10 to get the actual number of moles of $\ce{Cl-}$ needed (here, I multiply by 0.1 to achieve the same result):

$$\mathrm{mol}\,\ce{Cl-} = (2.5\times 10^{-6}\,\mathrm{mol}\cdot\mathrm{L}^{-1}) \times 0.1\,\mathrm{L} = 2.5\times 10^{-7}$$

We are given a solution that is 0.001 $\mathrm{M}$ $\ce{Cl-}$, so we can get the volume needed in liters:

$${2.5\times 10^{-7}\,\mathrm{mol}\,\ce{Cl-}\over 10^{-3}\,\mathrm{M}\,\ce{Cl-}} = 2.5\times 10^{-4}\,\mathrm{L}$$

Finally, convert the above quantity to $\mathrm{mL}$:
$$\left({1000\,\mathrm{mL}\over\mathrm{L}}\right)\cdot\,2.5\times 10^{-4}\,\mathrm{L} = 0.25\,\mathrm{mL}$$


What is the reaction of pentane-2,4-dione with chlorine in presence of NaOH? We know that in haloform reaction alpha acidic hydrogen are substituted by halogen atom. But in the above case we find two acidic hydrogens in between the two keto groups (due to -R effect) which are more acidic than the hydrogens on the terminal methyl groups.

So what should be the products of this reaction?

I assume that there is sufficient base and chlorine and that the conditions are vigorous enough to get beyond dichloride 4, which, being a non-enolic β-diketone at C3, is labile toward Haller-Bauer-type cleavage. Acetylacetone 1 is highly enolic and readily deprotonated. The first two chlorinations lead to dichloride 4. Cleavage of dichloride 4 with base gives enolate 6 which can be protonated by water, or to save space, chlorinated to provide the trichloroketone 8, which is an intermediate in the haloform reaction of acetone. Cleavage with base leads to chloroform (10) and a second equivalent of sodium acetate (7).


Epoxide Opening with Sodium Hydride I understand that in general, when in acidic media, the nucleophile tends to attack the more substituted side of the epoxide due to electronic reasons. And when in basic media, the nucleophile attacks the less substituted side of the epoxide due to steric reasons.

What if the hydride anion attacks an epoxide? I'd imagine the hydride anion to be quite small, and able to attack both the more substituted and less substituted sides of an asymmetric epoxide. What are the experimental results?

Also I have my doubts about hydride anion even being able to open an epoxide; I am told by Clayden that hydride anion never acts as a nucleophile because its 1s orbital is too small.

in acidic media, the nucleophile tends to attack the more substituted side of the epoxide due to electronic reasons

This is an $\ce{S_{N}1}$ like reaction. First we would protonate the epoxide oxygen.  If we examine the case of an unsymmetrical epoxide and draw the resonance structures for this situation (see below) we would expect the secondary carbocation structure to count more towards the actual structure than the primary carbocation structure.  Therefore, we would expect a nucleophile to react preferentially at the tertiary epoxide carbon, quite selectively.

  in basic media, the nucleophile attacks the less substituted side of
  the epoxide due to steric reasons.

This would be a traditional $\ce{S_{N}2}$ reaction.  Attack at the least substituted (sterically most accessible) carbon would be expected.

  What if the hydride anion attacks an epoxide? I'd imagine the hydride
  anion to be quite small, and able to attack both the more substituted
  and less substituted sides of an asymmetric epoxide. What are the
  experimental results?

The hydride anion is not free, but complexed, so it has reasonable steric bulk - enough for stereoselective reactions to occur.   Further, the reaction between LAH and epoxides is quite facile, no doubt accelerated by the release of steric strain in the 3-membered epoxide ring system. A traditional $\ce{S_{N}2}$ reaction takes place.  In the figure below, the alcohol shown is formed in ca. 90% yield.  Because of the "backside" attack nature of the $\ce{S_{N}2}$ reaction axial attack is strongly preferred in cyclic systems.


Intermolecular Forces for Ionic Compound? So in class we have learned London dispersion, dipole-dipole, ion-dipole and hydrogen bonding for intermolecular forces. Our teacher always uses covalent molecules as examples. So I was wondering which intermolecular forces ionic bonds have, if so, how are they formed?

Ions in ionic compounds are held together by electrostatic attractions, i.e. the idea that "opposite charges attract".

The strength of an electrostatic attraction is given by Coulomb's law:

$$F = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}$$

where $q_1$ and $q_2$ are the charges on the two ions and $r$ is the distance between them. In a completely ionic bond, $q_1$ and $q_2$ are multiples of the elementary charge, $e = 1.602 \times 10^{-19} \text{ C}$; however, no bond is completely ionic.


Reactions of differents halohydrins with tert-butoxide I have two compounds (1R,2S)-3,3-dimethyl-2-bromocyclohexanol and (1R,2R)-3,3-dimethyl-2-bromocyclohexanol, both of them are treated with $\ce{KO^tBu}$ (a strong but bulky base) to form two different isomers of $\ce{C8H14O}$ which is not an alcohol.

I though that (1R,2R)-3,3-dimethyl-2-bromocyclohexanol is going to form an epoxide due the configuration of the halogen and the hydroxyl (although I don't know if it is required an specific configuration) and (1R,2S)-3,3-dimethyl-2-bromocyclohexanol is going to have a E2 reaction due steric hindrance but that will end with an alcohol anyway.

Chair conformation RR-1 (shown as its alkoxides) is in equilibrium with conformation RR-2, which is the unfavorable partner owing to the strong 1,3-diaxial Me/alkoxide interaction. It is this conformation that leads to the epoxide 1. This route is possible if the rate of formation is much greater than the rate constant for RR-1 --> aldehyde 2. RR-1 has concentration on its side and anti-periplanar alignment of bonds for ring contraction. I prefer ring contraction but this does not preclude epoxide formation.
The RS isomer has two chair conformations, RS-1 and RS-2.  Again, RS-2 is an unfavorable conformation (vide supra) but has proper bond alignment for ring contraction. Conformation RS-1 has the necessary bond alignment for E2 elimination to form cyclohexanone 4 via the enol 3.
A priori, it is not easy to predict with certainty what the product distribution will be. The epoxide can only arise from the RR isomer while aldehyde 2 can arise from either bromohydrin. Organic chemistry is an experimental science.


Hygroscopic behaviour of DMSO - how bad is it? For a fluid flow experiment I am using DMSO (dimethylsulfoxide), because of its low volatility, reasonably high surface tension, low viscosity and relative safety.

In the MSDS of DMSO I find that the liquid is hygroscopic, meaning it adsorbs moisture from the surrounding environment. Unfortunately, I cannot find anywhere how much water will be absorbed (mass%) or how quickly this will happen, is it saturated in minutes, hours, days?

Therefore, my question is twofold: (i) how much water will DMSO at maximum absorb and what is the rate at which this absorption occurs? (ii) with the absorbed water, how badly do the physical properties (viscosity, surface tension, density) of the (now) mixture change?

Being curious about the hygroscopicity I have tried the experiment using pure, dried DMSO in our lab where the temperature was $21^\circ \text{C}$ and the relative humidity about $60\%$.

Have a look at the two images below. This is a DMSO droplet originally with $V=50\; \mu\text{L}$. You can clearly see that the volume of the droplet has increased substantially in 20 minutes from the different height.

Using a crude approximation, assuming that both shapes are caps of a sphere, I calculated that the change in height from 100 px, to 112 px is equivalent to a volume increase of $15\%$, which is much too big for an accurate flow experiment.


How can BN be paramagnetic? I read something recently that said "BN has been observed in the gas phase, is paramagnetic, and has a vibrational frequency lower than N2. These properties can be explained by the molecular orbital diagram of BN".

However, when I drew such a diagram, I failed to see how it described any of these properties, particularly paramagnetism, because there were no lone pairs of electrons. Any ideas?

This is the general MO diagram you need to fill with the valence electrons of BN

Boron has 3 valence electrons, and nitrogen has 5 valence electrons, this makes 8 electrons.

You have to start filling the orbitals from those with lowest energy to those with higher energy.

So, 2 electrons on $\sigma_\mathrm{2s}$, two electrons on $\sigma_\mathrm{2s}^*$, two electrons on $\sigma_\mathrm{2p}$. You have now 2 electrons left, and two orbitals of the same energy.

In this case, you need to follow Hund's rule, which states that  if two or more orbitals of equal energy are available, electrons will occupy them singly before filling them in pairs. So you will have to put one electron on the $\pi_x$ orbital and another one on the $\pi_y$ orbital.
So you end up with 2 umpaired electrons, and paramagnetism of the molecule is explained.


When does ipso attack occur in electrophilic aromatic substitution? My textbook doesn't mention the conditions required for an ipso attack to occur. It just describes some examples where groups like $\ce{-C(CH3)}$ and $\ce{-SO3H}$ are replaced by $\ce{Br}$ using reagents like $\ce{Br2}$/$\ce{AlCl3}$.
Can someone describe the required conditions in detail?

In ipso substitution first step is usually attack of the nucleophilic species at the ipso carbon of the aromatic
ring (the carbon bearing the leaving group in this case).

Scheme 1

The first step is followed by elimination of the leaving group and regeneration of the aromatic ring.

Scheme 2

This mechanism greatly resembles the tetrahedral mechanism . The attacking species forms a bond
with the substrate, giving an intermediate, such as 1, and then the leaving group
departs. We refer to this mechanism as the SNAr mechanism.The reaction between 2,4,6-trinitrophenetole and methoxide ion involves ipso substitution.Intermediates of this type are stable salts, called Meisenheimer or Meisenheimer–Jackson salts,7 and many more have been isolated .

Scheme 3

In the reaction of dinitro compound  with piperidine,same ipso substitution operates.

Scheme 4

In the reaction between picryl chloride and hydroxyl ions same ipso substitution operates.

Ref

1.MARCH’S ADVANCED ORGANIC CHEMISTRY REACTIONS, MECHANISMS,AND STRUCTURE
SIXTH EDITION  ,Michael B. Smith ,Jerry March

2.Organic Chemistry, Volume 1, 6/E By Finar


Are acid chlorides really acidic? If I test a pure solution of an acid chloride with a pH-meter, will I get a reading of pH 7?

When I add water to it, I know $\ce{HCl}$ gas will form, but this gas will dissolve in water to give an acidic solution.

My main confusion is that we call ethanoyl chloride an "acid" chloride, but is its pure solution acidic?

An acid chloride is a covalent molecule.  It has no pH.  If you put a pH electrode into a reactive acid chloride you will dehydrate and acylate its hydrated glass surface ($\ce{#Si-OH}$) and damage its future response when properly used.  pH electrodes are stored sitting in buffer or saline solution.

An acid chloride is an oxo-acid whose acidic $\ce{-OH}$ has been replaced with $\ce{-Cl}$.  Hydrolysis restores the oxo-acid, displacement makes derivatives. Hydrochloric acid is not an acid chloride.


Gibbs free energy ΔG=ΔG°+RTln(Kp) different expressions for ΔG ΔG=ΔG°+RTln(Kp)

If I understand it correctly, the ΔG in this equation gives the "distance" in Gibbs free energy of a chemical system at a particular moment in a reaction (encoded by the Kp) to the gibbs free energy of the chemical system at equilibrium.
So when we write this in symbols
for a general equilibrium reaction A + B <=> C + D:

ΣG(products at eq) + ΣG(reactants at eq) - (ΣG(products at k) + ΣG(reactants at k)) with k denoting a random point with certain concentrations in the reaction and eq denoting the equilibrium point.

However, ΔG is also ΣG(products at k) - ΣG(reactants at k).

So how are these two expressions for ΔG related? Basically how would you prove that

ΣG(products at eq) + ΣG(reactants at eq) - (ΣG(products at k) + ΣG(reactants at k))
= ΣG(products at k) - ΣG(reactants at k)

$K_p$ is normally reserved for equilibrium thus only apples when $\Delta G=0$ and conventionally $Q$ is used otherwise. In a plot of free energy $G$ vs extent of reaction $\zeta$, the slope is $(\partial G/\partial \zeta)_{T,p} =\Delta G$, and equilibrium occurs when the slope is zero then $\Delta G^\mathrm{o} =-RT\ln(K_p)$.  (The extent of reaction is zero with only reactants present, and 1 when one mole of reactants has become product.)

Thus to speak of 'distance' does not mean the energy difference from equilibrium as $\Delta G$ is the slope not the value of the energy $G$ vs $\zeta$.

In the sums you ask about, always calculate products - reactants. The only experimental quantities you have are the concentrations or pressures so it would be possible to calculate $\Delta G$, using your first equation, if the values are known at equilibrium and at some other point. This seems to be what you ask about.


Why is N-acetyl alanine methyl amide considered a dipeptide? I'm going through an Amber Molecular Dynamics tutorial here and am a bit confused about why in step 8, we create a sequence as:

foo = seq { ACE ALA NME }

and this is called a dipeptide. I think there's only one amino acid, alanine, isn't there?

The nomenclature here is based on counting the number of backbone peptide bonds rather than the number of aminoacids. This does not apply the usual naming convention.
If you look at the wikipedia page on "peptide (amide) bond", it states

A peptide bond is an amide type of covalent chemical bond linking two consecutive alpha-amino acids from C1 (carbon number one) of one alpha-amino acid and N2 (nitrogen number two) of another along a peptide or protein chain.
It can also be called an eupeptide bond to separate it from an isopeptide bond, a different type of amide bond between two amino acids.

Furthermore,

When two amino acids form a dipeptide through a peptide bond it is type of condensation reaction.
[...] The two joined amino acids are called a dipeptide.

Note that you can have aminoacids connected through their side chains. Glutathione (GSH) comes to mind as a good example of this:

It contains three aminoacids (Gly, Cys and Glu). Would you call this a "tripeptide"?
The wikipedia does:

It is a tripeptide with a gamma peptide linkage between the carboxyl group of the glutamate side chain and the amine group of cysteine, and the carboxyl group of cysteine is attached by normal peptide linkage to a glycine.

What about the bonds in GSH? The wikipedia goes on:

[...] the tripeptide glutathione is synthesized in two steps from free amino acids, by two enzymes: glutamate–cysteine ligase (forms an isopeptide bond [emphasis mine], which is not a peptide bond) and glutathione synthetase (forms a peptide bond).

So, here they call GSH a tripeptide based on the number of aminoacids, not the number or even type of "peptide bond".
Hmm, pretty confusing. Perhaps then it is best to go to the original literature on the subject, linked in the wikipedia:

"Nomenclature and Symbolism for Amino Acids and Peptides. Recommendations 1983". European Journal of Biochemistry. 138 (1): 9–37. 1984.

"Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)". Pure and Applied Chemistry. 66 (5): 1077–1184.

According to reference 1:

Part 1, Section C : PEPTIDE NOMENCLATURE
3AA-11. DEFINITION OF PEPTIDES
A peptide is any compound produced by amide formation between a carboxyl group of one amino acid and an amino group
of another. The amide bonds in peptides may be called peptide bonds. The word peptide usually applies to compounds whose
amide bonds are formed between C-1 of one amino acid and N-2 of another (sometimes called eupeptide bonds), but it includes
compounds with residues linked by other amide bonds (sometimes called isopeptide bonds). Peptides with fewer than about 10-
20 residues may also be called oligopeptides ; those with more, polypeptides. Polypeptides of specific sequence of more than about
50 residues are usually known as proteins, but authors differ greatly on where they start using this term.

There is some degree of ambiguity regarding the numerical prefixes to the term "peptide", which are not discussed explicitly. But the above article provides a solid guideline about what you should call a peptide, namely two or more aminoacids linked through amide bonds.
So what about AcAlaNHMe (the capped alanine peptide in question)? One reason it is called a dipeptide (and not a peptide or tripeptide) is to emphasize that it contains two peptide (or amide) plains with intervening $\phi$ and $\psi$ dihedral angles which confer the potential conformational flexibility of a regular tripeptide. This is discussed for instance in this article.


Is there a metal for which its gas is necessarily a plasma? Related to this question:
Gaseous metals?

The top answer clarifies that even with metals, the gaseous state is distinct from a plasma.

However, my question is, is this actually true in all cases?
Or is there an element and a pressure for which heating it to a gas will necessarily also be hot enough for it to immediately become a plasma?
Or is there a situation in which heating can bring about the transition from solid or liquid straight to plasma while skipping gas?

In principle, all that is required is for a substance to evaporate from a liquid into a plasma directly is for its ionisation energy to be significantly smaller than the average kinetic energy of its particles at its "boiling point". Or in other words, as a very rough estimate this is true when $\mathrm{k_BT_{boiling}≳EI}$.

Going further down in the periodic table, elements tend to have higher boiling points and lower ionisation energies. After reaching the superheavy element region (Z>100) things get much more complicated due to strong relativistic effects, but the elements from Rf to Hs (Z=104 to 108) are expected to have the highest boiling points at ambient pressure of any known chemical substance, possibly well past $\mathrm{6000\ K}$. However, their calculated ionisation energies of $\mathrm{6-8\ eV}$ suggest they would only ionise considerably at temperatures 10 times higher still ($\mathrm{1\ eV/k_B= 11605\ K}$), meaning there is still a long way to go.

As others have mentioned, increasing the pressure is a reliable way to push boiling/sublimation points higher, up to where they can match and surpass the energy regime required for ionisation. However, this would require absurdly high pressures; for example, the boiling point of tungsten at $\mathrm{1000\ bar}$ is still a mere $\mathrm{9000-10000\ K}$ (ref). Thus, a very crude extrapolation would suggest the direct conversion of liquid (or more likely solid) tungsten to a plasma would happen around $\mathrm{90000\ K}$ under a pressure of $\mathrm{20\ Mbar}$ (20 million atmospheres!).


Comparing SN2 reaction rate for phenacyl chloride and methyl chloride KI in acetone undergoes $\mathrm{S_N2}$ reaction with (1) and (2). Compare the rates of the reactions.

Like many kinds of these questions there are many factors involved. One is the steric effect, and, since steric effects plays a central role in inversion of reactant and there is a very bulky group attached to (2), I think (1) should be more reactive. Also, (2) will serve as a better $\mathrm{S_N1}$ substrate as the carbocation generated will be resonance stabilised.
And so using these two reasons I thought (1) is more reactive than (2), but, according to the answer, (2) is almost 500 times more reactive than (1), and I am not able to see the reason for this mammoth difference.

First, let me point out that a rate difference of 500 is really not that large.  There are solvolysis reactions with relative rate differences on the order of $\mathrm{10^{10}}$ or greater (1).

The transition state for an $\mathrm{S_{N}2}$ reaction involves hypercoordinate (or hypervalent) bonding.  The transition state is, more or less, a trigonal bipyramid structure and the bonding at the central carbon atom ($\ce{Nu-C-X}$) involves a 3 center-4 electron bond (see the above link and links therein for more detail on this concept). With 5 electron pairs being shared by this central carbon, anything that removes electron density from the central carbon atom will prove beneficial and accelerate the reaction.

$\mathrm{S_{N}2}$ Reaction Pathway

In your molecule #2, a benzoyl group is attached to the central carbon involved in the $\mathrm{S_{N}2}$ process.  You are correct that such a group is larger than a hydrogen and should decrease the reaction rate due to steric effects. However the size of the benzoyl group is not exceptionally large and the steric effect will likely be small. More importantly, the benzoyl group will also exert an electronic effect. The central carbon in an $\mathrm{S_{N}2}$ reaction is $\mathrm{sp^2}$ hybridized [e.g. leading to a trigonal bipyramid structure as mentioned above] and has a $\mathrm{p}$ orbital.  This $\mathrm{p}$ orbital can interact with the adjacent carbonyl group through resonance.  This resonance effect will tend to remove (delocalize) electron density from the central carbon atom.  Also, the carbonyl group will inductively remove electron density from the central carbon atom.  These electronic effects will remove electron density from the central carbon atom thereby accelerating the reaction as observed.

(1) For example, the relative rates of solvolysis of 7-tosyloxynorbornane and anti-7-tosyloxynorbornene in acetic acid are $\mathrm{1:10^{11}}$


R or S configuration? The respective configutations of a and b are...?

I tried to find the answer for a (for me is: 2S, 3S) but I'm not sure of my results, can someone help me?

Here is my sage advice to you, Silvia95. Do not redraw structures!!! Use the method described here. Transcribing structures to place the lowest priority group in the rear will invariably lead to errors. Leave the structure where it is and use your hands as tools to determine chirality. Make a "thumbs up" gesture with either hand and point the thumb from the asymmetric carbon to the lowest priority group. Your fingers should point in descending order from highest priority to the next to lowest priority. If not, you are using the wrong hand. Right hand, R; left hand, S.C2 of dibromide a has the hydrogen in the rear. The fingers of your left hand show Br>C3>C1. C2 in dibromide b has the hydrogen in the front. Now the thumb of your left hand points toward you and Br>C3>C1. Using the same technique for C3 of a and b, your right hand fingers reveals Br>C2>C4. These two structures are identical, (2S,3R)-2,3-dibromopentane. If you rotate either structure by 180o about a vertical axis in the plane of the screen, the two structures will superimpose.


Is 44.5 mg/L a measurement of weight or volume? I have a test result that I am trying to fully grasp. I need to determine how much of a contaminant I need to reach that amount in 4,800 gallons. Making the conversion to liters is no problem, nor is the math necessary to figure how much I need by weight to get this much contaminant.

The one thing I am unclear on is my assumption that $44.5~\mathrm{mg~L^{-1}}$ is by weight.
In other words, if I have a result that measures $50~\mathrm{mg~L^{-1}}$ of a metal (nickel, for instance), that means if I were to be able to extract the metal and weigh it, it would weigh $50~\mathrm{mg}$.

Any help you can offer would be greatly appreciated.

$50~\mathrm{mg~L^{-1}}$ is an expression of concentration by mass. It means that if you took one litre of the solution and extracted all the metal, the resulting metal would weigh 50 milligrams.

The other common option is to express concentration in terms of moles per litre ($\mathrm{mol~L^{-1}}$) which means that if you extracted all the solute from one litre of solution, you would have 50 moles of solute.


What is the plastic material of the CD cases? What is the material of which the transparent part of CD cases are made of? It is nicely rigid and really lightweight.

These are called slimline jewel cases.

From the Wikipedia article on Optical Disc Packaging:

  The bulk of slimline cases are made with translucent or transparent polystyrene, and are available in multiple colors. A stronger alternative is made from semi-opaque, semi-flexible polypropylene which is strong enough to protect the disc, but flexible enough not to break easily.

Here is an image of a polystyrene case to compare to yours:


Meaning of Partial Half-Life for Two Parallel Reactions Consider two parallel reactions:
$$\ce{A   ->  B }\tag1$$
and
$$\ce{A ->  C}\tag2$$
What is the meaning of partial half life of the equation $(1)$ and that of equation $(2)$ with respect to above reactions?
It seems to me that it means the half life when one of these reactions are taking place individually, one-at-a-time, and not simultaneously. For example, partial half life of the equation $(1)$ is when half of $\ce{A}$ converts to only $\ce{B}$ and not to $\ce{C}$.

The partial reaction half-life relates to the reaction speed constant by the same way as for the "normal" reaction half-life.
If there are 2 parallel reactions of the first order, $\ce{A -> B}$ and $\ce{A -> C}$, and if there is the reaction rate for the former:
$$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}=k_{\ce{B}} \cdot [\ce{A}]$$
then the partial half-life w.r.t. $\ce{A -> B}$ is:
$$t_{1/2,\mathrm{B}}=\frac{\ln{2}}{k_{\ce{B}}}$$
Analogically the similar for the other reaction:
$$\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t}=k_{\ce{C}} \cdot [\ce{A}]$$
$$t_{1/2,\ce{C}}=\frac{\ln{2}}{k_{\ce{C}}}$$
The overall half-life w.r.t. $\ce{A -> X}$ is then
$$t_{1/2}=\frac{\ln{2}}{k_{\ce{B}}+k_{\ce{C}}}=\frac{1}{\frac 1{t_{1/2,\ce{B}}}+\frac 1{ t_{1/2,\ce{C}}}}$$
----  Responses to comments
The partial half-life is the extrapolated time after which all $\ce{A}$ would have decayed, if it had been decaying by the current and constant rate of given reaction and only by this reaction. But the main meaning is as a kind of reciprocal value of the reaction constant.
If we draw the chart of the partial reaction rate, then it's tangenta at $t=0$ will cross $x$-axis at the reaction partial half time. It is the same as if it was the only reaction. It does not say half of $\ce{A}$ decays in this partial half time.
If $\Delta t \ll \min{(t_{1/2})}$, then it does not matter it is just partial halftime, as we can neglect decay of the other parallel reactions.


Calculating dew point and bubble point of gas mixtures How do you calculate approximately the triple point and critical point of any given CH-based gas mixture?

I have developed a software which uses iterations to find bubble point and dew point of any CH-based gas mixture. It uses Peng-Robinsson EOS and Van Der Vaals mixing algorithm.

When calculating the dew point pressure for a certain temperature, it's difficult to know the starting range for the iteration. First I used False-Position method in finding the root, but now I have switched to Bisection. Although it's slower, it's more safe as it's not affected by function derivatives.

What would be needed is as approximation of the range where the liquid/vapour phases are found. Setting the iteration's range to start below triple point pressure, or above critical point pressure will cause the iteration to diverge and never find a root.

In the picture above, the C is the critical point. There seems to be some easy methods to calculate the critical temperature, but not so with the critical pressure.

For bubble point it would give me a good indication of where the max range is.

But I realised that the critical temperature point won't help me finding the min and max range for dew point calculation. As you can see in the picture, the point M (max temperature point) would be a good indication of where the max range is for dew point.

Anyway I found a much simpler solution. I consider the bubble point and dew point calculation functions as a black box.

As I'm only seeking the bubble point / dew point for a specified temperature, I step the possible pressure range (1 kPa - 10 MPa), with steps of 1 kPa. Then I calculate the numerical derivate of using current step and previous step.

This function derivate is consistent and negative in the range of the function domain where a bubble or dew point value is valid.

Outside the function domain the derivate is close to zero (+-1E-15).

So this gives me a good starting point for the root finding algorithm.

I also changed from False Position method to Bisection method, as it's less sensitive to derivates.

I hope this will help someone else with similar problem.


Convert grams of water to steam I'm working through problems in my textbook. How much heat is required to convert 2.55 g of $\ce{H2O}$ at 28 °C to steam at 100 °C?

In simpler problems I've used the fact that $\Delta H_\text{fus} = 6.02\ \mathrm{kJ/mol}$ for $\ce{H2O}$ at a melting point of 0 °C to do these kinds of calculations. But here the temperatures are different. How can I approach this problem? Is there some other conversion factor I should be aware of?

You need the vaporization $\Delta H_\text{vap}$ and you need to use this equation: $$Q = m~c~\Delta T$$

Where $Q$ equals the energy to heat $1\ \mathrm{mol}$ of water up to $\mathrm{100~^\circ C}$. $m$ is the mass of water, $c$ is the specific heat capacity and $\Delta T$ is the difference between the initial and final temperature.

In this stage you have the energy that is used to heat the liquid water up to $\mathrm{100~^\circ C}$ but the water is still liquid and you need steam water.

So you need to calculate the energy for phase transition from liquid to steam and the equation for this is: $$\Delta H_\text{vap} \cdot n = Q$$

$Q$ is the necessary energy for the phase transition, $n$ is the amount of substance and $\Delta H_\text{vap}$ is the energy used for phase change of $1\ \mathrm{mol}$ of the substance.

You also need to convert the mass to amount of substance (or use latent steam heat instead) and to calculate the $Q$ in the two stages.

This is the energy used for heat and change the phase.


Molar mass of a mixture of two solids I wonder, if I wanted to know the molar mass of a some white crystalline solid that is a mix of $\ce{C6H12O6}$ and $\ce{C12H22O11}$, would I just do this :

$$170.1559 \; \ce{g/mol} + 343.2965 \;\ce{g/mol} = 522.4524 \; \ce{g} / 2 \ce{mol}$$

Thus : $261,2262 \; \ce{g/mol}$.

Would this be a correct operation ? Or it wouldn't make any physical sense ? Thank you!

Sounds like a chemical mixture of sucrose and glucose.  In which case, you only talk about the molar mass of the two sugars separately.  The mass or number of moles of the mixture can be determined by knowing the ratio of sucrose and glucose in the mixture.  FYI... the molar mass of glucose is 180 g/mol (not 170).


Relative molecular mass of an unknown gas in a mixture A gaseous mixture contains oxygen and another unknown gas in the molar ratio of 4:1 diffuses through a porous plug into 245 seconds, under similar conditions same volume of oxygen takes 220 seconds to diffuse. What is the molecular mass of the unknown gas?

I used the Graham's law of diffusion to find the molecular mass of mixture.
i.e.
$$\frac{r_\text{mix}}{r_{\ce{O2}}}=\frac{220}{245}=\sqrt{\frac{M_{\ce{O2}}}{M_\text{mix}}}$$
$$M_\text{mix}=39.6$$

But what's next? How do I calculate the $M_\text{gas}$?

After getting relative molecular mass ie $39.6$,

$$\mathrm{M_{net} = \frac{(n_1\times M_1 + n_2\times M_2)}{(n_1 + n_2)}} \tag{i}$$

Given that the molar ratio of gases is $4:1$, the molar ratio of oxygen is $\mathrm{4x}$ and that of the unknown gas is $\mathrm x$. Plug $\mathrm{n_1 = 4x,\ n_2 = x,\ M_1= 32,\ M_2 = M,\ M_{net} = 39.6}$ in equation (i) and get $\mathrm M$.


Model of an ideal liquid in chemistry In chemistry there are ideal models/concepts of ideal crystal, ideal gas and even ideal solution, but there is no model/concept of ideal liquid.

In physics (fluid dynamics, to be precise) there is a model of perfect fluid, a fluid without internal friction and, as a consequence, no viscosity, and thermal conductivity. The model covers macroscale physical properties and doesn't really touch what's happening on the "chemistry level".

Why chemistry lacks a proper model of ideal liquid when a seemingly more complex model of an ideal solution has been developed? Isn't there a need for one, and if so, is there a concept of ideal fluid that comes the closest to be an alternative of, let's say, an ideal gas law, but for liquids?

The nearest to an ideal fluid is a hard sphere fluid, but this is removed from the ideal gas or even solution concept of ideality in a critical way.

Ideal (also called perfect) gases are ideal because they lack intermolecular interactions. A statistical description that ignores the intermolecular potential suffices to describe an ideal gas. A first extension toward nonideality, following the vdW equation, describes particles in a gas as hard spheres (having an excluded volume).

The property of reduced compressibility encoded by a hard sphere potential also provides the simplest model for a liquid. The hard sphere potential suffices to describe a simple liquid, and it is about as ideal an abstraction of a liquid as you can get. However, while it shows a liquid-solid phase transition, it does not display a liquid-gas transition. In that sense it is not a universal model, it does not capture some key properties of liquids, in the same sense that the ideal gas law is not universal because it too does not address some simpler key properties of real gases (such as excluded volume). However, introduction of an additional attractive term as in the full vdW model, bringing us further away from ideality, allows us to observe a critical point below which a phase transition clearly occurs between liquid and gas. I should add that the hard sphere potential can be expressed in reduced units, making it universal within the scope of that limited model. It is a suitable first approximation for understanding some properties of nonpolar liquids, for instance condensed noble gases.

I would add that Raoult's law is another ideal model for a liquid, and consistent with the ideal gas concept (or rather Dalton's law). It is also a limiting law (or model), like the ideal gas, and applies to dilute solutions.


Comparison of acidities My book says

  To compare the acidities of two compounds, compare the stability of the anions formed by loss of a proton with respect to the original compound.

Suppose we have acetic acid and ethyl alcohol. Now on losing a proton, they form the respective anions:

My book again says

  The acetate anion is resonance stabilized whereas in the ethoxide ion the charge is localized on the oxygen atom. Hence, acetate ion is more stable as compared to the ethoxide ion and hence, acetic acid is the stronger acid.

The problem is we have only compared the stabilities of the anions with respect to each other and not with respect to the original compounds. So how do we know that it is correct?

Once you understand that the acetate ion is more stable the the ethoxide ion, we understand the fact that after releasing The acidic hydrogen the final product formed is more stable for acetic acid as compared to ethanol.
This would lead to the conclusion that acetic acid is a better acid as compared to ethanol.

While comparing the acidity of two species, only the comparative study of conjugate base is enough, since its simply the comparison between compounds.

Comparing the stability with the original compounds will lead you to the conclusion that whether the compound itself has a tendency to act as an acid or base.
Alternately one could compare the stability of the original compound and its conjugate(Which is the acid dissociation constant) and compare its value with the acid dissociation constant of other compunds to compare its acidity.

In any case comparing the stability of the conjugate ions will tell you which one of them is comparatively more acidic/basic than others. It will not let you conclude whether its the nature of the compound to naturally act as an acid/base or not. For that one should compare the stability with original compund or just check its dissociation constant.


Can all species that undergo aldol reactions also undergo Cannizzaro reactions? Can all species that undergo aldol reactions also undergo Cannizzaro reactions? Ater all, the nucleophile can always attack the carbonyl group.

No. The Cannizzaro reaction only works for aldehydes with no $\alpha$-protons, benzaldehyde being a classic example.

For a start, ketones and any other species that can form stable enolates, such as esters or carboxylic acids, can undergo the aldol reaction or similar reactions like the Claisen condensation. However, the Cannizzaro reaction is limited to aldehydes because only they can disproportionate to form alcohols and carboxylic acids.

Secondly, the Cannizzaro reaction involves strongly basic conditions, typically using sodium or potassium hydroxide. Under these conditions an aldehyde with $\alpha$-protons will undergo deprotonation to form an enolate much more readily than nucleophilic attack on the the acyl carbon by a hydroxide ion.


Why does the fructose monomer in sucrose appear different from isolated fructose? I understood that the representation of fructose is:

But during the formation of sucrose:

How did fructose, whose representation is given in the first picture, turn out to be different in the second. I tried to visualize myself rotating the molecule but the position of oxygen changes and it doesn't match with the one in sucrose.

Ah, a (fairly) common conundrum that assails us Chemistry students when we start Biochem. ;-)

At first glance, the fructose molecule in your first picture, and that in the second picture appear to have different conformations:

Believe me, they're the same. "How?" you ask? Stay with me here (it requires a little imagination):

Draw a vertical line/axis on your first diagram along the plane of the furanose ring, connecting the oxygen atom at the apex, right down to the middle of the C3-C4 bond. Then rotate the molecule 180 degrees counter-clockwise.

Now compare this with the fructose residue in sucrose.

(I lack the necessary Org. Chem. sketching tools to depict this... apologies)

If you weren't able to wrap your head around "rotation" bit, then, let's assign locant numbers to the carbon atoms in the two diagrams and then compare them:

   I've drawn them tiny, my bad

Must've clicked by now ;)


electrolysis - selection of what is going to be on anode and cathode I have a question about electrolysis. How do we know what will be on anode and cathode if we have for example NaCl diluted in water. How do we know if it's gonna be Na+ or H+??

It depends on its oxidation/reduction potentials. For example, in the electrolysis of a NaCl solution (in water), the reduction potential of water to H2 is -0.83 V while the reduction potential for Na+ to Na is -2.71 V. As the potential for water is greater, water will be reduced instead Na+.
Another example, a solution of CuSO4 in water, the potential reduction for Cu2+ to Cu is +0.34 V, greater than the -0.83 V for water, so the Cu2+ will be reduced.


Do Grignard reagents react with amides? Clearly the amide will be unreactive given the poor quality of the leaving group and poor electrophilcity due to delocalization. However, does it react with organolithiums for example?

Amides will react with three equivalents n-butyllithium to give nitriles via geminal lithium oxyimide intermediates (J. Org. Chem., 1967, 32 (11)). n-butyllithium is a strong enough base to deprotonate the molecule twice at nitrogen and once alpha to the carbonyl group. Grignard reagents may well behave similarly.

I do not think this is a particularly useful transformation though because n-butyllithium is a rather indiscriminate reagent. There are a wide range of other methods available - see this paper if you're interested.

In the special case of a N,O-dimethylhydroxyamide, you can make ketones using either.


anode and cathode in lithium ion battery? By definition, the anode is the electrode where oxidation takes place, the cathode is the electrode where reduction takes place.

When I am charging the Li-ion battery, positive electrode undergoes oxidation and hence I may call it the anode. When I am discharging the Li-ion battery, negative electrode undergoes oxidation and I may call it the anode. Is this correct?

Do the anode and cathode interchange for charging and discharging?

This is indeed correct, since, as you said, the cathode is per definition the site for reduction and the anode for oxidation, both names have to be "switched" when viewing electrolysis (charging of the Li-ion battery) and galvanic cells (discharging of the Li-ion battery). What does not change is the location of the + and - pole.

A good way to remember the site of oxidation and reduction is the word pair an-ox /cat-red.


Is the correct zeroth-order wave function $\phi^{(0)}$ an eigenfunction of the perturbation Hamiltonian $H'$? Consider the perturbation treatment of He $1s2s$ excited state configuration. The quantum mechanical Hamiltonian for the He-atom is:
\begin{aligned}
\hat{H} = \sum_{i=1}^2 h_i + \frac 1 {r_{1 2}} \\
h_i = -\tfrac 12 \nabla_i^2 - \frac 1 r_i
\end{aligned}
In the following perturbation treatment, the unperturbed Hamiltonian $\hat{H}^0$ is $\sum_{i=1}^2 h_i$, while the perturbation Hamiltonian $\hat{H}'$ includes the inter-electron repulsion $r_{1 2}^{-1}$.
The four unperturbed degenerate wave functions $\psi^{(0)}$ are shown below as Slater determinants:
$$
\begin{aligned}
&\psi_1^{(0)} = \left|\begin{array}{} 1s(1) \overline{2s}(1) \\ 1s(2) \overline{2s}(2) \end{array}\right| \equiv \left| 1 \overline{2} \right\rangle & \psi_2^{(0)} = \left| \overline{1} 2 \right\rangle \\
&\psi_3^{(0)} = \left| 1 2 \right\rangle &\psi_4^{(0)} = \left| \overline{1} \overline{2} \right\rangle
\end{aligned}
$$
The correct zeroth-order wave functions $\phi^{(0)}$ for the perturbation Hamiltonian $\hat{H}'$ are linear combinations of unperturbed eigenfunctions such that $\phi^{(0)} = \sum_i c_i \psi_i^{(0)}$. Solving for $\phi^{(0)}$ using the linear variational method is equivalent to diagonalizing the perturbation Hamiltonian $\hat{H}'$ in a basis of $\psi^{(0)}$. The eigenvectors and eigenvalues of the $\mathbf{H}'$ matrix, whose elements are $\left\langle \psi_m^{(0)} \middle| \hat{H}' \middle| \psi_i^{(0)} \right\rangle$, are the correct zeroth-order wave functions and the corresponding first-order corrections to the energy, respectively.
Because $\left\langle \psi_{m \neq 3}^{(0)} \middle| \hat{H}' \middle| \psi_3^{(0)} \right\rangle = 0$ (the Condon-Slater rules), $\psi_3^{(0)} = \left| 1 2 \right\rangle$ is already the eigenvector of the perturbation Hamiltonian $\hat{H}'$ with the eigenvalue $E_3^{(1)} = \left\langle  1 2 \mid \mid 1 2 \right\rangle = \left\langle  1 2 \mid 1 2 \right\rangle - \left\langle  1 2 \mid 2 1 \right\rangle$. Could you suggest whether it is correct to go on to state the eigen-equation of $\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$ ?

No, writing $\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$ will be incorrect. I will give you two explanations for the same, one intuitive and other mathematical.

Intuitive Explanation:
Assuming that you can write $\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$. But, $\hat{H}_0\psi_3^{(0)} = E_3^{(0)} \psi_3^{(0)}$, and in an overall sense, you can write
$$\hat{H}\psi_3^{(0)} = (E_3^{(0)}+E_3^{(1)}) \psi_3^{(0)}$$
This essentially means that electron-electron repulsion is keeping the wavefunction same as that of single electron Hamiltonian. Now, since in single electron Hamiltonian, each electron does not "see" the other electron, and they are allowed to be in the same spatial position. Now, on bringing the $1/r_{12}$ term inside the Hamiltonian, the energy goes to $\infty$ for such a case, and hence a singularity is created in the system.
So, it means that the $\psi_3^{(0)}$ is not an eigenvector of total Hamiltonian, $\hat{H}$. Since $\psi_3^{(0)}$ is defined to be eigenvector of $\hat{H}_0$, so the initial assumption that $\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$ is incorrect.

Mathematical Explanation:
You have given $\left\langle \psi_{m \neq 3}^{(0)} \middle| \hat{H}' \middle| \psi_3^{(0)} \right\rangle = 0$, which is correct. Along with that, we know  $\left\langle \psi_{m}^{(0)} \middle|  \psi_n^{(0)} \right\rangle = \delta_{mn}$. However, you could write the given statement $\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$ provided $\psi_n^{(0)}$ would be a complete basis set (for explanation, see below). However, $\psi_n^{(0)}$ is not a complete basis set, since for a He atoms, there are infinite number of ways to put two electrons in an orbital, since there are infinite number of orbitals. Since you are treating the orbitals as hydrogenic, it is not guaranteed that the orbitals will be orthogonal to each other once the $1/r_{12}$ term is pulled in.
Even for a hypothetical case if we consider only 1s and 2s orbitals are present and are orthogonal to other orbitals even after the presence of the $1/r_{12}$ term, still they would not be a complete basis set. The reason is simple: You have considered only $1s^12s^1$ configurations, and not $1s^2$ and $2s^2$.

For complete basis, proof:
Assuming $\psi_n$ is an othronormal and complete basis set of dimension $M$, one can
write: $$\hat{H}'\psi_3^{(0)}=\sum_{i=1}^M c_i \psi_i^{(0)}$$ Now,
$$\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)} \Rightarrow C_m=0
> \forall m\neq 3$$ $$\Rightarrow \hat{H}'\psi_3^{(0)}= c_3
> \psi_3^{(0)}$$
However, this would have been true if, along with other conditions,
$\psi_n^{(0)}$ would have been a complete basis set.

So, from the given information, it is impossible to conclude that $\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$ mathematically, and from a chemistry standpoint, $\hat{H}'\psi_3^{(0)} \neq E_3^{(1)} \psi_3^{(0)}$.


What is the procedure to create stock solutions? Until now to execute this recipe, each time i had to count those milligrams/Liter. An enough difficult procedure.

I am posting here this question to help me understand the procedure of creating stock solutions. So the next time i will create that plant substrate, to take a specific volume of that stock solution.

Thank you.

Since it contains agar, you wouldn't be able to make a stock solution with the agar in it; but you can make a stock solution of the other items and add the agar when you need the working solution. Also, since there is a fair amount of sugar in the recipe, it would be best to add that to the working solution as well.

You should be able to make a 100 times concentrate (leaving out the sugar and agar) and then sterilize it in an autoclave. When you need to make a working solution, weigh 5 g of agar and 20 g of sugar, add 10 mL of stock solution, and about 900 mL of water. Boil the solution to dissolve the agar, cool (to about 50°C) and add water (before it sets) to get 1000 mL of solution.


No sign flipping while figuring out the emf of voltaic cell? I learnt that for a voltaic cell, the value for the $E_\text{cell}^\circ$ when the reaction is spontaneous is given by

$$E_\text{cell}^\circ = E_\text{cathode}^\circ - E_\text{anode}^\circ, \label{eqn:1}\tag{1}$$

so that the difference in the right gives us a positive value for $E_\text{cell}^\circ$.

But suppose we are given two half-reactions:

$$
\begin{align}
\ce{Ag+(aq) + e- &→ Ag(s)} &\qquad E^\circ &= \pu{0.80 V} \\
\ce{Sn^2+(aq) + 2 e- &→ Sn(s)} &\qquad E^\circ &= \pu{-0.14 V}
\end{align}
$$

When finding the overall spontaneous reaction, we must flip the second reaction, multiply it by $2$, and then add it with the first to get our desired equation.

But when determining the $E_\text{cell}^\circ$, why don't we negate the minus sign of the second half-reaction and make positive, before we put it in $\eqref{eqn:1}$ to figure out the $E_\text{cell}^\circ$? Shouldn't we do that because we reversed the second equation?

My book tells me to keep the $E_\text{half-cells}^\circ$ as they are written in the tables and simply put them in $\eqref{eqn:1}$. But why?

Take a look at the two half reactions:

$$
\begin{align}
\ce{Ag+(aq) + e- &→ Ag(s)} &\qquad E^\circ &= \pu{0.80 V} \\
\ce{Sn^2+(aq) + 2 e- &→ Sn(s)} &\qquad E^\circ &= \pu{-0.14 V}
\end{align}
$$

If there is an electron for grabs (like the ones in the wire of a voltaic cell), $\ce{Ag+(aq)}$ and $\ce{Sn^2+(aq)}$ are competing for it. Whichever half reaction has the higher (more positive) reduction potential will win. If the reduction potentials are equal, it is a draw and the reaction is at equilibrium. So we are taking the difference of the reduction potentials to see in which direction the reaction will go.

  No sign flipping while figuring out the emf of voltaic cell?

Take a look at the equation you are using to figure out the emf. You are already treating the oxidation half reaction differently than the reduction half reaction because there is a negative sign in front of the anode reduction potential.

$$E_\text{cell}^\circ = E_\text{cathode}^\circ - E_\text{anode}^\circ$$

If you switch the anode and cathode half reaction, you would get the opposite sign for the emf. (Not that the reaction would go in that direction.)


The sponge isn't foaming up, why? When I add dish soap to a sponge it is able to foam up. But if I leave the sponge alone for an hour, it doesn't foam up, not even if it is wetted.

Why? The sponge should still contain dish soap an hour later.

Tap water contains calcium and magnesium ions; more so if the water is hard. These ions can bind to soap over time and make an insoluble soap curd.

Is the sponge clean? If not, another reason is that the surfactants in the soap that make the foam might be forming micelles around whatever grime is in your sponge, so that there aren't many left to make the foam.


Quantization and Bohr's model According to quantization it's said that emitted or absorbed energy is quantized.

Then, when it's said in bohr's model an electron changes its orbit (Let's say it goes to a higher energy shell from $n=2$ to $n=3$) The energy difference between the orbits is $100~\mathrm{kJ}$.  Now, according to quantization is the $100~\mathrm{kJ}$ itself a quantum. Or the $100~\mathrm{kJ}$ will be absorbed in quantized way?

The energy difference between $n=2$ and $n=3$ in Bohr's model is
\begin{equation}
\Delta E=13.6\ \mathrm{eV} \times \left(\frac{1}{4}-\frac{1}{9}\right) = 1.88\ \mathrm{eV} = 3\times10^{-22}\ \mathrm{kJ} = 181.4\ \mathrm{kJ/mol}
\end{equation}
Thus, for this excitation to occur, a single energy quantum (photon) with an energy of $1.88\ \mathrm{eV} = 3\times10^{-22}\ \mathrm{kJ}$ is required.

The excitation does not happen if:

the photon has more or less energy than $1.88\ \mathrm{eV}$ (The energy difference needs to be matched exactly!)
the same amount energy is distributed over multiple photons (The energy needs to be provided by a single photon!)

As already has been pointed out, the $181.4\ \mathrm{kJ/mol}$ itself is not a quantum, but one mole of quanta, where each quantum has $1.88\ \mathrm{eV} = 3\times10^{-22}\ \mathrm{kJ}$ of energy. As each atom can only absorb this energy once (assuming it does not relax back to $n=2$ by emitting the energy), to absorb so much energy will also require one mole of atoms.


Reaction Penetrating Metal Sheet I'm writing a novel set 28 years in the future in which the protagonist attempts to manipulate a communications satellite by directing a swarm of nanorobots (or possibly micro-robots, depending on plausibility).

An initial wave of nanorobots is tasked with creating a path, such as a tunnel, through the satellite's casing so that a second wave can enter.

Inspired by real-life nanotechnology that delivers chemotherapy drugs only to leukemia cells, I thought to have each nanorobot carry a tiny amount of super acid, each depositing its payload onto the same spot of the metallic hull of the satellite. My thinking is that the vacuum of space would carry off the product of metal + acid, leaving fresh metal for the next microrobot to attack.

Is it plausible to erode a narrow path through metal in this manner? What acids would be well-suited to the task? How would I go about calculating the order-of-magnitude number of nano-robots needed to complete a path through a given thickness of (let's say) aluminum?

Depending on the material that the casing is made out of, you could have a mercury or gallium payload.  They both are capable of significantly weakening the structures of some metals (such as aluminum).  See the following video of gallium weakening a coke can.  Mercury can also weaken the structure, especially in conjunction with gallium as seen here.  Mercury will form an amalgam with aluminum and significantly weaken it, as will gallium.  You might conceivably have nanobots bring a payload of the two metals to weaken the structure, especially since they are not nearly as reactive with other things.

Unfortunately, this would only work if the structure you are trying to damage is primarily aluminum, but might give a starting place for non-acidic solutions.


Difference between thermal reactions and photochemical reactions In a photochemical reaction, everything starts with absorption of a photon. I.e. a ground state reactant is excited to the first excited state. Sometimes the reaction barrier on the ground state surface is too high; perhaps lower barriers exist in other dimensions that lead to undesirable products or dissociation. Perhaps the relevant barrier is much lower on the excited state surface, which makes the reaction possible in the excited state.

Is the difference, then, that thermal reactions take place exclusively on the ground state surface, while a photochemical reaction takes place, at least in part, on the excited state surface?

Yes, you are right thermal reactions take place exclusively on the ground state surface, while a photochemical reaction takes place, at least in part  on the excited state surface.

But there is another difference between them as
photochemical processes offer an advantage over thermal methods by forming thermodynamically disfavored products. Therefore, they overcome large activation barriers in a short period of time, and allow reactions that were inaccessible by thermal processes.


Why does roasting requires lower temperature than melting point Why do processes like roasting and calcination require a temperature lower than the melting point? Since intermolecular forces are lower in a liquid won't it be easier to oxidise (roasting) / decompose (calcination) in this state than in a solid?

Reason 1:

For example if you have an porous ore and you melt it, the surface area will decrease dramatically. If you just roast it, the surface area will stay about the same.

A higher surface area will aid calcination proceses as more oxygen gets into contact with the surface and will eventually react.

Reason 2:

A higher temperature means more energy to heat the ore and this means it costs more money.

Reason 3:

Some oxides will volatilize when temperature is high (e.g. litharge).


Are burns from Lye or Potash that bad? I read that the chemicals potassium hydroxide and sodium hydroxide are extremely corrosive and will cause severe burns but I also see videos of people saying that if you get lye on you, then just wash it off and you’ll be fine. Is that actually ok. The fact that the stuff is sold in hardware stores makes it seem even less dangerous than the warnings make it out to be. Which is it?

As noted in the comments, lye burns are very damaging to skin and especially mucous membranes. This damage can be drastically reduced if the lye is washed off as quickly as possible.
One reason that these solutions are considered particularly dangerous is that there is no discomfort immediately upon contact (in contrast to acids, which tend to produce burning sensations immediately). Instead, the affected surface simply feels a bit soapy.  As a result, the presence of lye on skin may not be noticed until it is too late to prevent damage by washing.
On a side note, the "it's sold in hardware stores" argument is not a good test for chemical toxicity. There are numerous highly toxic chemicals that can be purchased readily.


How does charge density affect the rate of a decomposition? For example, for the thermal decomposition of carbonates of group 2 metals, how does the charge density decreasing down the group affect the rate of this decomposition?

Increase in charge density increases the polarizing capacity of the atom, decreasing the bond strength, thus facilitating easy decomposition


Do 1s & 2s orbitals overlap of different atoms while forming sigma bond? I Do 1s & 2s orbitals overlap of different atoms while forming sigma bond?
I think they should overlap as there will not be enough energy difference between them.

Orbital combinations involving inner shell electrons can indeed interact with outer-shell counterparts having the same symmetry.  But, the inner shell orbitals are so deeply buried inside their respective atoms that the amount of overlap, and therefore the effect on bond/orbital energies, is very small.  We find that in general, molecules are well modeled, without unneeded complexity, by including only outer shell orbitals.  Moreover, any overlap involving inner shells, or even with outer shells, usually does not involve just $s$ orbitals.  There can be $p$ orbital combinations and even $d$ orbital combinations having the same symmetry as the $s$ orbital ones, so the higher angular momentum orbitals also mix in.  Only hydrogen and helium atoms with just $1s$ electrons can avoid that.

It isn't just our description of chemical bonding that exploits this property of inner shell electrons.  We sometimes identify elements in a sample by hitting the samples with X rays that knock out inner shell electrons, and measuring the spectrum of absorbed energies.  The energy required to knock out inner shell electrons from an atom is a characteristic of that element, and lack of strong overlap with other atoms makes that energy essentially invariant.


what is the exact meaning of oxidation state? can you mention the concept of oxidation state without its simple definition i-e it is the apparent charge +ve or -ve.....
I found these in wikipedia but didn't understand."Conceptually, the oxidation state, which may be positive, negative or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component. This is never exactly true for real bonds.". what does it mean? please clear with example if possible.

Oxidation state is a formal number ascribed to atoms, to tell if it has more or less electrons than in its natural state.

For single atoms it is fairly straightforward, as electrons can only be removed or added in discrete units. Therefore, an atom or ion has an oxidation state that equals it electric charge.

But electrons are not small particles, orbiting the molecule. In many ways an electron is better described as a cloud of negative charge hovering around the molecule and chemical bonds.

When different atoms are connected in a molecule, this cloud is not evenly distributed between the atoms. Example: In Carbonmonoxide (CO) the Oxygen attracts most of the electron-cloud. But not all, if so the molecule would part into ions.

In reality, the oxidation state of the atoms in CO should be calculated with decimals instead of whole numbers, but for practical reasons we ascribe the whole of an electron in a bond to one of the atoms.

In Carbonmonoxide the atoms are bonded by a double-bond (4 electrons) and if all of these “belongs” to oxygen it will have two more than the basic atom and thus the oxidation state is -2.


Why do amorphous substances have different values of physical properties in different directions? Today my teacher taught about amorphous and crystalline substances. Crystalline substances are said to have different values for physical properties (like speed of light, electrical resistance, etc) in different directions. Crystalline substances are said to have orderly arranged constituents, however this orderly arrangement is not said to be same in different directions. Thus, they are said to show different values for physical properties in different directions.
  Amorphous are said to be isotropic in nature, i.e they have same value for all the physical properties in different directions. I was stuck here, amorphous substances have disordered arrangement of constituents. So, they should strictly have different values in different directions. Is it we speak about average values of physical properties in case of amorphous substances? or is it that, they strictly have same values for the physical properties in all directions?

Yes, you’re exactly right that we are talking about an average. If we could somehow look at a small enough portion of an amorphous solid, we would see differences between the properties in different directions. Indeed, a small enough portion of, say, amorphous ice is a single water molecule, which is certainly not isotropic: we can talk about the direction perpendicular to the molecular plane or whatever.

But when we talk about real, physical materials, as usual in chemistry, we are talking about Avogadro’s number (plus or minus a few orders of magnitude) of molecules. So taking the spatial average of an amorphous structure makes perfect sense, and the law of large numbers means that the tiny differences between properties in different directions will be utterly undetectable.

(As an aside, it’s worth noting that the perfect crystalline structure we imagine is also an average over every unit cell in the material – it’s just that because each unit cell is oriented in more or less the same way, different directions are not “mixed” as part of this average, so crystals can have anisotropic properties.)


Do secondary alcohols prefer to undergo SN1 or SN2 reactions when reacting with a strong acid? For example, let us say we have 2.2-dimethycyclohexanol reacting with $\ce{HBr}$. We know that the oxygen gets protonated and then leaves as water, but the question does it leave on its own or does a backside attack happen? If its $\mathrm{S_N1}$ we have 1-bromo-1,2-dimethylcyclohexane as the major product. If its $\mathrm{S_N2}$, we have 1-bromo-2,2-dimethylcyclohexane as the major product. This is assuming we are at 0℃.

I believe that its $\mathrm{S_N1}$ because bromide is a weak base.

Actually, secondary alcohols undergoes $\mathrm{S_N1}$ or $\mathrm{S_N2}$ depending upon the substrate you are using. For e.g. In your case(as provided above) the substrate has steric hindrance nearby alcoholic group and the carbocation will be stable due to the migratory aptitude of the methyl group. Hence, here should be $\mathrm{S_N1}$ and you have to see the structure of substrate to decide $\mathrm{S_N1}$ or $\mathrm{S_N2}$ for secondary alcohols.
Also, alcohols are not good leaving groups also! But after protonation the water is a very good leaving group.


Derivation of mean speed from Maxwell–Boltzmann distribution I found that if a velocity of a gas follows the Maxwell–Boltzmann distribution, the mean velocity is given by
$$\langle v \rangle = \sqrt{\frac{8RT}{\pi M}}$$
where $R$ is the gas constant, $T$ is temperature, and $M$ is molar mass. How would one obtain this result from the Maxwell–Boltzmann distribution?

If one has a probability density function $P(x)$, then the expectation value of a quantity $f(x)$ is given by
$$\langle f \rangle = \int f(x)P(x)\,\mathrm{d}x$$
evaluated over the limits of the probability density function, i.e. if your PDF runs from $-\infty$ to $\infty$, then those are your limits of integration.
In our case, the PDF is the Maxwell–Boltzmann distribution (denoted $P(v)$ here) and the quantity we want to find the expectation value of is simply the velocity, $v$. Since velocity can only be positive,* the limits of integration are from $0$ to $\infty$.
Therefore, the mean velocity $\langle v \rangle$ is given by the integral
$$\langle v \rangle = \int_0^\infty v P(v)\,\mathrm{d}v$$
where $P(v)$ is the Maxwell-Boltzmann distribution
$$P(v) = \left(\frac{m}{2\pi kT}\right)^{3/2}4\pi v^2 \exp{\left(-\frac{mv^2}{2kT}\right)}$$
So:
$$\begin{align}
\langle v \rangle &= \int_0^\infty v P(v)\,\mathrm{d}v \\
&= 4\pi\left(\frac{m}{2\pi kT}\right)^{3/2} \int_0^\infty v^3 \exp{\left(-\frac{m}{2kT}v^2\right)}\,\mathrm{d}v
\end{align}$$
The integral can be evaluated using integration by parts repeatedly. The process is not interesting and you could consult a table of standard integrals to find the result:
$$\int_0^\infty v^3 \exp{(-\alpha v^2)}\,\mathrm{d}v = \frac{1}{2\alpha^2}$$
Setting $\alpha = m/2kT$,
$$\begin{align}
\langle v \rangle &= 4\pi\left(\frac{m}{2\pi kT}\right)^{3/2} \frac{4k^2T^2}{2m^2} \\
&= \sqrt{\frac{8kT}{\pi m}} \\
\end{align}$$
$m$ here refers to the mass of one molecule, whereas $M$ in your question refers to the molar mass of the compound. They are related by $M = N_\mathrm{A}m$, where $N_\mathrm{A}$ is the Avogadro constant. Since $R = N_\mathrm{A}k$, you can multiply top and bottom by $N_\mathrm{A}$ to obtain the desired result
$$\langle v \rangle = \sqrt{\frac{8RT}{\pi M}}$$

Footnote
* If the assertion that velocity is non-negative is confusing, it's worth noting that the $v$ in the Maxwell–Boltzmann distribution $P(v)$ is not the vector quantity $\vec{v} = (v_x, v_y, v_z)$, but rather the magnitude of the velocity $v = |\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$. Part of the derivation of the Maxwell–Boltzmann distribution involves converting from the vector quantity to its magnitude: that's where the $3/2$ exponent comes from (it's actually $1/2$ per dimension in three dimensions), and also where the factor of $4\pi v^2$ comes from ($4\pi v^2$ is the formula for the surface area of a sphere: loosely speaking, this factor represents the "collection" of all points $(v_x, v_y, v_z)$ which have the same magnitude $v$). A fuller explanation may be found in any typical physical chemistry textbook.


Determining the group number of an element from its successive ionization energies The successive ionization energies of an unknown element X are:
  1st - 9534; 2nd - 11790; 3rd - 13683; 4th - 15309; 5th - 16458.
  To what group number does X belong?

There aren't any significant jumps so that means X is in groups 5-8. After that I don't know.

There are other little jumps to look out for, like the removal of the first electron from  a filled subshell. Consider the ionization energies (in eV) for the third period elements:

$$
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline
\text{electron}&\ce{Ar}&\ce{Cl}&\ce{S}&\ce{P}&\ce{Si}&\ce{Al}&\ce{Mg}&\ce{Na}\\\hline
\ce{3p}&1521&-&-&-&-&-&-&-\\\hline
\ce{3p}&2666&1251&-&-&-&-&-&-\\\hline
\ce{3p}&3931&2298&1000&-&-&-&-&-\\\hline
\ce{3p}&5771&3822&2252&1012&-&-&-&-\\\hline
\ce{3p}&7238&5159&3357&1907&787&-&-&-\\\hline
\ce{3p}&8781&6542&4556&2914&1577&578&-&-\\\hline
\ce{3s}&11995&9362&7004&4964&3233&1817&738&-\\\hline
\ce{3s}&13842&11018&8496&6374&4356&2745&1451&496\\\hline
\ce{2p}&40760&33604&27107&21267&16091&11577&7733&4562\\\hline
\end{array}
$$

Notice how, in addition to the big jump between $\ce{3s}$ and $\ce{2p}$, there is a small jump between $\ce{3p}$ and $\ce{3s}$ for Ar, Cl, S, P, and Si. If you have five ionization energies, and you don't see the large jump, look for the small jump. If you don't see the small jump, what group is X in?

There is another even smaller jump to look for. Half-filled subshells are also slightly more stable than other configurations (except filled subshells). Thus, it should require a slightly higher increase in ionization energy to remove an electron form any configuration with $\ce{np^3}$ than from $\ce{np^2}$ or $\ce{np^4}$. Let's look at the step wise increase in ionization energy (in other words, how much higher is the ionization energy for the second electron than the ionization energy for the first electron and so on?).

$$
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline
\text{electron}&\ce{Ar}&\ce{Cl}&\ce{S}&\ce{P}&\ce{Si}&\ce{Al}&\ce{Mg}&\ce{Na}\\\hline
\ce{3p}&-&-&-&-&-&-&-&-\\\hline
\ce{3p}&1145&-&-&-&-&-&-&-\\\hline
\ce{3p}&1265&1047&-&-&-&-&-&-\\\hline
\ce{3p}&1840&1524&1252&-&-&-&-&-\\\hline
\ce{3p}&1467&1337&1105&895&-&-&-&-\\\hline
\ce{3p}&1543&1383&1199&1007&790&-&-&-\\\hline
\ce{3s}&3214&2820&2448&2050&1656&1239&-&-\\\hline
\ce{3s}&1847&1656&1492&1410&1123&928&713&-\\\hline
\ce{2p}&26918&22586&18611&14893&11735&8832&6252&4066\\\hline
\end{array}
$$

You can see that the increase in ionization energy for the third electron to be removed from argon is larger than the increase for the second or fourth electrons. Your numbers don't have this little increase in them at all. Have you considered that X might be a transition metal?


How to calculate the pH of a buffered solution with Henderson Hasselbalch? Lets say you have a buffered solution containing $0.50~\mathrm{M}$ acetic acid ($K_a = 1.8 \cdot 10^{-5}$) and $0.50~\mathrm{M}$ sodium acetate, and you want to calculate the $\mathrm{p}\ce{H}$ of this solution.

This can be done in two ways: using the standard equilibrium chart (initial, change, equilibrium), or using the Henderson Hasselbalch equation.

My question is why do these two give the same result.

The Henderson Hasselbalch equation is derived starting with the equilibrium equation:

$K_a = \frac{\ce{[H+][CH3COO^{-}]}}{\ce{[CH3COOH]}}$ and then the $\ce{[H+]}$ is found, and then the $\log_{10}$ is taken. However, doesn't this assume that the system is already at equilibrium and that the concentrations of $\ce{[CH3COO^{-}]}$ and $\ce{[CH3COOH]}$ are the concentrations at equilibrium?

However, that is clearly not the case as shown by the ICE chart. However, both give the correct pH. Why is this?

The main equilibrium in such a solution is the following:

$\ce{CH3COOH +CH3COO^- <=>CH3COOH + CH3COO^-}$.

This should be your ICE chart's equation too. I suspect that the equation you have for your ICE table is the following:

$\ce{CH3COOH +H_2O <=>CH3COO^- + H_3^+O}$

You can certainly solve the second equation with an ICE table by plugging in the initial concentration of the conjugate base, $\ce{CH3COO^-}$.

  However, doesn't this assume that the system is already at equilibrium
  and that the concentrations of $\ce{[CH3COO-]}$ and $\ce{[CH3COOH]}$
  are the concentrations at equilibrium?

Yes. The concentrations (or more accurately, activities) used in the H-H equation are equilibrium concentrations.

In the case of a 0.5 M acetic acid/acetate ion buffer, we can, for most purposes, conclude that the initial molarity of the main system components - acetic acid and acetate ion will remain constant. This is because the main acid-base reaction in the system is the following:

$\ce{CH3COOH +CH3COO^- <=>CH3COOH + CH3COO^-}$.

As we can see, the above equation is symmetric. There is no net change in system composition resulting from the above equation and the given molarities.

We also have these equilibria:

$\ce{CH3COOH + H2O- <=>H_3^+O + CH3COO^-}$
$\ce{H2O + CH3COO^- <=>CH3COOH + HO^-}$

But both acetic acid and its conjugate base, acetate ion, are weak. So these won't do much to change the system composition. Plus these aren't the favored reaction in the system; that would be the one above (the one between the strongest acid and strongest base in the system).

If you wanted to solve this system analytically - i.e. you want an even more accurate measure of the system's pH, then you would have to employ the law of conservation of mass and invoke the concept of solution electro neutrality.

Mass conservation:

$1.0 M$ $\ce{=[CH3COOH] + [CH3COO^{-}]}$

Solution electroneutrality:

$\ce{[HO^-{}] + [CH3COO^{-}] = [H_3^+O]}$.

These two statements, coupled with $\ce{K_a(CH3COOH)}$ and $\ce{K_b(CH3COO^{-})}$, and $\ce{K_w}$, give you a system of equations in which you can solve for $\ce{[H_3^+O]}$.


Gibbs Free energy of a reversible reaction From what I understand, it sounds like the Gibbs Free Energy change of a reversible reaction at equilibrium is zero. However, since I know that Gibbs Free energy change depends on temperature, does this not imply that equilibrium can only ever be reached at one very specific temperature? This doesn't sound right, as I know that equilibrium can be established at lots of different reaction temperatures, although the position of equilibrium changes with temperature of course. I asked three chemistry teachers at my school, who all seem just as baffled as I am about this.

In general, any expression for $G$ will by necessity have a dependence on $T$, if only through the explicit incorporation of temperature in $G = H - TS$.

Looking at the Calphad (Calculation of Phase Diagram) community, they go further. Referring to A.T. Dinsdale, "SGTE Data for Pure Elements", CALPHAD 15(4) 317-425 (1991), they state, for the purposes of defining $G$ for the elements:

  The Gibbs energy is represented as a power series in terms of temperature $T$ in the form:
  $ G = a + bT + cT \ln(T) + \Sigma d_{n}T^{n}$

From the definition of $G$, $S$, and $H$ one can then get:

$S = -b-c-c\ln(T)-\Sigma nd_{n}T^{n-1}$ and

$H = a-cT-\Sigma (n-1)d_{n}T^{n}$

So, one sees, given the chosen representation for $G$, that both $H$ and $S$ should depend on $T$ unless $G$ can be represented in a very simple form (only $a$ and $b$ being non-zero). Since Dinsdale uses a more complex power series, one can rest assured that $G$, even for the elements, tends to be rich in temperature dependence.


Recreational drugs by half life I am interested in the Biological Half Life of substances. In the previous wikipedia link, I've already found the half lives of common pharmaceutical products. I can't seem to find similar data for recreational substances, does anyone know these half lives?

Wikipedia reports biological half-lives for several recreational drugs I just tried (cocaine, THC, MDMA). In each case, I went to the page for the particular substance, and Wikipedia keeps an information box at the top of the article containing the pharmacokinetic data. THC's half-life has a citation to the scientific literature, while the other two I checked do not. I think it might be slightly harder in general to get good data for the purely recreational drugs. I would expect drugs without an accepted medical benefit to be less well-studied. To find what's there, though, it looks like you will be able to look up each active ingrediate individually. Depending on what you need, Wikipedia might be okay. If it's not, follow the Wikipedia citation back to the literature if it's there, or try PubMed.


What happens when a piece of copper is placed in 1M HCl? I know that the answer is that nothing happens, but the explanation is confusing me in some places. Firstly, we are given that:
\begin{align}
\ce{Cl2 + 2e- &-> 2Cl-} & (E &= \pu{1.36 V})\\
\ce{Cu^{2+} + 2e- &-> Cu(s)} & (E &= \pu{0.34 V})\\
\ce{2H+ + 2e- &-> H2} & (E &= \pu{0 V})
\end{align}

What's confusing me is that in the solution, we'll have $\ce{H+}$, $\ce{Cl-}$ and $\ce{Cu(s)}$, which are all products in the equations (except for $\ce{H+}$). I get that for the reverse equation (oxidation), the oxidation potential is given by the negative of the reduction potential. What I'm wondering is, if we look at copper and chlorine first, is the only option for the $\ce{Cl-}$ in solution to be oxidised (with $E = \pu{-1.36 V}$)? Also, can the copper also only be oxidised (with $E = \pu{-0.34 V}$)? And if both of these are true, then does that mean that these two can't react at all, not because of the total $E$ being negative, but because they can both only be oxidised?

Can I just clarify, without going into the detail of the specific problem, if you have Cu(s) in solution (like you do here) can it ONLY be oxidised? So essentially you're looking for whether there's anything in solution that can oxidise Cu?

Let us take a longer approach to address your query. For electrochemical problems like these, first you have to make a list of starting materials and possible products. Do a thought experiment even before even looking up the tables. It is long way but it will help you in doing future problems.

Your question is: What happens when a piece of copper is placed in 1M HCl?

Do a thought experiment +  use some basic intuitive chemistry.

Making a list of starting materials, a) Cu metal, b) H$^+$, c) Cl$^-$.

Now there are only two possibilities for each substance.

a) Copper Cu can be oxidized to Cu$^+$ or Cu$^{2+}$ or reduced to Cu$^-$. Basic chemistry tells us that metals like to form cations so Cu$^-$ can be eliminated. Similarly, there is possibility for Cu$^+$ or Cu$^{2+}$. However, say, you are just interested in Cu$^{2+}$.

Now you have the problem well defined: Will Cu -> Cu$^{2+}$?

b) Follow the same reasoning: H$^+$, it can be oxidized to  H$^{2+}$ or reduced to H$_2$. Basic chemistry will tell you that H$^{2+}$ is not possible. Now your question is more well defined: Will H$^+$ -> H$_2$

c) In the same way, ask the same question for chloride ion. It can be oxidized to Cl$_2$ or reduced to  Cl$^{2-}$. Basic chemistry would tell you that Cl$^{2-}$ is not feasible. So your only concern is: Will Cl$^-$ -> Cl$_2$

Now you can ask only two questions:

Can copper metal reduce H$^+$ to H$_2$?

or

Can copper metal oxidize Cl$^-$ to Cl$_2$? This can be easily eliminated because in order for copper to oxidize it must reduce itself further, which is not possible.

At this stage you can utilize the electrode potentials. Recall cathode refers to reduction and anode refers to oxidation
Ecell= E$_{cathode}$-E$_{anode}$

Ecell= E(for hydrogen half cell bc it is being reduced) - E (copper half cell)

Ecell= 0.00- (+0.34) = - 0.34 V

The negative sign shows this not possible under these conditions.

Do the same for chloride ion.

Ecell= E(Reduction of Cu to Cu$^-$) - E (chlorine half cell)
Ecell= Undefined- (+1.36) = Undefined because Cu$^-$ does not exist in solution.

In short nothing will happen to copper in HCl.

P.S. Practically, HCl slowly dissolves copper in the presence of oxygen. This not relevant here.


Determining structure using HNMR and IR I have a homework problem where I'm to draw the structure of a compound of molecular formula $\ce{C10H12O2}$ using the following IR data and $^1$H NMR spectrum:

IR absorption at 1718 cm$^{-1}$

NMR Data:

Triplet (3H) at ~1.4 ppm

Singlet (3H) at ~2.4 ppm

Quartet (2H) at ~4.3 ppm

Doublet (2H) at ~7.2 ppm

Doublet (2H) at ~7.9 ppm

The two doublets lend to a disubstituted benzene. The IR absorption lends to a C=O.

I came up with this structure:

My thoughts were that the singlet owing to $\ce{CH3}$ would be shifted to the left of the other $\ce{CH3}$ due to the C=O.

However, I'm incorrect. This is the correct answer:

What makes this answer more correct than mine (other than the fact that it's an ester and we happen to be studying esters)? Thank you for any help you can offer.

Okay, following up on my comment, I found the spectrum for the "correct" answer:

As you can see in this IR the absorption of the carbonyl peak is shifted down to 1718 as the problem statement claims.

Now, consider this IR spectrum for a conjugated aldehyde which is similar to a conjugated ketone:

In this spectrum, we see that the IR absorption for an aldehyde (normally around 1715) is shifted down below 1700.

Everything else in your answer seems right though. You always need to be careful to consider conjugation when looking at the peak of a carbonyl stretch.

If you want to look around at HNMR/CNMR/IR for basically any reasonable organic molecule, you should check out the SDBS Database. I actually found this to be a useful study tool and just kind of interesting to look around in.


Variation of cone angle with concentration of fuel-air mixture A stoichiometric fuel-air mixture flowing in a Bunsen burner forms a well-defined conical flame. The mixture is then made leaner. For the same flow velocity in the tube, how does the cone angle change?

The cone angle of the Bunsen flame is determined by the flame speed of the mixture and the average flow velocity. To first approximation one can write: $$ \sin \alpha = \frac{S}{U} $$
where $\alpha$ is the half cone angle, $S$ is flame speed and $U$ is flow velocity.

When the mixture is made fuel-lean, depending on how flame speed changes the cone angle will change accordingly.

Usually flame speed for fuel-lean mixture is less than that for stoichiometric mixture. Therefore we can expect the cone angle to decrease.


Stability of branched alkenes I'm having difficulty understand the relative stabilities of the compounds

  2,4-dimethyl-3-methylidenepentane < (3Z)-2,5-dimethylhex-3-ene < (3E)-2,5-dimethylhex-3-ene

which, for simplicity’s sake, can be described as

  1,1-diisopropylethene < (Z)-1,2-diisopropylethene < (E)-1,2-diisopropylethene

The compounds are listed in order of increasing stability.

Also, what does the $\ce{C=C-C}$ bond angle have do to with stability?

The Z isomer is destabilized by comparison to the E isomer due to steric strain. This is a consequence of the isopropyl groups being on the same side of the double bond in the Z isomer, resulting in electron-electron repulsion between the groups. This also, at least partially, accounts for the increase in the bond angle that should be observed in the Z isomer: increasing that angle also increases the distance between the isopropyl groups, minimizing the mutual repulsion between them (admittedly, this is at the expense of the other angles, but the bulkiness of the isopropyls by comparison to hydrogens necessitates that distortion from idealized geometry). A similar effect occurs with the geminal isopropyl groups in the 1,1- isomer.

Additionally, the degree of substitution of an alkene has a dramatic effect on stability. The presence of alkyl groups on the carbons of an alkene is generally stabilizing (as can be empirically confirmed by examining thermodynamic data, i.e., heats of hydrogenation and heats of combustion). Evidently, calculations grounded in molecular orbital theory indicate that the mixing of $\sigma$ type molecular orbitals of alkanes with the $\pi$ orbitals of less-substituted alkenes is typically energy-lowering overall, and that the $\pi^{*}$ antibonding orbitals are sufficiently close in energy to the $\sigma$ orbitals of alkanes that hyperconjugative stabilization occurs (albeit relatively small in magnitude, since the energies are not very close). In the 1,1- isomer, only one carbon of the alkene can be stabilized by hyperconjugation, while both carbons should be able to benefit from it in the 1,2- isomers. (Speculatively, I would also think that the extent of orbital overlap should be poorer in the 1,1- isomer, since the geminal isopropyl groups would likely be staggered with respect to one another, which might prevent proper alignment of the alkyl group orbitals with those of the alkene $\pi$ system.)


Why is a face-centred cubic unit cell not regarded as equivalent to a body-centred tetragonal unit cell? When studying crystal lattices, I decided to learn which combinations of unit cells and lattice systems lead to the 14 Bravais lattices. I learnt that, for example, no centred triclinic unit cell exists.

Later while searching the Internet and Stack Exchange, I found an answer saying that an end-centred cubic cell can be transformed into a (smaller) tetragonal unit cell as shown in th picture below.

Then I thought on and realised that a face centred cubic lattice can be transformed into a (smaller) body-centred tetragonal unit cell, and into an even smaller one, too.

So, why are some centred unit cells (for hexagonal, triclinic and rhombohedral lattices) considered non-existent, while others (e.g. face-centred cubic) are considered to exist even though they, too, can be reduced?

In an FCC lattice  a=b=c while in a body centered tetragonal lattice a=b but c is unconstrained. The first example you gave should be that an end centered tetragonal is equivqlent to a primative tetragonal lattice. There is no end centered cubic lattice as that would violate the minimum cubic symmetry of 3 four fold symmetry axes. however the transition from FCC to a body centered tetragonal lattice would be displacive not transformational.


How much heat is gained/released based off of equation? Using this equation,
$\ce{CaCl2(s) -> Ca^{2+}(aq) + 2Cl^{-}(aq)}$

and $\Delta H = - 325.0 \; kJ $

If you have 15.0 grams of the solid, how much heat is released or gained in kilojoules?

I'm going to assume that the question is presenting the given change in enthalpy as a molar enthalpy change (so as to say that -325.0 kJ are evolved per mole of CaCl2).

Under that assumption, answering the question is then a matter of using the a molar fraction of CaCl2 for the amount you have present.

CaCl2 has a molar mass of 110.98 g/mol so with 15.0 g of CaCl2, you'd have roughly 0.1352 mol of CaCl2. 0.1352 mol * -325.0 kJ/mol = -43.94 kJ. The negative here means that the CaCl2 will release 43.94 kJ of heat.


Why methoxy group acts similar to trifluoromethyl in nucleophilic aromatic substitution In an online test, the following question was asked:

The answer to this question was:

I totally agree with the answer, as $\ce{OCH_3}$ is an EDG, hence it will tend to form meta product in this reaction.

While reading NAS on Master Organic Chemistry, I came up to this article:

This article says that $\ce{OCH_3}$ behaves similar to $\ce{CF_3}$, but I can't figure out the possible reason. Any help is appreciated!

Related (unanswered) question: Nucleophilic substitution in aromatic compounds

Normally during a nucleophillic aromatic substitution reaction one of oxygen's lone pairs would delocalize across the ring, the existence of a triple bond here prevents this from happening so oxygen actually ends up acting like an EWG.


Is this approach to finding density in crystal lattice acceptable? In answering a question in an elementary chemistry text that asks for the density of gold ($\ce{Au}$) based on crystal dimensions and weight per atom, I found the approach I used was a little different from that of the book and wanted to see if this is reasonable.

Gold crystal cells are face-centered cubes of a given length edge and a given weight per atom. The book reasons that since each of the 8 corner atoms is shared by 8 cells and each of six face atoms is shared by 2 cells, each cell adds $8/8+6/2 = 4$ atoms to a crystal structure - which gives the correct density.

My approach was to count all the atoms in a cube of cells and disregard sharing, since as long as the count was accurate the average number of atoms per cell would be correct in the limit.

For example: in a cube of 8 cells, there are $3^3$ corner atoms and $3\cdot 2^2(2+1)$ face atoms. So the general formula is $(n+1)^3+3n^2(n+1)$ atoms in an $n \times n$ cube. But in $n^3$ cells the ratio $r$ of atoms per cell to cells is

$$r=\lim_{n\to\infty} \frac{(n+1)^3+3n^2(n+1)}{n^3}=4$$

  So this also gives the correct density, and my question is whether there is anything wrong with this approach?

(In an earlier version of this question I failed to count all the face atoms; hopefully this is easier to follow.)

[Edited to correct formula.]

Yes of course, if you do the math right, that limes gives the exact same number.

Don't forget to look out for atoms that are completely within the unit cell.


Boilng vodka for A/C cleaning I hope this doesn't sound too strange.
I have plenty of cheap vodka and want to clean the A/C on my car.

So I thought about boiling 1/2 litre of vodka in my car while I ran the A/C with recirculation on.
The alcohol should get liquid within the air conditioner and desinfect and clean the vaporizer.

I am just not sure if I create an explosive mixture inside the car this way.

David

Alcohol vapors can certainly form explosive mixtures with air, so filling the car with ethanol vapors is potentially dangerous. A source of ignition, for example a spark, could under the right circumstances ignite the vapors. If the fan motor of the AC system is a brushed motor it can create small sparks. So I would say there is a fire hazard.
But there are other problems with your suggested setup. The ethanol and water vapors would condense on every surface of your car's interior, the windows, the roof, the doors, etc. This would mean you loose a lot of alcohol, so you would need more. Ethanol might attack and start to dissolve some of the interior, glued parts may become loose. And of course the car would smell very strongly like vodka for a long time.
So I would suggest disassembling the dash cluster and whatever else you need to get to the AC system, and using the vodka in liquid form to clean it properly.


Why 2p Orbitals overlap more effectively than 3p orbitals? So, for example, looking at HF and HCl, we see that the 1s orbital of the H overlap with the 2p orbital of F more effectively. Which argument can i use to explain that? Density?

As the quantum number increases, the size of the orbital, or more precisely, the area in which electrons are found, increases. The same 1s orbital of H combines with greater overlap with 2p of F,  while the larger 3p of Cl can’t be approached to that extent because of its size.
For the sake of demonstration, consider a child shaking hand with the other child, which is a full overlap of palms, while a child shaking hand with an adult is not!


deduce chemical properties of a compound from elements If we know the properties of two elements e.g Oxygen and Hydrogen.

We know the following properties

sound speed
boiling point
melting point
refractive index
etc.

given that information, is it possible to deduce the properties of a compound such as water H2O without resorting to chemical experiments? i.e deduce its boiling point

is it possible to deduce the properties of a compound such as water H2O without resorting to chemical experiments?

Not yet, assuming you mean extrapolating such information from the molecule's structure

Physical and chemical properties are hard enough to extrapolate from a compound's structure, let along its composition. The entire field of theoretical chemistry has been working on predicting the structure-property relationship for several decades. Great strides have been made, but accurate physical property prediction has not been achieved (though I would love to be proven wrong). In most cases, very accurate computation of properties of all but the smallest molecules takes more computing time than you would need to do the experiment.

Your question implies an interest in deriving such properties from the formula. As Canageek said, this is not possible, given the phenomenon of isomerism. One example I just used with my class is that there are 22 known molecules sharing the formula $\ce{C6H12}$, which is a relatively simple formula. Two of the more common are1-hexene and cyclohexane, which you can see have very different structures and properties (chemical and physical).


Perchlorate salts: (explosive) reactivity with methane? I read everywhere about the explosive reactivity of perchlorate salts (e.g., $\ce{Na}$, $\ce{Ca}$, $\ce{NH4}$ [especially]) when mixed with organic materials or solvents, yet I find no information about its reactivity with methane.

Can anyone comment please?

In general, explosion of alkali metal perchlorates (at least in small quatities) is deflagration, with the perchlorate providing oxygen for combustion of another component of a mixture, rather than detonation (due to decomposition of the perchlorate). However, perchloric acid itself and other inorganic and organic perchlorates such as cis-bis-(5-nitrotetrazolato)tetraminecobalt(III) perchlorate can transition from deflagration to brissant detonation.

So, to answer your specific question, gaseous methane would burn more readily in the presence of Group I perchlorates, but the flame-front would be limited by the small contact area and low concentration of methane at the solid/gaseous interface. On the other hand, if one were to synthesize methyl perchlorate or  ethyl perchlorate, one might not be able to perform subsequent syntheses.

So, to avoid maiming or perhaps a terminal experiment, eschew perchlorates.