Enthalpy definitions. What are their main differences? The definition of enthalpy is properly described in here.

The principal objective of this question is mainly academic. In order to any students with internet connection can find a definition and an example of the enthalpy concept applications.

What are the main differences of:

Standard enthalpy
Enthalpy of atomization
Enthalpy of formation
Enthalpy of hydration
Enthalpy of reaction
Enthalpy of combustion
Enthalpy of solution
Enthalpy of lattice dissociation
Enthalpy of lattice formation
Enthalpy of mixture
Enthalpy of excess mixture
Enthalpy of (any phase transition)
Law of Hess
Law of Kirchhoff

Any other definition not listed here

Explanation of notation:
$H$ is the enthalpy of the system.
$\Delta$ means change of, so $\Delta H$ means change of the enthalpy

This symbol means standard condition, standard condition is defined as a pressure of $100\text{kPa}$ and reactants and products are in their standard state, or concentration of solutions are $1\text{M}$. However, since LaTeX doesn't have this symbol, I'll substitute it for $^\ominus$.
Enthalpy changes:
Enthalpy of dilution – The enthalpy change when a solution containing one mole of a solute is diluted from one concentration to another.
Enthalpy of ($n\text{th}$) electron affinity – The enthalpy change when $n$ electrons are added to one mole of gaseous atoms.
$$\ce{Li(g) + e-(g) -> Li-(g) +60\ \text{kJ}}$$
$$\ce{F(g) + e-(g) ->  F-(g) +328\ \text{kJ}}$$
Enthalpy of ($n\text{th}$) ionization – The enthalpy change when $n$ electrons are removed from one mole of gaseous atoms. It is always positive.
$$\ce{Li(g) +520\ \text{kJ} -> Li+(g) +e-(g)}$$
$$\ce{He(g) +2372\ \text{kJ} -> He+(g) +e-(g)}$$
Enthalpy of lattice dissociation – The enthalpy change when one mole of an ionic lattice dissociates into isolated gaseous ions.
An example for sodium chloride which have an enthalpy of lattice dissociation, also known as lattice energy, of $787\ \mathrm{kJ/mol}$
$$\ce{NaCl(s) +787\ \text{kJ} -> Na+(g) + Cl-(g)}\qquad\Delta H=787\ \text{kJ/mol}$$
Enthalpy of lattice formation – The enthalpy change when one mole of solid crystal is formed from its scattered gaseous ions.
$$\ce{Na+(g) + Cl-(g) -> NaCl(s)} +787\ \text{kJ}\qquad\Delta H=-787\ \text{kJ/mol}$$
Which means Enthalpy of lattice dissociation$=-$Enthalpy of lattice formation
Enthalpy of hydration($\Delta_\text{hyd}H^\ominus$) – Enthalpy change when when one mole of ions undergo hydration.
Enthalpy of mixing – The enthalpy change from a substance when mixed.
Enthalpy of neutralisation – The enthalpy change when an acid is completely neutralised by a base.
For a strong acid, like $\ce{HCl}$ and strong base, like $\ce{NaOH}$, they disassociate almost completely $\ce{Cl-}$ and $\ce{Na+}$ are spectator ions so what is actually happening is $$\ce{H+(aq) + OH-(aq) -> H2O(l) + 58\ \text{kJ/mol}}$$
However, using a weak acid/base will have a lower enthalpy of neutralisation as normally most of the acid/base does not disassociate.
For example, mixing ethanoic acid and potassium hydroxide only has a enthalpy of neutralisation of $-11.7\ \text{kJ/mol}$
Enthalpy of precipitation – The enthalpy change when one mole of a sparingly soluble substance precipitates by mixing dilute solutions of suitable electrolytes.
Enthalpy of solution – Enthalpy change when 1 mole of an ionic substance dissolves in water to give a solution of infinite dilution.
Enthalpy of solution can be positive or negative as when a ionic substance dissolves, the dissolution can be broken into three steps

Breaking of solute-solute attraction (endothermic)
Breaking solvent-solvent attraction (endothermic), eg. hydrogen bonds, LDF
Forming solvent-solute attraction (exothermic)

An example of a positive enthalpy of solution is potassium chlorate which has an enthalpy of solution of $41.38\ \text{kJ/mol}$
Enthalpy of (Solid$\rightarrow$Liquid: $\Delta_\text{fus}H^\ominus$, Liquid$\rightarrow$Solid:$\Delta_\text{freezing}H^\ominus$, Liquid$\rightarrow$Gas: $\Delta_\text{vap}H^\ominus$, Gas$\rightarrow$Liquid: $\Delta_\text{cond}H^\ominus$, Solid$\rightarrow$Gas: $\Delta_\text{sub}H^\ominus$, Gas$\rightarrow$Solid: $\Delta_\text{deposition}H^\ominus$) – The enthalpy change from providing energy, to a specific quantity of the substance to change its state.
$$\Delta_\text{fus}H^\ominus=-\Delta_\text{freezing}H^\ominus,\Delta_\text{vap}H^\ominus=-\Delta_\text{cond}H^\ominus,\Delta_\text{sub}H^\ominus=-\Delta_\text{deposition}H^\ominus$$
Standard enthalpy of atomization($\Delta_\text{at}H_T^\ominus$) – Change when a compound's bonds are broken and the component atoms are reduced to individual atoms at at $T^\circ K$.
$$\ce{S_8 -> 8S}\qquad\Delta_{at}H^\ominus=278.7\ \text{kJ/mol}$$
Standard enthalpy of combustion($\Delta_\text{c}H_T^\ominus$) – The enthalpy change which occurs when one mole of the compound is burned completely in oxygen at $T^\circ K$ and $10\ \mathrm{kPa}$.
$$\ce{H2(g) +\frac{1}{2}O2(g) -> H2O(g)}+572\ \text{kJ}\qquad\Delta_cH^\ominus = -286\ \text{kJ/mol}$$
Standard enthalpy of formation($\Delta_\text{f}H_T^\ominus$) – Change in enthalpy during the formation of one mole of the compound from its constituent elements, with all substances in their standard states, and at a pressure of $100\ \mathrm{kPa}$ at $T^\circ K$.
It can be calculated using Hess's law if the reaction is hypothetical. An example is methane, $\ce{C}$ and $\ce{H2}$ will not normally react but the standard enthalpy of formation of methane is determined by Hess's law to be $-74.8\ \text{kJ/mol}$
$$\ce{\frac{1}{2}N2(g) +\frac{1}{2}O2(g) -> NO(g)}\qquad\Delta_\text{f}H^\ominus=90.25\ \text{kJ/mol}$$
Standard enthalpy of reaction($\Delta_\text{r}H^\ominus_T$) – Enthalpy change that when matter is transformed by a chemical reaction at $T^\circ K$ and $10\ \mathrm{kPa}$.
$$\ce{H2(g) +\frac{1}{2}O2(g) -> H2O(g)} +572\ \text{kJ}\qquad\Delta_\text{r}H^\ominus = −572\ \text{kJ/mol}$$
$$\Delta_\text{r}H^\ominus=\sum H^\ominus_\text{products}-\sum H^\ominus_\text{reactants}$$
$$\Delta_\text{r}H_\text{forward}=-\Delta_\text{r}H_\text{backwards}$$
Laws:
Le Chatelier's Principle – When an external change is made to a system in dynamic equilibrium, the system responds to minimise the effect of the change.
Yellow $\ce{Fe^3+}$ reacting with colorless thiocyanate ions $\ce{SCN-}$ to form deep red $\ce{[Fe(SCN)]^2+}$ ions:
$$\ce{Fe^3+(aq) + SCN-(aq) -> [Fe(SCN)]^2+(aq)}$$
When $\ce{NH4Cl}$ is added, $\ce{Cl-}$ reacts with $\ce{Fe^3+}$ ions to form $\ce{[FeCl4]-}$ ions. By the Le Chatelier's Principle, when $\ce{Cl-}$ is added into a solution of deep red $\ce{[Fe(SCN)]^2+(aq)}$ ions, the equilibrium will shift to $\ce{Fe^3+(aq) + SCN-(aq)}$, turning the solution pale red.
Kirchhoff's Law – Enthalpy of any substance increases with temperature.
Hess's Law – Total enthalpy change of a chemical reaction is independent of the number of steps the reaction takes.
Henry's Law – Amount of dissolved gas is proportional to its partial pressure in the gas phase.
Processes:
Constant Entropy – Isoentropic
Constant Pressure – Isobaric
Constant Volume – Isovolumetric
Side note: This is probably not a complete list as I may have missed some


Do electron donating groups make a molecule nucleophilic? Do electron donating groups make a molecule nucleophilic?
One group, the alkyl group in this example, is electron donating in $\ce{CH3CH2O-}$. Will the molecule be nucleophlic?

Do electron donating groups make a molecule nucleophilic?

Sometimes.

Sure, electron donating groups can make a molecule more nucleophilic by virtue of increasing the negative charge density on the nucleophilic atom.

However, nucleophilicity is a kinetic property, and having a lot of electron donating groups attached can significantly affect the ability of the nucleophile to approach the electrophile. This reduces nucleophilicity. An example would be tert-butoxide. Yes, oxygen with a negative charge is generally nucleophilic, and yes, the tert-butyl group is an inductive donor, but tert-butoxide is not very nucleophilic because of steric hinderance.

  One group, the alkyl group in this example, is electron donating in $\ce{CH3CH2O-}$. Will the molecule be nucleophlic?

Maybe. Nucleophilicity also depends on the solvent. A charged ion such as ethoxide ion would be well-solvated in a polar solvent, such as water, and less well-solvated in a non-polar solvent, such as hexane.

Solvation impacts reaction kinetics, and in turn, impacts nucleophilicity. Consider the halides for example. In polar protic solvents, iodide ion is the strongest nucleophile. Protic solvents are not able to solvate iodide ion as well as they can solvate the fluoride ion, which has a much higher negative charge density. In polar aprotic solvents, however, fluoride ion is generally a stronger nucleophile than iodide ion because fluoride ion is not as strongly solvated and it also has a great negative charge density.

In sum, nucleophilicity is not an inherent property. It depends on a variety of factors - both external and internal to whatever molecule or ion that you are considering.


Why is the difference between atomic radii of Na and K greater than that of Rb and Cs? I am aware that this question has already been asked here: Why does the "gap" between atomic radii get smaller down the periodic table?.

However, I was wondering if there is a way to answer this question using simple periodic trends instead of a formula to calculate atomic radius.

Since there is not only an s and p-block between Rb and Cs, but also a d-block, wouldn't the increase in shielding between Rb and Cs be higher than that of Na and K? Thus, shouldn't Rb and Cs have a larger difference in atomic radii?

I got this question in my chemistry class while learning about periodic trends, and it apparently also appeared on the 1993 AP Exam, so I would think that this problem could be explained using simple periodic trends.

Electronic orbitals do not come in fixed sizes.  A given orbital will shrink due to greater nuclear charge as more protons are addded to the nucleus.  Going from $\ce{Rb}$ to $\ce{Cs}$ you are adding more orbitals than when you go from $\ce{Na}$ to $\ce{K}$.  But you are also adding more orbital-shrinking protons to the nucleus.

The shrinkage effect on each orbital with more protons gets stronger as we go to heavier atoms, when relativistic effects kick in.  Put in very rough terms: when you solve the Schroedinger equation for electronic wavefunctions the size of the orbital would decrease if the mass of the electron were increased.  In nonrelativistic quantum mechanics this does not occur, but in relativistic mechanics the energy needed to squeeze the wavefunctions (which in this case comes from electrostatic attraction) acts like mass, and thus allows for even more shrinkage.  For alkaline earth metals we end up with almost no change at all going from $\ce{Ba}$ to $\ce{Ra}$.  See https://periodictablegroups.wordpress.com/alkaline-earth-metals-2/.


Is there a point where further dilution of ethanol is not exothermic? Diluting concentrated ethanol with water is exothermic. Diluting 95% refined spirit into 40% vodka makes the resulting product definitely warm. Dilution of 60% absinthe with water that is cool, but warmer than "nearly freezing" results in a lukewarm, unappetizing drink. I haven't noticed the effect when diluting weaker alcohols though.

Is there a threshold of concentration where further dilution of water-ethanol mix doesn't produce heat? Or is it merely an asymptotic relation, where at lower concentrations the effect becomes hard to notice?

Mixing ethanol and water is exothermic all over the range at about room temperature. Data are available, e.g., here excess enthalpy ethanol-water

However, at relatively high temperature and at least around azeotropic composition,  the mixing should become endothermic otherwise the ethanol-water azeotrope should boil at higher temperature instead of having Bp of 78.1 °C.


Rate of reaction- Arrhenius equation Can the Arrhenius equation be used if the reactants are in the solid state and if I can use it for any state, how do I do so?

Clear distinction has to be made between Arrhenius-equation and transition state theory. The formulas for TST can be derived, as opposed to Arrhenius equation, which by itself, is not based on a solid physical derivation. Basically any process can be modeled by an Arrhenius equation that has the formula:

$$ r \propto \exp{(-E_a/RT)} $$

Any molecular process can be reasonably approximated by this formula if it is a statistical transition between states with different energy. The reason for this is that the statistical probability of a state if proportional to the Boltzmann factor $g \exp{(-E/RT)} $, which is readily Arrhenius-like.

So what are some examples for Arrhenius-type behaviors? For example, if you look up the term Arrhenius-type equation on ScienceDirect you can see that research papers have been published on solid strain rates, membrane permeabilities, solubility rates and corrosion.

So, as you can see, an Arrhenius-type equation can describe a huge number of physical/chemical processes, so it is quite reasonable to use it for solid state reactions. The question is only how to find the energy of activation and the pre-exponent factor. For this, there is no general recipe. The best thing to do is to use them as empirical model parameters to be measured by experiments.


MO scheme for coordination complexes (tetrahedral) - Heteroligand vs. homoligand How is the MO diagram drawn for a tetrahedral complex with heteroligands (e.g. $\ce{[NiCl2(PPh3)2]}$) different from a tetrahedral complex with the same type of ligand? Do we consider the same "a1 and t2" SALCs as in the case of same ligand type tetrahedral?

Bis(triphenylphosphine)nickel chloride can have be tetrahedral or square planar depending on conditions. In any case the SALCs for a tetrahedral complex with heteroligands will depend on the point group of the complex. For [NiCl2(PPh3)2], the point gorup is C2v and not Td like for same ligand tetrahedral. You need to check the character tables


In Statistical Transition State Theory, why is the concentration of the transition state for the forward and reverse reaction identical? For example on Wikipedia, and also in my lecture notes, it is assumed that for a reaction
$$\ce{A + B<=> AB^{\ddagger} <=> P}$$
at equilibrium,
$$[\ce{AB^{\ddagger}_f}] = [\ce{AB^{\ddagger}_r}],$$
but I don't see how the equation follows from the assumption, since there is no direct relation between transition state concentration and total reaction rate. And if it is, how can they simply be equal instead of depending on some constants?

In transition state theory, it is assumed that the forward and the reverse reaction occur via the same transition state. This is implied by assuming a quasi-equilibrium between the reactants and the products, which adheres to a common Boltzman statistics. Therefore the labelling $\ce{AB^\ddagger_r}$, $\ce{AB^\ddagger_f}$ is artificial, because it is the same transition state:
$$\ce{AB^\ddagger_r} = \ce{AB^\ddagger} = \ce{AB^\ddagger_f},$$
and therefore the concentration is identical, too:
$$[\ce{AB^\ddagger_r}] = [\ce{AB^\ddagger}] = [\ce{AB^\ddagger_f}].$$


How do I utilize Henry's Law Constant? I have tried for several hours to answer this question with no luck. I believe I may be missing a couple of steps or I am not utilizing the given information properly. Nonetheless, some guidance will be greatly appreciated.

The concentration of $\ce{N2}$ in the ocean at 25C is 445 $\ce{\mu}$M. The Henry's Law constant for $\ce{N2}$ is 0.61 x $\ce 10^{-3}$ mol $\ce L^{-1}$ atm $\ce L^{-1}$.

Part A:

Calculate the mass of $\ce{N2}$ in a liter of ocean water.

4.45 x $\ce 10^{-4}$ mol/kg = molality

$\ce{N2}$ in moles = 28.014 g/mol

Using Molality Equation

Molality = $\frac{amount of solute (mols)}{mass of solvent(kg)}$

I want kilograms, thus:

4.45 x $\ce 10^{-4}$ $\frac{mol}{kg}$ = $\frac{28.014 g}{mol}$

Where do I go from here?

Part B:

Calculate the partial pressure of $\ce{N2}$ in the atmosphere

Use C = K x $P_(gas)$

4.45 x $\ce 10^{-4}$ = (0.61 x $\ce 10^{-3}$ mol $\ce L^{-1}$ atm $\ce L^{-1}$)x

x = 0.73 atm

Part A: Why do you want to use the molality?! You have the concentration i.e the number of moles in one liter. So the mass of nitrogen in one liter is simply:

Number of moles in one liter $\times$ molar mass.

Part B: It's correct.


Resonance of lone pair of nitrogen in 4‐nitroaniline and piperidin‐2‐one While working out some questions on comparing the basic nature of various compounds, I came across the following two compounds:

In case of A I could see several resonating structures, and also the −M effect of the nitro group. So justifying based on the number of resonance structures, I could tell that the lone pairs are more delocalized in A than in B. Hence B is more basic.
However, on verifying my answer I found that A was more basic.
The only reason I could come up with is that in compound A, majority of the resonance includes a negative charge on carbon, but in compound B, the negative charge lands on the more electronegative oxygen, which better succeeds in stabilizing it.
But it doesn't seem right for me.
Can anyone please tell me the correct reason for this?

Considering that you're referring to the difference in the number of resonating structures, you're right in that aspect. 4‐Nitroaniline does have more resonating structures than piperidin‐2‐one. But in organic chemistry, you should be knowing that for a phenomenon to happen, there have to be several contributing factors.
While this includes the number of resonating structures as one, you have to take into consideration the other factors as well. And this includes the other factor you have mentioned in your answer — the position of the negative charge.
Since carbon is an atom which has an electronegativity of 2.55 (According to the Pauling scale), a negative charge on it would make it highly unstable, lowering the resonance energy, thus reducing the resonance stabilization in it. Nitrogen and oxygen have electronegativity of 3.04 and 3.44, respectively, thus a negative charge on those atoms would increase the resonance energy, thus increasing the resonance stabilization.


How does one recognized Td/Oh symmetry in molecules? The identification of point groups of a molecule is usually done following a strict scheme, either manually or algorithmically. In all textbooks I could find, however, the first step of the scheme is actually not explicited: in this example scheme drawn from Housecroft and Sharpe,

you can see that there is a very unhelpful step “Does this molecule have $I_h$, $O_h$ or $T_d$ symmetry?”. It is implied that one would recognize e.g. an icosahedral molecular compound on sight. However, I wonder: how can one strictly identify these “special” point groups? What set of rules is to be followed (again, either manually or algorithmically)?

You can detect $O_{h}$, $I_{h}$ and $T_{d}$ symmetry by checking that a molecule has all of the subgroup symmetries of these point groups.

According to this untitled document, which I presume1 is by W.C. Trogler, the elements are as follows:

$T_{d}$: $E$, $4C_{3}$, $3C_{2}$, $3S_{4}$, $6\sigma{_d}$

$O_{h}$: $E$, $3C_4$, $4C_3$, $6C_2$, $4S_6$, $3S_4$, $i$, $3\sigma{_h}$, $6\sigma{_d}$

$I_{h}$: $E$, $6C_{5}$, $10C_{3}$, $15C_{2}$, $i$, $6S_{10}$, $10S_6$, $15\sigma$

Obviously, you don't need to check for the identity symmetry $E$.

If you're a visual sort of person a symmetry spot check is to mentally superimpose a tetrahedron, cube or dodecahedron over the molecule and see whether the view down the surface normal of each face is identical. Cubes and octahedra are duals of each other, as are dodecahedra and icosahedra. The tetrahedron is self-dual.

Interestingly, H&S does not list chiral forms of these point groups, probably because they are so rarely encountered, however researchers have come up with molecules that obligately satisfy $T$, $I$ and $O$ symmetry2 (I have not yet read the paper).

(1) I would be indebted to anyone who can furnish me with a full citation to this work, and confirm the authorship.

(2) Narasimhan, S. K., Lu, X. and Luk, Y.-Y. (2008), Chiral molecules with polyhedral T, O, or I symmetry: Theoretical solution to a difficult problem in stereochemistry. Chirality, 20: 878–884.


How to calculate S² value of a broken-symmetry wave function? $S$ represents spin, signifies the number of unpaired electrons in the system. For example, if the number of unpaired electrons is $1$, then $S=1/2$. $S^2$ is calculated as $S(S+1)$.

From what I have read, the $S^2$ value of a broken-symmetry singlet (contaminated by a triplet) is $1.0$, which is calculated to be the average of the singlet and triplet $S^2$. Similarly, the $S^2$ of broken-symmetry doublet (contaminated by the quartet state) turns out to be $1.75$ (average of a doublet and a quartet). I would like to know how these average values are being calculated. I understand that these are weighted averages (as $1.75 \neq 0.5\cdot(0.75+3.75)$, where $0.75$ and $3.75$ are $S^2$ values of doublet and quartet, respectively), but I don't know how that weighting is being done. I would be grateful if someone could provide a detailed explanation (mathematical derivation)  of this.

The spin operators for the $p^\textrm{th}$ MO are defined as follows:
\begin{align}
S_{x,p} &= \frac{\hbar}{2} \sum_{\tau,\tau'} c_{p,\tau}^\dagger \hat{\sigma}^x_{\tau,\tau'} c_{p,\tau'}, \\
S_{y,p} &= \frac{\hbar}{2} \sum_{\tau,\tau'} c_{p,\tau}^\dagger \hat{\sigma}^y_{\tau,\tau'} c_{p,\tau'}, \\
S_{z,p} &= \frac{\hbar}{2} \sum_{\tau,\tau'} c_{p,\tau}^\dagger \hat{\sigma}^z_{\tau,\tau'} c_{p,\tau'}, \\
\end{align}
where $\sigma_{\tau,\tau'}^{x}$ denotes the matrix elements of the Pauli matrix $\sigma^x$ and so on. Any spin-observable can then be constructed from these basic blocks. For instance, $S^2$ operator takes the form
\begin{equation}
S^2 = \sum_{p \in MO} \frac{3}{4}\left( c_{p,\uparrow}^\dagger c_{p,\uparrow} +
c^\dagger_{p,\downarrow} c_{p,\downarrow} - 2c^\dagger_{p,\uparrow} c^\dagger_{p,
\downarrow} c_{p,\downarrow} c_{p,\uparrow} \right )
\end{equation}
All one need is to evaluate the expectation value of these operators over the wavefunction of interest.


How to find the band gap for a compound? I have processed a si/polymer hybrid using a silane coupling agent which I need to determine the energy gap / band gap.  Would you please let me know the procedure to go about determining the band gap?

There are two ways in which the band gap of a material: using optical spectroscopy and using electrochemistry. The way you go about analyzing the data to determine the band gap is independent of material, but the process of collecting the data is very specific to your material, and you should consult papers where others have made similar measurements on similar systems.

Optical spectroscopy.

You need to collect an absorbance spectrum of your material. Given what I can gather from your fabrication description, you will probably need to collect said spectrum in the solid state. The exact means of doing that vary somewhat by spectrometer model. Find someone who knows your spectrometer very well and ask them how to do it (or read the manual).

The absorbance process measures the energy difference between the ground state and the excited state as electromagnetic radiation is absorbed. Since there are a number of vibrational states associated with both the ground state and the excited state, the absorbance spectrum appears as a blob instead of a nice line. The "true" band gap, or $E_{00}$, the energy difference between the lowest vibrational states of the ground and excited states, is difficult to measure. However, the most commonly accepted approximation is the onset of absorption from the low energy side. The onset value is the intersection of the extrapolations of the linear parts of the spectrum. See the image below.

Once you have the wavelength (x) at that point, you can convert it into energy using the Planck relation:
$$E(\text{J})=\dfrac{hc}{\lambda(\text{m})}$$
If you want your energy in electronvolts, you can convert directly:
$$E(\text{eV})=\dfrac{1239.84187 \text{ eV nm}}{\lambda(\text{nm})}$$

Electrochemistry, specifically voltammetry

Voltammetry gives you the ability to determine the peak oxidation potential, $V_{\text{ox}}$, and the peak reduction potential, $V_{\text{red}}$, of your material. Both cyclic voltammetry and differential pulse voltammetry can give you these data, and both should work on solids, though you may want to find someone who has experience in such things.

The assumption that we operate under when using electrochemistry to determine HOMO-LUMO or bad gap, is that when we oxidize a molecule or material, we take the first electron out of the HOMO (or valence band). When we reduce the molecule or material, we put an electron into the LUMO (or conductance band). Thus, $V_{\text{ox}}$ tells us something about the HOMO (valence band) and $V_{\text{red}}$ tells us something about the LUMO (conductance band). The energy gap (in electronvolts) is simply the difference between the two potentials (in volts) multiplied the charge on the electron $e=-1 \text{ eV/V}$.
$$E(\text{eV})=e(V_{\text{red}}-V_{\text{ox}})$$


Nomenclature of a branched alkane Here I need to give the systematic name of this molecule so I decided to first find the longuest chain of carbons which seems to be the one I put in red. Then I have 2 ethyls at 3 and 4 and I have an isobutyl at 6 so would the name be

3,4-diethyl-6-isobutyldecane ?

The name ‘3,4-diethyl-6-isobutyldecane’ that is proposed in the question is unambiguous and describes the correct structure. However, it corresponds to obsolete IUPAC recommendations.

The retained prefix ‘isobutyl’ was still contained in the 1993 recommendations but is explicitly no longer recommended in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).

According to current IUPAC recommendations, the preferred prefix for the group $\ce{(CH3)2CH-CH2-{}}$ is ‘2-methylpropyl’.

Therefore, the preferred IUPAC name (PIN) for the compound given in the question is 3,4-diethyl-6-(2-methylpropyl)decane.


Why are sp hybridized carbocations high energy? You don’t see carbocations on double bonds very much, and here's a good reason: compared to sp3, there is more s character in the orbitals, so the empty orbital is held more closely to the nucleus. This is destabilizing if we’re talking about positive charge.

(source)

This doesn't really make sense. Positive charge arises from the lack of electrons. How does holding an empty orbital (regardless of whether that makes any physical sense) closer affect your stability if the empty orbital is ... empty?

So, what is the exact reason sp2 (aren't most carbocations sp2 though?) and sp carbocations are generally less stable than their sp3 counterparts?

Also, how can we have an sp3 carbocation? Huh?!

The only reason I can think of is that it's energetically expensive to make an sp carbocation ... pulling electrons which are held more tightly to the nucleus away is probably harder than pulling away electrons which are not as close to the nucleus.

I can't think of any reason why the end products would be any less/more stable ... if anything an sp carbocation should be more stable than an sp2 carbocation in that in the sp carbocation the electrons are held closer to the nucleus and therefore better stabilizing the nucleus.

Let's just tackle the obvious part first: There is no sp3 carbocation. One reason for this is the same, as for why higher substituted carbocations are more stable than lower substituted. Electrons are more stable in orbitals with high s-character.
In $\ce{\overset{\oplus}{C}(CH3)3}$ the cationic carbon is sp2 hybridised while the empty orbital is 100% p character. All s character is used in occupied orbitals.
In $\ce{H2C=\overset{\oplus}{C}-CH3}$, one would assume from the common depiction, that the central cationic carbon is also (roughly) sp2 hybridised. My first draft of answering did also assume this. A quick DF-BP86/def2-SVP calculation revealed something different. The $\angle(\ce{CCC})$ bond angle comes out to be almost exactly $180^\circ$. The $\pi$ bond is obviously only of p character. The underlying $\sigma$ bonds can now contain the whole s character, making it sp hybridised, and as a result the p orbital of the cationic carbon atom is empty. There will be interactions with some carbon hydrogen bonds. (Hyperconjugation, see below.) This effect is most likely much less than in the prior case, causing it to be less stable. In addition to this, the molecule is much more constrained in its geometry. This linear arrangement is probably also only true for this very simple arrangement. Dependent on the moieties attached to it, a bent structure might be more stable and a different bonding angle will be observed. When that happens, there will be some s character in the empty orbital.
In the benzyl cation, due to symmetry restrictions, $D_\mathrm{2h}$, the empty orbital must be almost exactly sp2. Hence the underlying $\sigma$ bonds will roughly take away 67% s character. The empty orbital will therefore have a very high s character, about 33%, as a result. There is almost no bond widening possible to decrease this and successively increase s character in the $\sigma$ bonds. So the s character found in empty orbitals effectively destabilised the remaining occupied orbitals. (One could also think about it as some kind of ring strain.)
In $\ce{H3C-C#\overset{\oplus}{C}}$, symmetry is even higher, $C_\mathrm{\infty{}v}$ for the triple bond. The underlying $\sigma$ bond will have slightly more s character than in an ideal sp orbital. The empty orbital will, however, be binding a lot of s character. This is effectively destabilising the occupied orbitals again. Since the empty orbital contains the most s character in the orbital, it will also have the most destabilised $\sigma$ system of the series.

In summary, the more s character is in unoccupied orbitals, the less s character can be in occupied orbitals. This is destabilising these orbitals, causing the carbocation also to comparably less stable.
The quote expresses this feature in a quite confusing way.

Note
There are of course other reasons for higher substituted carbocations to be more stable than lower. One of the main effects is hyperconjugation of the $\sigma(\ce{C-H})$ bond with the empty p orbital.


Bond angles in sulfate My chemical bonding professor says that the $\ce{O-S-O}$ bond angles in $\ce{SO4^2-}$ are ideal (109.5°).

Why? Is this because all bonds are equivalent, and electron distribution is shared among all oxygens?

Yes. All the bonds are equivalent as in the 4 resonating structures, each with 1.5 S and O bonds. So the resonant hybrid has four 1.5 double bonds (sigma and pi) with -0.5 formal charge on each O. This is why, the ion has a perfect tetrahedral geometry and shape as it is very symmetric and equivalent.
Hope it helps.

The picture below on the right is what is thought to be the most accurate representation nowadays. However, the resonance structure on the left (and all of its equivalent representations) are more than likely acceptable for introductory chemistry classes.

Note that no matter which resonance structure you think is a better representation of the sulfate ion, the bonds are all equivalent and therefore sulfate ion has the ideal tetrahedral bond angle of 109.5 degrees.


What is Bent's rule? I'm all bent out of shape trying to figure out what Bent's rule means. I have several formulations of it, and the most common formulation is also the hardest to understand.

  Atomic s character concentrates in orbitals directed toward electropositive substituents

Why would this be true? Consider $\ce{H3CF}$.

Both the carbon and the fluorine are roughly $\ce{sp^3}$ hybridized. Given that carbon is more electropositive than fluorine, am I supposed to make the conclusion that because carbon is more electropositive than fluorine, there is a great deal of s-character in the $\ce{C-F}$ bond and most of this s-character is around the carbon?

Or is this a misunderstanding of "orbitals directed toward electropositive substituents"? The fluorine is $\ce{sp^3}$ hybridized and these orbitals are "directed" toward the carbon in that the big lobe of the hybrid orbital is pointing toward the carbon. So does electron density concentrate near the fluorine? Because that would make more sense.

And this s-character concentrated toward the fluorine has the effect of what on the bond angle? I understand that the more s-character a bond has, the bigger the bond angle - consider $\ce{sp}$ vs $\ce{sp^2}$. But since the $\ce{C-F}$ bond now has less s-character around the carbon, the $\ce{H-C-F}$ bond angle can shrink, correct?

That's a good, concise statement of Bent's rule.  Of course we could have just as correctly said that p character tends to concentrate in orbitals directed at electronegative elements. We'll use this latter phrasing when we examine methyl fluoride below. But first, let's expand on the definition a bit so that it is clear to all.

Bent's rule speaks to the hybridization of the central atom ($\ce{A}$) in the molecule $\ce{X-A-Y}$.

$\ce{A}$ provides hybridized atomic orbitals that form $\ce{A}$'s part of its bond to $\ce{X}$ and to $\ce{Y}$.  Bent's rule says that as we change the electronegativity of $\ce{X}$ and \ or $\ce{Y}$, $\ce{A}$ will tend to rehybridize its orbitals such that more s character will placed in those orbitals directed towards the more electropositive substituent.

Let's examine how Bent's rule might be applied to your example of methyl fluoride.  In the $\ce{C-F}$ bond, the carbon hybrid orbital is directed towards the electronegative fluorine.  Bent's rule suggests that this carbon hybrid orbital will be richer in p character than we might otherwise have suspected.  Instead of the carbon hybrid orbital used in this bond being $\ce{sp^3}$ hybridized it will tend to have more p character and therefore move towards $\ce{sp^4}$ hybridization.

Why is this?  s orbitals are lower in energy than p orbitals.  Therefore electrons are more stable (lower energy) when they are in orbitals with more s character.  The two electrons in the $\ce{C-F}$ bond will spend more time around the electronegative fluorine and less time around carbon.  If that's the case (and it is), why "waste" precious, low-energy, s orbital character in a carbon hybrid orbital that doesn't have much electron density to stabilize.  Instead, save that s character for use in carbon hybrid orbitals that do have more electron density around carbon (like the $\ce{C-H}$ bonds).  So as a consequence of Bent's rule, we would expect more p character in the carbon hybrid orbital used to form the $\ce{C-F}$ bond, and more s-character in the carbon hybrid orbitals used to form the $\ce{C-H}$ bonds.

The physically observable result of all this is that we would expect an $\ce{H-C-H}$ angle larger than the tetrahedral angle of 109.5° (reflective of more s character) and an $\ce{H-C-F}$ angle slightly smaller than 109.5°  (reflective of more p character).  In terms of bond lengths, we would expect a shortening of the $\ce{C-H}$ bond (more s character) and a lengthening of the $\ce{C-F}$ bond (more p character).


How to determine bond length and angle in a conjugated pi-system (polymethine)? I have a polymethine molecule with let's say 9 $\ce{C}$ atoms.

$\ce{H-CH=CH-CH=CH-CH=CH-CH=CH-CH3}$

How can I calculate the length of the whole molecule?

I need this number to compute the wavelengths the molecule will absorb (model of linear potential well).

So what are the average bond lengths and bond angles in this molecule and how are they calculated?

EDIT: Another example for polymethine (just the red part of it):

So what are the average bond lengths and bond angles in this molecule
  and how are they calculated?

1,3-butadiene should serve as a reasonable model for your work.

We can find the bond lengths in the literature.  The carbon-carbon double bond length is 1.338 Å, typical for a double bond (reference).  The carbon-carbon single bond length is 1.454 Å, shorter than expected due to resonance.  All of the carbons are $\ce{sp^2}$ hybridized, which means that all of the bond angles should be ~120°.

Using this information and vector addition you can determine the length of any polymethine.


Will charcoal briquettes (not activated) absorb water out of the air in the house in a subtropical climate? I live in Florida, an incredibly wet environment where it's very difficult to get moisture out of house without using a dehumidifier 24/7, which is too expensive.  Would charcoal briquettes be of any use?

The body of your question shows that your primary concern isn't charcoal's hygroscopic character per se but it's suitability as a practical means to lower the relative humidity inside your house.

Let's start with it's hygroscopic character: It seems that charcoal indeed displays hygroscopic properties (the propensity to absorb water to itself) to some degree (at least to the extent that it affects it's burning properties).

The subject was treated by this paper, wherefrom the following citation was selected as part of this answer to your question:

  The charcoal samples placed in the environment with high relative humidity showed an
  increase in humidity by the heightened final carbonisation temperature, with the highest average detected for charcoal produced
  at 750 °C. This was the reason why the charcoal produced in
  this temperature showed higher moisture related to higher porosity
  presenting a larger area available for water condensation inside the
  pores.

  The difference between the moisture adsorbed by the charcoal packed in
  two environments can be attributed to more water molecules present in
  the humid atmosphere (RH = 98%) and a greater ease of inducing the
  porous region of the charcoal. This means that the more humid the
  environment, the greater the adsorption of moisture by charcoal.

  These two different moisture conditions represent different possible
  situations of charcoal storage where high relative humidity is not
  ideal.

Now if one of the reasons you would like to use charcoal over other means of lowering the humidity inside your house is that you were planning to re-use the charcoal later for it's primary purpose (as a fuel), this means that it wouldn't suit that purpose very well, at least if you don't plan to dry your charcoal as an intermediary step in the process (dehumidify > dry > burn). Water requires a lot of energy to evaporate which would waste fuel if you wanted to use it directly as such after using it for your house.

Now for absorbing humidity, there are other (much better) substances you could use out there: You could for instance use the very safe and well known SiO2 form frequently referred to as "silica-gel", which works very well at higher RH levels, and can be conveniently "recycled" by ~ 1h pass in the oven at ~100-120C. The main weakness of silica gel in your case is that it quickly loses efficiency at higher temperatures that are not uncommon in Florida. If you'd like to be a slightly less conventional, one could recommend Calcium Chloride (CaCl2), a fairly cheap substance which will hold up to ~4 molecules of water for each of it's own up to a temperature of 45°C, and at the very minimum 2 molecules of water for each of it's own at any temperature naturally encountered in Florida.

I hope this helps!


Trans vs. cis, or Z vs. E I am new to organic chemistry and am confused regarding the following question:

Does the geometrical isomerism of 3-methyl-2-heptene fall into the category of trans vs. cis, or the category of Z vs. E? And what is the rule that determines whether it falls into the first or the second category?

My textbook is not clear on this.

Many thanks!

(Z) is the same like cis

(E) is the same like trans

BUT cis and trans is usually only for double bonds with 2 different substituents. Confusion can arise when, for instance, you have 2-halobut-2-enes, because some take the highest priority when determining if this is cis or trans whereas others may take the identical groups.

You can use (Z) and (E) for more than 2 substituents, for those with higher priority. Like in your example.

In general it is better to use (Z) and (E) nomenclature to avoid these ambiguities.


Why are there quantised energy levels in the vibrational energy of a molecule? Why are the energy levels for vibrational energy in a molecule discrete as opposed to continuous?
I don't understand how a vibration can't have continuous amounts of energy that depend on the frequency of any infrared photon.

Can someone combine this question onto my stack exchange account: olly_price, because I can't sign in on this computer.

"Why" is a good question, and one that science has yet to fully answer. We generally have a good understanding of "how" things work at the subatomic level, based on over a century of observation followed by theory and math backed up by experimentation. However, the really basic, naively simple questions, like "where do particles get the charges, spins and masses that define them" are the questions scientists are asking now.

As for the discrete quantum energy states theorized by quantum theory, those are the result of a lot of math that explains what we see much to well to be coincidence. Basically, an electron can be thought of as orbiting its nucleus, but it's really more like a vibration. Similar to a vibration, there are frequencies above the "fundamental" that produce "harmonics"; standing waves with wavelengths of a perfect integer fraction of the fundamental, and perfect integer multiples of frequency, which are self-stabilizing. Electrons can only exist in these discrete energy states because if they didn't, they'd "break orbit" and go zooming off as free radicals, or have an unstable or unenergetic path that sends them crashing into the nucleus.

What's weird is that that doesn't ever happen as far as we know; not only do electrons exist in stable states, they only exist in these states, and we can't get them to exist in any energy state between two discrete states.

Max Planck first posited these quantum energy states existed while attempting to solve a conundrum of classical physics. Back to the vibration analogy, consider a guitar string. By damping the string at precise points on the string, we can induce clear harmonics. however, the higher the harmonic, the harder you have to pluck to produce a wave of the same amplitude. As the wavelengths get smaller, the energy you have to put into them to obtain a constant amplitude (and thus the energy inherent in the wave; Newton's third law) increases exponentially, diverging to infinity as the wavelengths approach zero. Now, this was obviously wrong; observed visible spectra of a glowing piece of metal as it is heated clearly shows a "peak" in the frequency of light produced, and at frequencies above this peak, amplitude of emitted light drops off sharply.

Planck solved this "ultraviolet catastrophe" by describing what was going on as similar to this string having a number of springs of different length attached to it, with weights on the other end of each spring. As the string vibrated, it in turn induced vibrations in the springs (known technically as "harmonic oscillators"), and as the frequency increases, it approaches, matches, and then falls away from the harmonic frequencies of the springs, increasing their relative amplitudes and then making them fall off as the vibrations begin to cancel out, while simultaneously increasing the energy imparted to a different string whose harmonic was being excited. The relative harmonic frequencies of the springs are what determine the colors of light seen, and the total energy of the string is divided among the springs, keeping the same total energy as frequency increases instead of diverging.

To make the math work, Planck had to describe the springs as each receiving an integer portion of the energy of the entire system, and described these energy levels as proportional to the length of the spring. The proportionality is defined by a slope known today as Planck's Constant. Thus, Planck asserted that all energy existed in discrete amounts, which he called "quanta", hence "quantum theory".

While this worked, it seemed to do so by convenience, until Einstein then further developed photon theory: light was made of of photons, "packets" of energy that could be quantified in proportion to the photon's frequency, based on Planck's Constant. This in turn explained puzzling properties of the "photoelectric effect", namely that the threshold where electrons began to be ejected from metal when exposed to light (forming an electrical charge) was dependent not on intensity of light, but on the frequency of the light. Einstein's photons explained that, because an electron was likely to only be hit by a single photon of light, no matter how many there were, the energy imparted to the electron could only come from that single photon, and as a photon's energy increases with frequency, there is a minimum frequency required to impart enough energy with a single photon to eject an electron.

Planck's constant, and the highly definite nature of how atoms interact with photons, continues to show up in math that accurately describes things we observe about matter, such as how materials emit very specific wavelenghs of light while absorbing all others. That was explained by the Bohr model of the atom, which was the first model to introduce quantum theory to atomic structure. The Bohr model was based partially on a mathematical model first proposed by Johann Balmer and then improved on by Johannes Rydberg, showing that the emission spectra of hydrogen was predictable by an equation with integer solutions. Bohr showed that the energy states of electrons, which can be altered by absorption of photons, must exist in certain integral states, defined by an equation derived from Rydberg's but that also includes, you guessed it, Planck's Constant. The famous "double-slit experiment" showing that light is both wave and particle, and the resulting mathematical models by Louis de Broglie showing the movement of electrons as a series of stable standing waves of vibration in the electromagnetic field of the nucleus (electrons, in fact, are similar to photons in that they are as much a wave as a particle), reproduced the hypothetical energy states of Bohr's model by a different method, and in math, when you get independent confirmation of your results by a completely different mathematical model, you know you're on to something.

Now, that's the theory to date (well, actually I stopped about 100 years' worth of research ago, but them's the basics of the discovery of discrete quanta). What we are still trying to explain is why these quanta actually exist. A spaceship can have an infinite number of stable orbits around the Earth, and there are countless stable orbits of planets and planetoids around the Sun; why then does an electron stabilize so deterministically in these discrete states? Our best answer is that the electron's wave motion is only stable in these discrete states; otherwise the wave would interfere with itself or others and cancel out. However, we have never seen this "turbulence" happen and can't seem to force it to happen; it seems to be a fundamental quality of matter.


How ‘lively’ would this ‘health’ potion be? I’ve stumbled across a rather peculiar recipe for a allegedly ‘health’ potion in a computer game, composed of an unspecified amount of each of the following:

rust (iron oxide)
saltpetre (potassium nitrate)
oil of vitriol (sulphuric acid)
aqua vitæ (ethanol)

To make things clear: I’m not suicidal, I do in no way intent to actually mix that stuff (the initial tagging was maybe a bit misleading), but I’m interested in what would happen if someone actually would.  Drinking such a potion vile concoction is of course out of the question because of the sulphuric acid alone.

From what I know from highschool chemistry (and a generous helping of Wikipedia), saltpetre and sulphuric acid would be used to make explosives by way of nitrosulphuric acid.  Oversimplifying, by adding cellulose or tuluol one would end up with nitrocellulose or TNT, respectively.

But I lack the terminology to go further than that.  What would be the result of adding alcohol to that, and then iron oxide, or the other way round?  Is there some known mixture that would fit, however roughly, along those lines?  From the context of where that formula came from that wouldn’t surprise me.  It may of course as well be that the formula was simply chosen because of it’s obvious unhealthyness.

I hope this clarifies the intent of my question.  The idea may have come from a pseudo-fantasy/alchemy setting, but I’m interested in the ‘real’ chemistry behind it.

Explosives: It is true that sulfuric acid and sodium/potassium nitrate are used to make various explosives (by creating nitric acid), but the key is something to nitrate. The worst this mixture could generate is nitrated ethanol, and I was unable to find any source suggesting that could be done with nitric and sulfuric acid alone. Nitrating a molecule that small with that simple of a structure seems unlikely without extreme conditions or reagents. If it were possible, the most likely product does seem to be something unpleasant.

Poisons: What definitely would be produced is nitric acid, and the production of nitric acid via sulfuric acid and a nitrate salt produces nitrogen dioxide, a poisonous and caustic gas. This is some seriously bad stuff you do not want to come in contact with.

Based on that, I'm going to conclude this mixture is probably not explosive, but definitely could become so if an easily-nitrated compound was a contaminant. Those are more common than you might imagine. It definitely would create a nasty gas.


Is there an easy way to remember charges on ions? I have a chemistry test coming up and I might need to know the charge that goes with the different ions like $\ce{SO4}$ has $-2$, $\ce{NO2}$ is $-1$ and  $\ce{PO4}$ is $-3$.  Is there an easy way to remember this by looking at the periodic table or something like remembering that the second row elements are all going to be $\ce{something-O3}$  and the next row is all $\ce{something-O4}$ for the -ate endings (except $\ce{Cl}$ which is the odd man out). It will make writing out the different acid combos easier, if I can remember the charge difference that needs offsetting by the hydrogen.

I'm just trying to lay down some associative triggers for myself.

I don't believe there is any simple way to determine what they are except by just plain memorization. I have given some tricks below that have helped me.

When I write down my poly-atomic ions, I generally put them in similar groups in a certain order. From least amount of oxygen to most amount oxygen.

hypo (name of ion) ite -> least oxygen
(name of ion) ite -> one more oxygen than above
(name of ion) ate -> one more oxygen than 'ite'
per (name of ion) ate -> one more oxygen than 'ate'

Understanding the above helps most with 'ite' and 'ate'. 'Ate' generally being 1 more than 'ite' for oxygen.

Then I list them from least charge to most charge. I order them this way:

Nitr ite $\ce{NO2-}$
Nitr ate $\ce{NO3-}$
Sulf ite $\ce{SO3^2-}$
Sulf ate $\ce{SO4^2-}$

You can see that 'ate' has one more oxygen, and that I have listed them in charge order. Sometimes looking at the trends in oxygen and charge will help you memorize. I have had to remember them by writing them down several times and now I am comfortable.

Here is another example: Hypochlorite is
$\ce{ClO^-}$

So, chlorite should be?

   $\ce{ClO2-}$, because it has one more oxygen, and

Chlorate should be?

   $\ce{ClO3-}$, because it has one more oxygen than the above, and

$\ce{ClO4-}$, will be called?

   Per chlor ate, because it has the maximum amount of oxygen.

The above is not a hard and fast rule. One exception I can think of is manganate and permanganate.

Manganate is $\ce{MnO4^2-}$ but permanganate is $\ce{MnO4^-}$; same amount of oxygen, just a different charge.

Trying to figure out any of the above from a periodic table didn't make a difference to me because my professor tests us without it. Yikes!


Why is CF3COOH exceptionally acidic? The pKa values of almost all carboxylic acids lie much above 0. But this is violated by trifluoroacetic acid(-0.25). How can this be justified?

Try drawing the structure of the conjugate base $\ce{CF_3COO^-}$. In general, the more stabilized the negative charge of the conjugate base is, the more the equilibrium favors that form, thus the more the acid dissociates, thus the "stronger" the acid is.

So for a stronger acid (lower $\ce{pK_a}$), the negative charge must be more stabilized. When comparing $\ce{CF_3COOH}$ (TFA) to $\ce{CH_3COOH}$ (we'll use acetic acid to represent a "typical" carboxylic acid), TFA is stronger than acetic acid because its conjugate base can better stabilize the negative charge. If we draw the structures of the conjugate bases (I encourage you to do so), we will see that they both can stabilize the negative charge via resonance in the carboxylate functional group. However, TFA also has three highly electronegative fluorine atoms which withdraw electron density "through the single bonds" via induction.

As stated above, the $\ce{-CF_3}$ moiety is an electron withdrawing group, by the inductive effect. So our negative charge (rather, the electron density it represents) will be slightly drawn through the bonds toward the fluorines, thus it is further delocalized and thus more stable.

Note that we do not have convenient structural diagrams to show induction like we do resonance, but you should be able to imagine the electrons "sloshing" more towards one side of a bond/functional group. It's the same concept as a good old polar bond like $\ce{H-Cl}$; we say Cl "hogs" the electron density and pulls it away from H. In the same way $\ce{-CF_3}$ will draw electron density toward itself.

So to summarise, in general:

Stronger acid ~ more stable conjugate base. Whenever you want to qualitatively rationalize the relative strengths of two acids, draw their conjugate bases and determine which is more stable. (If your acid was neutral, its conjugate will be negative, and it will be easier to use this method than for a positive acid/neutral conjugate.) Stabilizing factors are size and electronegativity of the atom bearing the charge, resonance delocalization of the charge, hybridization of the orbital holding the charge, induction delocalization of the charge, and hyperconjugation which donates electron density and may destabilized a negative charge. SERHIH, roughly in that order of significance.

In this example, TFA and acetic acid are tied at size & electronegativity because both place the charge on an oxygen atom, tied at resonance because both may use the carboxylate functionality, tied at hybridization because the atoms in a carboxylate are necessarily $\mathrm{sp^2}$, but now induction is present in TFA and not acetic acid, so we expect TFA will be a stronger acid.


Why is the Gibbs free energy for phase changes zero at constant temperature and pressure? I get that $\Delta G$ measures the spontaneity/capacity of a system to do non-mechanical work, and that if:

$\Delta G > 0$, the reaction is not spontaneous

$\Delta G < 0$, the reaction is spontaneous

$\Delta G = 0$, the reaction is at equilibrium

So why is Gibbs free energy zero for phase changes at constant temperature and pressure?

If it is negative or positive why we say that freezing water releases heat to the surrounding and it is equal to $\Delta H$? If at equilibrium  $\Delta H$ is used to reform liquid how we say it is released to the surrounding. I know I miss something, anyone can tell me?

The initial and final thermodynamic equilibrium states of your system are as follows:

State 1:  $\pu{1kg}$ liquid water at $\pu{0 ^\circ C}$ and $\pu{1 atm}$

State 2:  $\pu{1 kg}$ water ice at $\pu{0 ^\circ C}$ and $\pu{1 atm}$

You want to find the change in enthalpy, entropy, and Gibbs free energy between these two states.

Now, we know that, to get the change in entropy between two thermodynamic equilibrium states of a closed system, we need to identify a reversible path between these states and then calculate the integral of $\mathrm dq/T$ for that path.  For the present situation, this can be accomplished by putting the system into contact with a constant temperature reservoir at a temperature only slightly lower than $\pu{0 ^\circ C}$.  As the liquid water freezes, its molecules lock into place (losing potential energy) and this results in a release of heat to the reservoir.  So the temperature of the system never deviates significantly from $\pu{0 ^\circ C}$. Since the process is at constant pressure, the change in enthalpy is equal to the heat transferred from the surroundings to the system:$$\Delta H = q$$where both $q$ and $\Delta H$ are negative for this process.  Since the process is also at constant temperature, the change in entropy, which is given by the integral of $\mathrm dq/T$, is:$$\Delta S=\frac{q}{T}=\frac{\Delta H}{T}$$Thus, for this change from State 1 to State 2, $\Delta S$ is also negative.  If we now use the change in enthalpy and the change in entropy to calculate the change in free energy between the two thermodynamic equilibrium states, we obtain:
$$\Delta G = \Delta H-T\Delta S=\Delta H-T\frac{\Delta H}{T}=0$$


Is tautomerism in ketimines possible? In a reaction between an aliphatic primary amine and a ketone, I could expect a ketimine as a product or intermediate. My question regards tautomerization of this intermediate: has tautomerization between the two possible imine structures ever been observed, or is even possible? I tried to search imine tautomerism and transamination, but I can't find what I'm looking for. I imagine this would be a simple experiment to conduct, but I can't find anything in the literature on this.

I found something very similar to what I sought in Dose, Nature 179, 734 (1957) though it involves aldimines and a decarboxylation step. The author gives a very well explained mechanism for the process and reports varying transamination ratios of valine based on the substitution of the aldehyde. An example:


3-ethyl-2-methylpentane-2,3-diol with sulfuric acid I was studying pinacol pinacolone reaction when I came across this. I believe the hydrogen cation will attack the hydroxyl group of carbon attached to two ethyl groups (since this carbocation is more stable due to the inductive effect), then methyl group will migrate and hence we get 3-ethyl-3-methylpentan-2-one, but the book says the answer is 2-ethyl-2-methylpentan-3-one. Is the answer in the book wrong or I am making a conceptual mistake?

Edit: This answer is based on a typo in the original question referring to a different starting compound than the OP intended.

A quick analysis of this reaction gives me the following:

3-ethyl-2-methylpentan-1,2-diol can be protonated by acid to produce cation A or B. Either of these cations can lose water to form a carbocation

The carbocation formed from A is a tertiary carbocation C that might not be likely to rearrange, but we'll consider the possibility in a bit.
The carbocation formed from B is a primary(!) carbocation D that seems unlikely to form.

  Pinacol rearrangement from cation A

A pinacol rearrangement from cation A through carbocation C forms an aldehyde: 3-ethyl-2-methylpentanal.

  Pinacol rearrangement from cation B

Having established that carbocation D is unlikely to form, we might suspect no rearrangement, but the rearrangement can occur in a concerted fashion with the loss of water. The end result is a ketone: 4-ethyl-3-hexanone. These kinds of concerted rearrangements to avoid primary cations are also known to occur in Friedel-Crafts reactions.

Neither product seems to match what you provided as the book's answer. Is it possible you miss-named the product?

  Which product is more likely the major product?

Rather than tell you, I will give you a hint:

One rearrangement produces a thermodynamically more stable cation than the other.


Meaning of Cu1.5Al in Cu-Al phase diagram (I understand all other indexes incl. fractions like Cu0.75Al0.25) I understand Cu0.75Al0.25 - that's just 75% Cu, 25% Al. I understand integer indexes - that's a normal chemical formula after all.

However, I have no idea how to interpret Cu1.5Al, I've never seen a fraction in a chemical formula, and if it was a composition-percentage it should be like in the 0.75-0.25 example?

Look at between 50% and 60%, below 630°C in the diagram below.

I read this page but still don't understand it.

As the page you referenced mentioned, things can get a bit tricky.

I will start with your example of $\ce{Cu_{0.75}Al_{0.25}}$ - that is not just $75%$ Cu and $25%$ Al - that phase has a particular crystal structure, with the prototype (basically the first identified alloy with that crystal structure) of W (tungsten). So, it has the arrangement of atoms in the unit cell in the same places as pure tungsten, but $1/4$ of them are Al, and $3/4$ of them are Cu. There is no ordering of these atoms on the W lattice sites, or else the symmetry would be different, and the prototype might just be $\ce{Cu3Al}$.

For $\ce{Cu_{1.5}Al}$ it is similar, but perhaps even more confusing.  The 'Published phase label' is $\ce{Cu_{1.5}Al}$, so that is what it is usually called in the literature.  The 'formula' is $\ce{Cu_{1.7}Al}$, which just means when people started referring to the phase they didn't have an accurate composition measurement.  Then, the 'Prototype' is $\ce{Co_{1.75}Ge}$.  That means that the first similar crystal structure was measured in the Co-Ge system.  The non-integer subscript(s) indicate that the unit cell has vacancies in the ideal structure (without looking it up, I would speculate that the Co sites are $7/8$'s occupied).

To further complicate matters, if you look at the Co-Ge phase diagram you will find that the phase label is now $\ce{Co_{1.67}Ge}$, Formula is $\ce{Co_{1.5}Ge}$, and the prototype is still called $\ce{Co_{1.75}Ge}$.

So: crystal structures can be complicated, identification of composition has been getting better over time (folks have been looking at the Al-Cu system for a long time!), and if you really want to know what is going on you probably have to dive into the x-ray diffraction literature and look at the published atom positions in a given phase.


Can you discriminate a monooxygenase from other enzymatic mechanisms by the requirement of NADPH? I am working on a project about CYPs 450. It seems that most of them function as monooxygenases but there are other categories as well.

Assuming that the question makes sense, is it correct to say that CYP 450 monooxygenases always require the presence of a regenerating system (NADPH) as opposed to other CYPs 450 which are NOT functioning as monooxygenases?

I would say this is not a correct statement.  The main reason being that just because an enzyme is not a monooxygenase doesn't mean it doesn't require a regenerating system for NADPH.
Examples would be many dehydrogenase, hydroxylase and synthase enzymes. They also require NADPH regenerating systems to operate efficiently.  So, one of the CYP 450 enzymes may fall into any of these categories requiring NADPH/NADP yet not be a monooxygenase.


Why does calcium oxide react with sulfur dioxide? Which of the following compounds reacts with calcium oxide, $\ce{CaO}$?
  (a) $\ce{K2O}$
  (b) $\ce{Na2O}$
  (c) $\ce{SO2}$
  (d) $\ce{MgO}$

I thought that since sodium is higher in the reactivity series than Ca, then it would be B. However, the answer is C, and I'm not quite sure why. Can anyone enlighten me?

One way of thinking about it is that calcium is a metal and therefore forms a basic oxide, $\ce{CaO}$. On the other hand, sulfur is a non-metal and forms acidic oxides $\ce{SO2}$ (or $\ce{SO3}$). So, the reaction between $\ce{CaO}$ and $\ce{SO2}$ can be considered an acid-base reaction.

There is a somewhat obscure acid-base theory, the Lux–Flood theory, which is applicable to such reactions. In this context, bases are oxide donors (like $\ce{CaO}$) and acids are oxide acceptors (like $\ce{SO2}$).


Making a solution with two solutes I need to make a solution that has a 15% (wt/wt%) HCl acid concentration and a 6% (wt/wt%) NaI concentration. I'm not sure yet how much of this solution I will be required to make so a general recipie would be awesome.  I would also like to know how to make the concentrations wt/vol% as well.

Additional information:

I have a bottle of Muriatic Acid. Its label says the contents are HCl acid, 20 degree Baume (31.45% by weight). The product's MSDS sheet says its 65-75% water and 25-35% HCl, and its specific gravity is 1.16 (1.16 g/ml).

I have Sodium Iodide (a brick of it). Its label says 99+%, pure, anhydrous.

You need 15 g HCl and 6 g NaI per 100 g of the solution. Treating 31.45% (I don't know much about practice) as exact, there is 15 g HCl in 15 g / 31.45% = 47.69 g of your muriatic acid. So you need 6 g of sodium iodide, 47.69 g of your muriatic acid and the rest, i.e. 46.31 g, of pure water per 100 g of the solution.


Derivation of Henderson–Hasselbalch equation While deriving the Henderson–Hasselbalch equation why don't we consider the hydrolysis reaction?
For example if we consider an acidic buffer of acetic acid and sodium acetate, why do we not consider the hydrolysis of the acetate ion furnished by sodium acetate?

Just because it's negligible.

Well, I tried to make some calculations. I was too lazy to make them using all this internal stackexchange math system, so I hope you'll understand my handwriting.

I also need to say what conditions I used: activity=equilibrium concentration; I didn't care much about hydroxonium, because it doesn't change anything; I used 0.1M solutions of both acetate and acid just because I wanted to.

The conclusion is: really, if you have 0.1M solutions, a change in 10-10M will change absolutely nothing, so why suffer and not neglect this?

P. S. All calculations made in Excel 2013 :)


Fermentation of different types of sugar I'm searching for the difference in the fermentation of muscovado sugar, honey and sugar cane. I have used a specific microbiota to to ferment these three types of sugar and have to explain why I've got different results regarding acetic acid and ethanol production in the final beverage, as well as different amounts of sucrose, glucose and fructose in same.

Micro-organisms eat the sugars then excrete all sorts of compounds. Acetic acid and/or ethanol is commonly excreted by anaerobic organisms.  Anaerobic organisms can derive energy from multiple reaction paths, even within the same micro-organism (consider the case of facultative anaerobes.)

Since the same organism can use different paths to derive it is quite natural to assume what comes out is naturally affected by what goes in.


Guess the shape and phase of the MOs of butadiene using LCAO and Hückel's method? I used Hückel's method along with a Linear Combination of Atomic Orbitals (LCAO) to calculate an estimate for the orbital energies of cyclobutadiene ($\ce{C4H4}$) and butadiene ($\ce{C4H6}$). For butadiene, I ended up with the result

$E = \alpha \pm 1.62  \beta$ and $E = \alpha \pm 0.62 \beta$,

where $\alpha$ and $\beta$ are both negative (and are the Coulomb integral and the exchange integral, respectively). I'm fairly confident with the method and interpretation of these integrals so far, but I don't know how to use these results to "guess and draw" the shapes and phases of the molecular orbitals, and I'm also not sure how to tie in what $\alpha$ and $\beta$ represent.

As I understand things, in these molecules, hybridisation of the orbitals occur and $\mathrm{sp^2}$ orbitals form, taking up three electrons from each carbon atom and leaving one electron left over in the third p orbital. This p orbital is the one I assume is portrayed in the below images.

Could someone please give me a hint towards the following:

relating the molecular orbital shapes/phases with the Coulomb integral (overlap of one carbon's electron wavefunction with a neighbouring carbon's potential) and the exchange integral (the overlap of two electrons' wavefunctions in the field of one of the carbons
how to get any sort of information about the phases of the different orbitals. I understand that $E = \alpha + 2\beta$ will have the lowest energy, as both $\alpha$ and $\beta$ are negative, but how does that correspond to the phase configurations shown below? Why do the size of the circles in the first image vary?

Note: I've been sent here from the Physics Stack Exchange, please be nice. :) Thanks for any help! This is my first post so please also let me know if the posting etiquette passes muster.

Sources: First image, second image

The shape of the molecular orbitals is not guessed, it is determined by the LCAO-MO expansion coefficients which one needs to find out in the first place. The hint is that an eigenvalue equation is usually solved in two steps:

We start by writing down the eigenvalue equation in the matrix form and proceed then by writing down and solving the characteristic equation in which the secular determinant is equated with zero. By solving the characteristic equation we find the eigenvalues (energies).
Then we put eigenvalues (energies) back into the matrix eigenvalue equation and solve it for the coefficients (LCAO-MO expansion coefficients) that determine eigenvectors (molecular orbitals).

As far as I understand, the second step was not simply done yet. Note that any scalar multiple of an eigenvector is also an eigenvector with the
same eigenvalue, so that eigenvectors are not unique. In solving the matrix eigenvalue equation one will find not the actual values of the LCAO-MO expansion coefficients, but rather some relationship between them which could be satisfied by a number of different coefficients. In principle one is free to choose any set of coefficients satisfying the relationship, and usually there is some obvious choice.

For the simplest case in which just two coefficient are related with each other through some ratio the obvious choice is to take some coefficient to be one, then the value of the second coefficient is nothing but the value of the ratio. One could also normalize the wave function then if desired.

Also note that one can use symmetry to simplify the math. OP did not mention were the symmetry adapted linear combinations (SALCs) of atomic orbitals used at the first step, so I mention just in case.

P.S. Fully worked examples could be found in different places, for instance, here (PDF, 150 KB).


What copper oxide is being created upon electrolysis? Whenever using copper as anode, I end up with cyan dust that doesn't dissolve in water.

What's that (presumably) oxide? Is it poisonous? Can I dissolve it in something to perform electroplating with it?

It appears to be copper hydroxide or copper carbonate.  Copper oxide tends to be black/grey/brown.  Copper hydroxide is blue and copper carbonate is green.  Copper carbonate forms when the hydroxide is exposed to CO2.

If you would like to dissolve it, try an acidic solution.


Mr of Acid getting wrong answer I am preparing for A level chemistry exams. One of the questions I came across is

  Calculate the $M_\mathrm r$ of unknown monoprotic acid. The volume is $15\,\rm cm^3$ and concentration $10.95\,\rm g\cdot dm^{-3}$.

I know how to calculate the number of moles and I know the equation to calculate $M_\mathrm r$ but could anyone give me some guidance on how to calculate the mass from this data?

The concentration of your acid is given as a mass concentration.

Without knowing the molar mass, the molar concentration and the amount of substance in a volume of $15~\mathrm{cm}^{-3}$ are not accessible.

However, knowing that

$\mathrm{1~L = 1~dm^{-3} = 1000~cm^{-3}}$

and therefore

$15~\mathrm{cm}^{-3} = 0.015~\mathrm{dm}^{-3}$

the mass of the acid (in $15~\mathrm{cm}^{-3}$) is given as

$m = 10.95~\mathrm{g\,dm^{−3}\cdot0.015~\mathrm{dm}^{-3} = 0.16425~g}$


Do chemical garden coral formations leach toxins? Short version, I want to make a coral garden with sodium silicate and salts, and I would like to put that in my fish tank, but I'm concerned about the fish getting poisoned.

https://en.wikipedia.org/wiki/Chemical_garden
https://www.thoughtco.com/make-your-own-magic-rocks-607653

The above pages describe the chemicals needed. The articles say that once the corals have formed, they can be rinsed to remove the sodium silicate, and kept in fresh water. Most of the articles I've read say the same thing, in different ways. But none of them say whether the coral degrades, or if it is stable.

Are there experiments that I can do to test this? How can I be sure?

Experiments to test if toxins leach from silicates do exist; however, these experiments involve dangerous chemicals.  For that reason these tests should be left to qualified lab personnel and not described here.  Performing these experiments are expensive and would need to be performed on every combination of methodologies to make a "chemical garden".  Because so many combinations exists, quality control on such an experiment would be difficult.

Experiment aside, a "chemical garden" in a fish tank is a bad idea.  While silica is chemically insoluble,  it can be converted biologically into orthosilicic acid.  This form of silica is soluble. This process in nature is called biological weathering.

Chromium, lead, and perhaps some of the other compounds are carcinogenic to humans.  These compounds, even in an immobilized form, are not worth the risk and should be nowhere near a fish tank.


Equilibrium, Kp constant, Partial Pressures A system in equilibrium contains
$$I_{2_(g)}\ \text{and} \ I_{(g)}$$
under pressures $$P_1=0,21 \,\text{atm. and}\  P_2=0,23 \,\text{atm.}$$
While its temperature remains constant, we reduce the system's volume to half its original value.

I have to determine $P_1'$ when the system's equilibrium is reestablished.

I know I can calculate $K_p$ and I know I somehow have to use $PV =NRT$, but from there on I am lost, I cannot figure out how to come up with enough equations to find all of my unknowns. Can anyone show me an explicit solution to the problem?

What I tried, but got really nowhere:

$$2I_{(g)} \leftrightarrow I_{2_{(g)}}, K_p=\frac{P_1}{P_2^2}=3,97.$$
$$P_t=P_1 + P_2 = 0,44 \,\text{atm.}$$

When the new equilibrium is established :

$$Kp=\frac{P_1'}{P_2'^2}=3,97$$

$$\frac{2P_1 }{P_1'}=\frac{n_1}{n_1'}$$
and same with $P_2, P_2', n_2, n_2'$, but I don't know where to go from there.

The method referred to in the comment is not correct and is a common mistake. One may think,

$$p_1'+p_2'=p_t\frac{V_t}{V_t'}=2p_t=0.88 \rm atm,$$

but this is invalid since $n_1+n_2\neq n_1'+n_2'$, thus $p_tV_t\neq p_t'V_t'$

The correct way is to use the fact that the number of iodine atoms is constant, from which we get,
$$2n_1+n_2=2n_1'+n_2',$$
$$(2p_1+p_2)V_t=(2p_1'+p_2')V_t',$$
$$1.3 {\rm atm}=4p_1+2p_2=2p_1'+p_2',$$

Then you can go on by combining this with the equation you have already obtained from $K_p$ (listed in your question) and solve for $p_1'$ and $p_2'$.


Two different electronic configurations for cerium Depending on the textbook there are two different electronic configurations stated for cerium. On the one hand

$$[\ce{Xe}]\mathrm{4f^15d^16s^2}\quad$$

and on the other hand

$$[\ce{Xe}]\mathrm{4f^26s^2}$$

Are both configurations possible? If not, why so?

The filling of electrons follows Aufbau principle. However, introduction of $\mathrm{d}$ and $\mathrm{f}$ orbitals and resulting poor shielding affects predicted energy order, and it consists last 2 or three shells incompletely filled. The energy associated with this subshells is nearly same, and one of the subshell of outer shell is bigger in size than subshell of inner shell. So, to perform minimum repulsion between electrons, one or two electrons jump to a bigger subshell (with similar energy) so this exceptions occur.
In case of cerium, energy of $\mathrm{4f}$ and $\mathrm{5d}$ subshells are quite similar and $\mathrm{5d}$ subshell is bigger than $\mathrm{4f}$ subshell. So 1 electron in $\mathrm{4f}$ jump to $\mathrm{5d}$ subshell.


Are our hands really chiral? Everyone says that our left and right hand are mirror images of each other and are non-superimposable, unlike a sphere whose mirror image is exactly superimposable. But shouldn't our left and right hand be considered achiral?
If I move my hands in front of me such that palm of both the hands face towards me and move my say left hand horizontally such that it can be superimposed on the right hand clearly they can't be because the thumbs point in opposite directions.

But if I rotate my left hand by 180 degrees so that now the palm of my left hand faces away from me then both the left and right hand are superimposable. So shouldn't they be achiral? I know there are two more similar questions akin to this but none of the answers answer my simple question.

Holding your hands in this way merely proves that your hands are mirror images. If you take any object (chiral or not) and hold it up to a mirror, you can always align common features.

Imagine instead placing one of your hands inside the other. You may be able to align the overall thumbs and fingers, but they will be facing opposite directions, thus are not completely superimposable.

If you'd like further confirmation that your hands are chiral, try shaking someone's right hand with your left. Or try putting a left hand glove on your right. The interaction of mirror image forms (enantiomers) of a chiral object (like your hands) is different with the mirror image forms of another chiral object (like a glove or someone else's hand).


What is the proper nomenclature for CH₂O₂? Can you tell me what's the nomenclature of $\ce{C H_2 O_2}$?

What is its IUPAC name?

The compound you provided can be written more clearly as $\ce{HCOOH}$

IUPAC name : Formic acid

Systematic name :Methanoic Acid

Other names:
Aminic acid, Formylic acid, Hydrogen carboxylic acid, Hydroxymethanone, Hydroxy(oxo)methane, Metacarbonoic acid, Oxocarbinic acid, Oxomethanol


Which proton is more acidic in this compound? I'm having trouble determining which of these protons is more acidic. I was thinking the solution had something to do with the double bond in the aromatic ring, but I'm still not sure.

You could start by drawing the resonance structure of the naphthalene ring, this gives a compound in which the conditions on the two nitrogen atoms are interchanged, thus both nitrogen atoms are equivalent. As $\ce{N-H}$ bonded protons are more acidic than $\ce{C=C-H}$ protons and $\ce{N-C-H}$ protons, the answer must be the hydrogen atom bonded to the nitrogen atom, as both nitrogen atoms are equivalent, either of the hydrogen atoms is a valid answer (as they are equivalent).


Derivation of time probability distributions for single molecule kinetics If we have a protein $\ce{A}$ that can bind to protein $\ce{B}$:

$$\ce{A + B <=>[$k_\mathrm{fwd,1}$][$k_\mathrm{rev,1}$] AB}$$

and the bound complex $\ce{AB}$ can be converted into $\ce{C}$:

$$\ce{AB <=>[$k_\mathrm{fwd,2}$][$k_\mathrm{rev,2}$] C}$$

Assume we have a volume of 1 and some initial amount of $\ce{A}$ and $\ce{B}$ (which are $A_0,$ $B_0$) and that $C_0 = 0.$ We then measure the amount of time elapsed before the first molecule of $C$ is formed. The experiment is repeated many times, and the frequency of occurence of first $C$ during each time interval 0 to $t_1$, $t_1$ to $t_2$, $t_2$ to $t_3$, etc. is plotted, so that the result is a probability distribution of the time elapsed before the first molecule of $C$ is formed.

I would like to know what is the mathematical function that describes this distribution and how is it derived?

UPDATE: this is not a homework question and the point is not a specific answer. I would like references to literature/textbooks that derives these distributions analytically for simple examples like the above would be helpful. Have not seen a setup where distribution of duration is derived. Commenters say it's standard. Please give references.

To see how this calculation is done, let us consider the simpler case of just the second step shown above, ie $\ce{AB ->[k_{f2}] C}$. Since we are interested in the situation where no $C$ has yet formed, the reverse reaction can be ignored. I will assume that this reaction is first order in $AB$, so $k_{f2}$ has units of $s^{-1}$.

In this case, the rate constant $k_{f2}$ represents the average frequency of the reaction.

Now let's consider that we establish an infinite number of evenly spaced time intervals of length $\Delta t$. When we consider the probability distribution, we need to consider two factors:

1) The probability of reaction occurring in an interval $t_n + \Delta t$ and

2) The probability that the reaction has not yet occurred at $t_n$

The first value is simply the product of the rate constant $k_{f2}$ and the length of the time interval: $k_{f2}\Delta t$.

The second probability is slightly more complicated, since it is the chance of the reaction $not$ occurring in each of the $n$ intervals preceding $t_n$. That probability is given by $1-(k_{f2}\Delta t)^n$. Conveniently, as our time interval $\Delta t$ goes to $0$, this probability becomes a continuous function $e^{-k_{f2}t}$.

To get our final probability, we simply multiply these two probabilities:

$[\text{total probability of reaction occurring in interval }t+dt] = [\text{probability of occurring in interval }t+dt] \times [\text{probability that reaction has not occurred by time }t]=k_{f2}dt\times e^{-k_{f2}t}$

The probability as a continuous function of $t$ therefore can be written as

$p_{\ce{AB->C}}(t)=k_{f2}e^{-k_{f2}t}$.

As expected, integration from $t=0$ to $t=\infty$ gives an overall probability of $1$.

Reference: Adapted from the lecture notes published by Dmitrii Makarov here: http://makarov.cm.utexas.edu/resources/Lecture-notes,-tutorials-etc./single-vs-bulk.pdf

For the two-step process, if $\ce{A+B <=> AB}$ establishes equilibrium very fast relative to the $\ce{AB->C}$ reaction, then the above is a reasonable approximation (after adjustment of $k_{f2}$ to reflect the equilibrium concentration of $AB$). If the equilibrium is not established quickly, then you would use the same reasoning to determine the probability first of $AB$ having formed by time $t$ and then the probability of it reacting based on how long it has been in existence.

I have not read Makarov's textbook on single molecule kinetics, but I assume it covers these kinetics in much greater detail and might be a good resource:

Single Molecule Science: Physical Principles and Models
1st Edition by Dmitrii E. Makarov; CRC Press 2015; ISBN 9781466559516


Why is carbon tetravalent? How does the electron from the 2s orbital "jump" to the 2p, thus leaving 4 unpaired electrons to form four covalent bonds?

Also, does the octet rule not apply to carbon? If it does, how can it bond with only two electrons? It would lack two more.

Look up "orbital hybridization".  It says that atomic orbitals can be combined to make new orbitals for bonding purposes.  Carbon can take 1 s and 3 p orbitals to make 4 sp3 orbitals that can make 4 bonds.


Explanation for the reactions in a saltwater battery with zinc and copper electrodes I am a physicist, not a chemist. I'm trying to get a basic understanding of the reactions taking place in a battery using a saltwater electrolyte with copper and zinc terminals. I'm writing a general science level article about it. I have found that the reactions taking place at the terminals are:

\begin{aligned}
 \text{Zinc terminal:}&&\ce{Zn(s) &-> Zn^{2+}(aq) + 2e^-} &(\Delta G = -0.76\:\mathrm{V})\\
 \text{Copper Terminal:}&&\ce{Cu^{2+}(aq) + 2e^- &-> Cu(s)} &(\Delta G = -0.34\:\mathrm{V})
\end{aligned}

I think I understand the first reaction in terms zinc leaving the terminal and going into solution kind of as if it were dissolved $\ce{ZnCl2}$ (does that make sense?). But I need help understanding the second reaction. I've inserved a copper terminal into the solution. But not copper ions. Am I to understand it like this: when I put the copper terminal in the solution, copper ions immediately drift off, perhaps for the same reason the zinc ions do, but then deposit back onto the terminal in the presence of the electrons released when the zinc goes into solution?

I'd like an explanation somewhere between that it just happens and full on redox theory, if that is possible.

This answer deals with the situation, when but electrodes are in the same salt water solution.

Assuming your salt solution contains sodium chloride
$$\ce{NaCl (s) ->[\ce{H2O}] Na+ (aq) + Cl- (aq)},$$
then your reaction at the zinc terminal will certainly be
$$\ce{Zn (s) + 2Cl- -> ZnCl2 (aq) + 2e-}.$$
On the copper terminal water will be electrolysed
$$\ce{H2O + 2e- ->[\ce{Cu}] 1/2 H2 ^ + {}^{-}OH},$$
or using autoprotolysis
\begin{aligned}\ce{
2H2O &<=> H3+O + {}^{-}OH\\
H3+O + e- &->[\ce{Cu}] 1/2 H2 ^ + H2O
}.\end{aligned}

This will happen since sodium has a much lower electron affinity than water/ hydronium
\begin{aligned}
\ce{Na+ + e- &-> Na} &E^\circ&=−2.71\:\mathrm{V}\\
\ce{2H2O + 2e- &-> H2 + 2 {}^{-}OH}&E^\circ&=−0.83\:\mathrm{V}\\
\ce{2H+ + 2e- &-> H2}&E^\circ&=0\:\mathrm{V}.
\end{aligned}

Depending on what salt solution you are using, different cations might be reduced. You can find a collection of values at wikipedia.

In a battery (Galvanic cell) there are usually coupled half cells, and in this case this would probably be $\ce{Zn|Zn^2+||Cu^2+|Cu}$.

This means, that the Zinc terminal is in a zinc containing solution and the copper terminal is in a copper containing solution. In this case your assumption is correct
\begin{aligned}
 \text{Zinc terminal:}&&\ce{Zn(s) + ZnCl2 (aq)&<=> 2Zn^{2+}(aq) + 2Cl- + 2e^-} \\
 \text{Copper Terminal:}&&\ce{CuCl2(aq) + Cu (s) + 2e^- &<=> 2Cu(s) + 2Cl-}
\end{aligned}
There has to be a salt bridge to neutralise charge differences (ion flow).
Note that in that case zinc ions will go into solution, while copper ions will be removed from the solution. If the concentration of copper drops close to zero, then no current can be measured. The galvanic cell can be charged by "reversing" the process, apply opposite current. Then zinc ions will be removed from solution and copper will go into solution.
In reality, this is a equilibrium process.


Between ethene and propene which is more acidic and why? Between ethene and propene, which is more acidic and why? If they are they are the same then why so?
 Their $\mathrm{p}K_{\mathrm{a}}$ is the same $\mathrm{p}K_{\mathrm{a}}$ as ~44.

The difference between there acidity arises only due to the difference in inductive effect[or the electron donating capacity of the alkyl group] of the alkyl group attached to the $\ce{C=C}$ in both the cases.

By this site there exists a difference of total one unit

For more info on Inductive Effect

Relative acidity of a compound is checked by removing the acidic hydrogen and then comparing the stability of the anions.The more stable the anion is the faster it is formed and hence the acid is stronger.

The anions formed are $\ce{H3C-CH=CH-}$ and  $\ce{H2C=CH-}$ . In the former one the anion or the negative charge on the anion is destabilized by the electron donating ability or the inductive effect of $\ce{CH3-}$, hence it is less stable and less rapidly formed and hence less acidic.

  Why $\ce{H3C-}$ has electron donating effect?

In an alkyl group, for example, in methyl group carbon is linked to three hydrogen atoms. Carbon is more electronegative than hydrogen. So carbon pulls bonded electrons nearer to carbon and carbon acquires a slight extra negative charge (delta-) and correspondingly hydrogen lose hold over bonded electrons and acquires a slight positive charge (indicated by delta +). Thus carbon get a little more electronic charge which the carbon will push towards the group to which it is linked.When this electron rich carbon centre (methyl group) attached with other compounds it behaves as electron donor.
And thus methyl group acts as electron donating group.
Be in mind, the electronegativity difference between carbon and hydrogen is very less.

$\ce{H3C-CH=CH-}$ is less stable due to presence of above factor.

You can also use conjugate base method or charge holding capacity method to compare the acidity of compound to give the same result as above.


Order of reactivity of carbonyl compounds towards nucleophilic addition Arrange the following according to reactivity towards nucleophilic addition reaction:

I have already seen this post regarding the reactivity. It helped me to some extent.

$\ce{Cl}$ increases the positive charge on the carbonyl carbon increasing its reactivity.

So (B) > (A) > (the remaining 3)

Further, aldehydes are more reactive than ketones. But I should also consider steric factors. In compounds (C), (D), and (E), compound (D) has highest carbocation stability but also has highest steric hinderance.

Compound (C) has lowest steric hinderance but low carbocation stability as well. How can I arrance (C), (D), and (E)?

To all those who say steric effects dont play a role: This is taken from page 147 of organic chemistry book by Clayden:

  The same structural features that favour or disfavour hydrate formation are
  important in determining the reactivity of carbonyl compounds with other
  nucleophiles, whether the reactions are reversible or not. Steric hindrance and
  more alkyl substituents make carbonyl compounds less reactive towards any
  nucleophile; electron-withdrawing groups and small rings make them more reactive

Carbocation doesn't come into the picture at all, when we are discussing about nucleophilic addition reactions. Steric hindrance will also not play a role because the compounds are planar (attack is going to take place on $sp^2$ hybridised carbon). The nucleophile is going to attack only from the top and bottom* as shown:

The only factors that come into picture in the options are the electron withdrawing or releasing nature of the groups attached to the carbonyl carbon. If the group is $e^-$ withdrawing in nature, the deficiency on the carbon atom increases, making it susceptible to nucleophilic attack.

So, the correct order is : B > A > C > E > D

P.S.:* Since the attack is from top and bottom, both of which are equally likely, the resulting product is a racemic mixture (if the carbon on which attack happens is chiral after the attack).


Why is chloride oxidised instead of water in copper chloride electrolysis? For the Electrolysis of Copper Chloride:

Cathode: $\ce{Cu^{2+} + 2e- <=> Cu}$
Anode: $\ce{2Cl <=> Cl2 + 2e- }$

I am confused about the reaction taking place at the anode. Wouldn't $\ce{H2O}$ rather undergo oxidation since it has an increased reducing ability? eg. It is above $\ce{2Cl-}$ on the Table of Standard Reduction Potentials.

Am I missing something? Or do all halogens discharge more readily than water?

The standard redox potential of chlorine is $E^\circ = +1.358\ \mathrm{V}$; the actual potential depends on the concentration of $\ce{Cl-}$:

$$\ce{Cl2 + 2e- <=> 2 Cl- }\quad\quad E^\circ = +1.358\ \mathrm{V}$$
$$\begin{aligned}
E&=E^\circ+\frac{RT}{2F}\cdot\ln\frac{1}{\left[\ce{Cl-}\right]^2}\\
&=1.358\ \mathrm{V}-0.05916\ \mathrm{V}\times\log{\left[\ce{Cl-}\right]}
\end{aligned}$$

The standard redox potential of oxygen is $E^\circ = +1.229\ \mathrm{V}$; the actual potential depends on $\mathrm{pH}$:

$$\ce{O2 + 4H+ + 4e- <=> 2H2O}\quad\quad E^\circ = +1.229\ \mathrm{V}$$
$$\begin{aligned}
E&=E^\circ+\frac{RT}{4F}\cdot\ln\left[\ce{H+}\right]^4\\
&= 1.229\ \mathrm{V}-0.05916\ \mathrm{V}\times\mathrm{pH}
\end{aligned}$$

The redox potentials $(E_{\ce{O2}} < E_{\ce{Cl2}})$ suggest that oxidation of $\ce{H2O}$ to $\ce{O2}$ at the anode should be preferred over oxidation of $\ce{Cl-}$ to $\ce{Cl2}$.

However, depending on the material and shape of the anode, the overpotential of oxygen at the anode can be very large. Therefore, the oxidation of $\ce{Cl-}$ to $\ce{Cl2}$ at the anode is feasible.

Nevertheless, at high $\mathrm{pH}$ and at a low concentration of $\ce{Cl-}$, the oxidation of $\ce{H2O}$ to $\ce{O2}$ is still preferred over oxidation of $\ce{Cl-}$ to $\ce{Cl2}$. Hence, if also $\ce{H2}$ and $\ce{OH-}$ are produced during the electrolysis of aqueous solutions (this is less relevant for the reduction of $\ce{Cu^2+}$ to $\ce{Cu}$ which is considered in the question, but for example important for the chloralkali process)
$$\ce{2H2O + 2e- -> H2 + 2OH-}$$
the electrolysis of chloride cannot be made complete in a single step since $\mathrm{pH}$ is increased and the concentration of $\ce{Cl-}$ is decreased during the electrolysis:
$$\ce{2 H2O + 2 Cl- -> H2 + 2 OH- + Cl2}$$


How do I convert 4-(Methylthio)benzaldehyde to 4-(thioacetate)benzaldehyde? Sorry if I have the incorrect name of the final product. I'm trying to acylate the methyl mercaptan substituent to a thioester (thioacetate). I appreciate any help!

You could react 4-(methylthio)benzaldehyde with $\ce{HBr}$ to form a thiol, and then react the thiol with acetyl chloride in triethylamine to form 4-(thioacetate)benzaldehyde.


Calculating Final Temperature and Volume of Adiabatic Expansion A $\mathrm{5.00\ L}$ sample of $\ce{CO2}$ at $800 \ \mathrm{kPa}$ underwent a one-step (irreversible) adiabatic expansion against a constant external pressure of $100\ \mathrm{kPa}$. The initial temperature of the gas was $300\ \mathrm{K}$.

  An alternative path between the initial and final states consists of a reversible isothermal expansion from $5.00\ \mathrm{L}$ to the final volume $V$, followed by (reversible) constant volume cooling to the final temperature $T$.

  a) Give equations (in terms of $V$ and $T$, the final volume and temperature of the gas) for $\Delta U$, $Q$ and $W$
  for all three processes.

  b) Briefly state why $\Delta U$ is the same for both paths.

  c) Hence or otherwise calculate the final volume and temperature of the gas.

My Attempt

Part A

I can do this part. I am pretty sure I got the correct answers except for $\Delta U$ for the cooling process. Could you please check if my answers are correct.

Adiabatic expansion: $\Delta U = 33.33(T-300)$, $Q = 0$ and $W = 33.33(T-300)$

Isothermal expansion: $\Delta U = 0$, $Q = 4000\ \ln\frac{V}{5}$ and $W = -4000\ \ln\frac{V}{5}$

Cooling: $\Delta U = 33.33(T-300)$, $Q = 33.33(T-300)$ and $W = 0$

Part B

That is just because internal energy is a state function, independent of the path taken.

Part C

What confuses me here is the word 'hence', which implies that I need to use the fact that internal energy are equal for the two process. However I have no idea how to use this to find the final temperature and volume. I suspect that I might gotten the expression for $\Delta U$ wrong for the cooling process.

The final volume V and final temperature T can be determined exclusively from the information for the irreversible path if we assume that the gas continues to expand until it equilibrates at a final uniform pressure of $P=P_{ext}=100$kPa and a final uniform temperature of T.  Let $T_0$ be the initial temperature (300 K), $P_0$ be the initial pressure (800 kPa), and $V_0$ be the initial volume (5 L).  Then the number of moles is $$n=\frac{P_0V_0}{RT_0}$$
The work done on the surroundingsis given by:$$W=P_{ext}(V-V_0)$$
The change in internal energy is $$\Delta U=nC_v(T-T_0)$$
So, from the first law,$$nC_v(T-T_0)=-P_{ext}(V-V_0)$$
The final volume is given by: $$V=\frac{nRT}{P_{ext}}$$
These equations are sufficient to solve uniquely for the final temperature T and the final volume V exclusively in terms of the input data values.


How to decide what pathway a reaction will follow, elimination, addition to a carbonyl or nucleophilic substitution? There is a substrate such as this,

and it is treated with alcoholic KOH.

I know that a hydrogen in $\alpha$-position to the carbonyl group will be abstracted, leaving a carbanion. My query lies in deciding what pathway will the substrate follow from hereon. I see three options: An E1cB elimination of the bromide, aldol condensation by nucleophilic addition to the carbonyl group and SN2 with the 1 degree bromide.
Which pathway will be favored and why?

I'd be glad if someone could specify the conditions for one pathway being predominantly favored over another (even if I haven't provided information regarding that, like temperature, solvent, etc) or some empirical data on one pathway being faster or slower. Any degree of help will be appreciated, thank you.

Edit: The answer in the book where this question is from says that nucleophilic addition should predominate (providing neither justification nor data), which contradicts my guess that elimination would dominate because of it being an internal mechanism.

The nucleophilic substitution with another equivalent of the primary bromide will never be favourable as the nucleophilicity of the enolate generated is pretty low (the charge is delocalized!)
Even if E1cb was a possible route, you would end up with an extremely reactive vinyl aldehyde, which would likely participate in a Michael addition with an enolate in the reaction medium. So, isolating the product from this route would be very unlikely, even though I cannot back my claim up immediately.

If you ask me, both the Michael addition and the Aldol reaction should compete, given the nature of the base the problem has used. Had it been something like LDA (or any other directed Aldol conditions), the greater stability of the enolate would probably have precluded E1cb as a possible route.


How was the quantum three-body problem solved? What method was used to solve the quantum three-body problem?  I heard it was solved and was not sure about the status of the problem.

It never was solved, analytically (which means an exact mathematical solution)

It was solved approximately, using perturbation theories.  In the case of the helium atom, an estimate of the energy electron-electron repulsion is added to the calculation.  The appropriateness of the perturbation added is based on the comparison of the result to experimentally determined values.