How to apply oxidation numbers in the reaction to form hydrogen peroxide from a superoxide? In the following reaction:
$$\ce{2O2- (aq) + 2H+ (aq) <=> H2O2 (aq) + O2(g)}$$

I see where some oxygen atoms have lost electrons, but I do not see what has gained electrons.
On the reactant side, the superoxide has a $-1$ charge per oxygen atom. On the product side, half the oxygen atoms now have a $0$ charge, so they lost an electron each (right?). For it to be a true redox reaction, does not a species/atom have to gain electron(s)?

On the reactant side, the superoxide has a $-1$ charge per oxygen atom.

No, the superoxide has a $-\frac{1}{2}$ charge per oxygen atom.

The reaction is better thought of as two separate reactions:

\begin{align}
\ce{O2- &<=> O2 + e-}\tag1\\
\ce{O2- + 2H+ + e- &<=> H2O2}\tag2
\end{align}

This demarcation is particularly relevant in a biological context, because there is a class of enzymes Superoxide reductase (SOR) that only catalyze the reaction forming $\ce{H2O2}$ and another class of enzymes Superoxide dismutase (SOD) that catalyzes both.

Here, oxygen is the element which is simultaneously undergoing oxidation (equation (1)) and reduction (equation (2)). This is termed a disproportionation reaction.

See also: A special case of disproportionation (or 'dismutation') is 'radical disproportionation'


Why weren't the electrons attracted to doubly charged Helium particles in the Rutherford experiment? I have a chapter and in it there was a topic on the Rutherford experiment.

It is written that doubly charged helium particles or alpha particles were thrown towards a very thin gold foil and some of the alpha particles where bounced back because there were positive particles (protons) and positive and positive always repel each other.

If this was the reason for the bouncing back of the alpha particles, then why didn't the electrons, which are much smaller than protons, get attracted from alpha particles?
Especially when the alpha particles are doubly charged.

The gold foil experiment conducted by Rutherford (or more appropriately by Geiger and Marsden under Rutherford's direction) involves a stream of $\alpha$ particles (which are helium nuclei) bombarding a thin gold foil.

Why didn't some electrons go with the alpha particles because of electrostatic attraction? Some probably did. The experiment was equipped with a fluorescent zinc sulfide screen to detect alpha particles ($\ce{He^{2+}}$ ions). This screen might have also detected $\ce{He+}$ ions (but not necessarily in a way that was different than the detection of alpha particles), but probably not free electrons or helium atoms. The experiment was only set up to detect alpha particles, and so that's all that Rutherford, Geiger, and Marsden observed. We'll never know how many (if any) alpha particles picked up electrons in their experiment. We could, however, repeat the experiment, replacing the fluorescent screen with a mass analyze to detect at least the difference between $\ce{He^{2+}}$ and $\ce{He+}$.

More importantly, we known that many alpha particles passed through (and some were deflected) unchanged. Why didn't these particles pick up electrons? The answer to this question is another question: What is holding those electrons in the gold foil?

Gold atoms have 79 protons holding down those electrons. The first ionization energy of gold is 890.1 kJ/mol and the second is 1980 kJ/mol. However, these values are for isolated gold atoms. Gold (like all metals) is held together by metallic bonds with delocalized electrons occupying a valence band. Each electron is not being held by just one gold nucleus. Once the first few electrons are ejected or snatched by the alpha particles, the gold foil will have a net positive charge. It will be much harder to remove negatively charged particles from the positively charged foil.


Why does acetone have a lower boiling point than hexane? Why does acetone have a lower boiling point than hexane? I thought that since hexane is non-polar then it should have weaker intermolecular forces and a lower boiling point, but it doesn't. Why?

For reference, the boiling point of acetone is roughly 56 °C  while that of n-hexane is roughly 68 °C.

Hexane is a larger, longer-chain molecule so there are more London dispersion forces between n-hexane molecule which outweighs the dipole forces in acetone.


Conversion of an epoxide into an alcohol I need help changing an epoxide into an alcohol, i.e. the second step in the scheme below. I found a reaction that created two hydroxyl groups but cannot find one that only produces one:

Superhydride (lithium triethylborohydride) does this. JACS 95 (1973)  8486 reports complete selectivity for unsubstituted over disubstituted carbon with very high yield. HC Brown JOC 45 (1980) (1) reports complete selectivity for 2-butanol from 1,2 epoxybutane and 97:3 for the reduction of styrene oxide.


Process of a substitution reaction In this question I am not able to decide by what process this reaction is happening.
I first thought that one of the $\ce{H}$ from $\ce{CH3}$ will depart $\ce{Br}$ from $\ce{CBrCl3}$ but I am wrong. Would someone tell me what is the right process for this reaction?

This chemistry was first described in this 1960 JACS paper here.

The first step is cleavage of the $\ce{Br-CCl3}$ bond to give $\ce{Br^.}$ and $\ce{^.CCl3}$.
The second step is $\ce{Br^.}$ abstracting a proton from toluene to give $\ce{HBr}$ and a benzyl radical.
The third step is the $\ce{^.CCl3}$ radical abstracting a proton from toluene to give chloroform and a benzyl radical.
The fourth step is a benzyl radical abstracting $\ce{Br}$ from $\ce{Br-CCl3}$ to give benzyl bromide and $\ce{^.CCl3}$.
The fifth step is two $\ce{^.CCl3}$ radicals quenching to give hexachloroethane $\ce{Cl3C-CCl3}$.

So overall the products are $\ce{HBr}$, Benzyl bromide, Chloroform, and Hexachloroethane.
Reactions 3 and 4 are the major portion of the reaction so equimolar amounts of benzyl bromide and chloroform are formed along with small amounts of $\ce{HBr}$ and hexachloroethane (approx 20:1).


Is NMR possible using nitrogen as the nucleus? Is NMR spectroscopy performed using nitrogen nuclei? If yes, then where?

Of course yes -- because of the spin $\ce{^{15}N}$ and the much lesser used $\ce{^{14}N}$ nuclei posses:

The image is from here, the site provides some useful additional  overview to the source mentioned by @orthocresol.

P.S. If there were no NMR active nitrogen nuclei, 3D NMR would not this widespread in protein research:

(source)


is avogadro's law for the same gas? Avogadro's law is V1/n1 = v2/n2 - volume and number of  moles are directly proportional. My question is - can this law apply to two different gases - for example, Hydrogen with volume v1 and number n1, and Oxygen with volume v2 and number n2 - are they proportional?

Since Avogadro's Law is speaking about ideal gases then it's true for all gases that behave ideal.
Since ideal gases don't interact with each other and don't have any volume for single molecules (they only have a volume as a whole) Avogadro's Law holds for different gases.

Since for real gases the approximations of the ideal gas don't hold then Avogadro's Law doesn't hold. Yet a ratio similar can be achieved by dividing V over n of different gases to get the molar Volume and by using different equtions to describe real gases a ratio factor can be achieved for a given pressure and temperature.


Confusion about the graph plotted from the Arrhenius equation According to Arrhenius equation $k = A\mathrm e^{(E/RT)}$, where $k$ is the rate constant, $E$ is the activation energy, $R$ is the universal gas constant and $T$ is the temperature.

Taking log both sides,
$$\log k = \log A - \frac{E}{2.303RT}$$
Why is the graph plotted between $\log k$ and $1/T$ linear? We know $E =$ threshold energy—average energy of the molecules. Now average energy is given by $\sqrt{ (8RT/\pi M)}$, it is itself a function of temperature. So the slope is variable and hence should have been some other curve.

Arrhenius equation

$k = A \exp{\dfrac{-E}{RT}}$

where:

    k is the rate constant
    E is the activation energy
    R is the universal gas constant
    T is the temperature

The line

    Why is the graph plotted between $\log k$ and $1/T$ linear?

Taking log both sides,

$\text{log}(k) = \text{log}(A)-\dfrac{E}{2.303RT}$

A straight line for $y$ in terms of $x$ is:

$y = a + bx$

so:

  $y$ is $\text{log}(k)$
  $a$ is $\text{log}(A)$
  $b$ is $-\dfrac{E}{2.303R}$
  $x$ is $\dfrac{1}{T}$

so plotting $\text{log}(k)$ versus $\dfrac{1}{T}$ gives a straight line.

E "confusion"

    We know $E =$ threshold energy—average energy of the molecules. Now average energy is given by $\sqrt{ (8RT/\pi M)}$, it is itself a function of temperature. So the slope is variable and hence should have been some other curve.

As you note there is some temperature dependency. But from the Wikipedia article:

    "Given the small temperature range kinetic studies occur in, it is reasonable to approximate the activation energy as being independent of the temperature."

It is sort of circular reasoning but look at the Arrhenius equation in log form.

  $\text{log}(k) = \text{log}(A)-(\dfrac{E}{2.303R})\dfrac{1}{T}$

then

  $\dfrac{d(\text{log}(k))}{d(1/T)} = -\dfrac{E}{2.303R} $= constant

if $\text{log}(k)$ and $\dfrac{1}{T}$ are not linear over the temperature range of interest because of a temperature dependency, then the reaction isn't of the "Arrhenius equation" type. Such plots are in fact used to verify confirmatory to the Arrhenius equation over the temperature range of interest.


How Would A Vast Mixture (Like an Ocean) of Ammonia-Water Behave in Terms of Evaporation and Freezing? First, I'll go ahead and say I'm not very versed in chemistry. My question comes from my interest in planetary-sciences, specifically the speculation that there are likely exoplanets and moons with surface oceans of water-ammonia. However I can't find much hard info on how a vast body (i.e. ocean) of water-ammonia would behave differently than just water.

In short my question is how might a water-ammonia ocean differ from a "pure" water ocean in terms of:

Ice(s) formation
Evaporation
Changes in water/ammonia ratio based on the above two factors (ice formation and evaporation).

This seems largely like a chemistry question to me. If I'm wrong please let me know which Stack Exchange you think I should post in. :)

Parameters:

5% - 40% ammonia.
Temperature range: -80C to 15C
Atmospheric Pressure: 0.5 bar to 30 bar.

More Info:

I've heard that the freezing and boiling points of the mixed liquid would exist on some continuum between the freezing and boiling points of its constituent liquids (ammonia and water), varying depending on the exact mix ratio and of course pressure. Is that correct?

What I'd like to most wrap my head around is whether the water-ammonia mix would behave as a single uniform liquid, freezing and evaporating at a rate determined by the average of the mix's freezing/boiling point, or whether the ammonia molecules would start to evaporate before the water molecules, and the water molecules start to form ice before the ammonia molecules.

If the later behavior is correct I could imagine an ocean of water-ammonia varying substantially in its water/ammonia ratio from pole to equator thanks to the greater heat near the equator and the ammonia molecules evaporating.

Thanks!

Freezing

  I've heard that the freezing and boiling points of the mixed liquid would exist on some continuum between the freezing and boiling points of its constituent liquids (ammonia and water), varying depending on the exact mix ratio and of course pressure. Is that correct?

For freezing points, that is incorrect. Pure water freezes at 273 K and pure NaCl (table salt) "freezes" at 1074 K. However, if I dissolve NaCl in water, the freezing point of water will be lowered. In you water/ammonia scenario, water will freeze as pure water, but the freezing point will be lower. Ice formation would increase the concentration of ammonia in solution.

Evaporation

Water and ammonia mix in the liquid state, and everything mixes in the gas state. Evaporation (and condensation) depend on the ratio of components in the liquid state, the partial pressure of components in the gas state (or ratio of components and total pressure), and the temperature. At equilibrium at earth-like temperatures, there would be more ammonia than water in the atmosphere (the boiling point of ammonia is much lower than that of water, so the vapor pressure of ammonia is higher than that of water). A sudden drop in temperature would make it rain ammonia, mixed with a bit of water. So condensation and precipitation would increase the concentration of ammonia in the liquid state, and a sunny day would lower it.

  If the later behavior is correct I could imagine an ocean of water-ammonia varying substantially in its water/ammonia ratio from pole to equator thanks to the greater heat near the equator and the ammonia molecules evaporating.

You are making a lot of assumptions here: that the exoplanet rotates fairly quickly around its own axis, that its axis is roughly perpendicular to the star, and that the star is close enough to heat up the planet.

  What I'd like to most wrap my head around is whether the water-ammonia mix would behave as a single uniform liquid, freezing and evaporating at a rate determined by the average of the mix's freezing/boiling point, or whether the ammonia molecules would start to evaporate before the water molecules, and the water molecules start to form ice before the ammonia molecules.

No, they would do different things. Pure water at terrestial atmospheric pressure freezes at 0 $^\circ$C, while ammonia already boils at -33 $^\circ$C. Mixing them will lower the freezing point of water and increase the boiling point of ammonia, so there is no temperature where you would separate them completed by freezing all water and boiling off all ammonia. On the other hand, they don't freeze or boil in lock step. If you lower the temperature, water will form ice and the ammonia concentration in the liquid will increase. If you increase the temperature, more ammonia than water will turn into gas (evaporating or boiling), so the ammonia concentration in the liquid will decrease.


Ab Initio and Molecular Orbitals What is the difference between ab initio calculations and molecular orbital theory?

Ab initio means calculations that are strictly from the principles of quantum mechanics. This contrasts with a non ab initio method which can involve parameters that do not arise directly from principles (i.e. results from experiment and/or 'better' calculations). One typical example of something that is not ab initio is a functional for Density Functional Theory. Because the exact functional is unknown, different functionals are generated using large datasets of high level calculations. These are often good at reproducing experimental data in many situations however you cannot derive them directly from the postulates quantum mechanics.

Molecular Orbital Theory (which can be implemented with the Hartree-Fock method) is an example of an ab initio method. One can (very roughly) come up with MO theory by approximating each electron as only feeling the effect of the average positions of the other electrons. So, if an orbital has an electron in it all the other electrons will feel a repulsion from the space occupied by that orbital. This results in a set of coupled one particle problems the solutions (eigenfunctions) which can be solved to find the Molecular Orbitals for the system.

More generally, ab initio calculations also refer to the more accurate post Hartree-Fock methods such as Configuration Interaction (CI), and Coupled Cluster (CC).


condensation of tellurium on inside of ampoule? After heating an ampoule of tellurium around its melting point in a furnace for about a day, I've noticed drops of the element on the upper inside surface of the ampoule, as if it condensed there. I've also noticed similar behavior with other intermetallic compounds. Is it common for something like tellurium to condense and resolidify in drops when an ampoule of it is heated (like water as shown here)?

This is a general phenomenon. All substances at any temperature above $0\ K$ will shed a few atoms from the solid or liquid, forming a gas however dilute. The tendency for this to happen varies with temperature, and it can be measured by determining the equilibrium pressure of a substance inside an evacuated chamber. The key concept behind this is vapour pressure. There are several previous Chem.SE questions regarding vapour pressure, so you may find something useful, such as perhaps this.

As a rule of thumb, the higher the boiling/sublimation point of a substance, the lower its vapour pressure at any given temperature (in fact, the definition of the boiling/sublimation point is the temperature where the vapour pressure equals the pressure exerted by the surroundings onto the solid/liquid substance, which is in many cases equal to $1\ bar$). Tellurium is a relatively low boiling-point solid, with a vapour pressure of $1\ bar$ at $988\ °C$. According to this article, the vapour pressure of tellurium at its melting point ($450\ °C = 723\ K$) is approximately $70\ Pa$. This means there is a fair amount of tellurium vapour present inside the ampoule at this temperature. It is entirely conceivable that some of this vapour could deposit in small droplets on the upper surface of the glass ampoule, outside the main body of the liquid. Note that theoretically the deposition of tellurium on the upper surface should also happen at room temperature, though the equilibrium vapour pressure of tellurium at $25 °C$ is only about $5 \times 10^{-4} Pa$, which would make the transfer of a macroscopically visible amount of tellurium an extremely slow process.


Precipitate Sodium Hypochlorite from tap water before distillation Is there a fairly simple way to precipitate Sodium Hypochlorite from tap water before I distill it? Or, second best, a way to neutralize it?

Sodium hypochlorite is not carried over into water by distillation.  Instead, at high (~100°C) it decomposes, primarily to sodium chloride (i.e. table salt) and oxygen gas, but also possibly to trace amounts of chlorate.

You could neutralize the hypochlorite by mixture with stoichiometric amounts of hydrogen peroxide, which cause immediate release of oxygen gas, a reaction which occurs even at room temperature.


Non-radiative transitions or what features of rhodamine result in it being so highly fluorescent? Rhodamine is a fluorescent dye. What features of the molecule result in it being so highly fluorescent ?

I assume the extended conjugated system with a stabilized first excited state is necessary, to absorb light in the visible region of the spectrum, and I would also think a rigid, planar structure is preferable for good Franck-Condon factors for absorption. Are there other factors involving the energy/stability of the $S_{1}$ state I should be aware of ?

Rigidity is definitely an important factor. At $295\ \mathrm{K}$, indeno[2,1-a] has a fluorescence quantum yield $\Phi_F$ of 0.94, (DOI). In an organic glass at $77\ \mathrm{K}$, even (E)-stilbene, which hardly shows any fluorescence at room temperature, exhibits  a $\Phi_F$ of 0.75.

Vibrational stretching of $\ce{C-C}$ bonds in substituents can significantly contribute to radiationless deactivation of an $S_1$ state and therefore reduce the fluorescence quantum yield $\Phi_F$. Admittedly, toluene isn't famous for its fluorescence ($\Phi_F = 0.15$), but the fluorescence of tert-butyl benzene is even lower ($\Phi_F = 0.03$).

Apart from radiationless deactivation through internal conversion (IC), an excited singlet state can be depopulated by intersystem crossing (ISC) to a triplet state. Under certain conditions, this is even possible for compounds that usually have very low triplet quantum yields.

Halobenzenes added to a solution of fluorescing anthracenes in dioxane (DOI) or ethyl iodide added to solutions of pyrene in alkanes (DOI) have shown to quench the fluorescence normally observed in the absence of the halogenated compounds. The latter exhibit a external heavy atom effect.

The lack of $S_1 \rightarrow T_1$ intersystem crossing is probably the most important factor that makes rigid compound show large $\Phi_F$ values.

It is striking that strongly fluorescent compounds typically have an extended $\pi$-system. Here, the $S_0\rightarrow S_1$ excitation is a $(\pi,\pi^*)$ transition.

On the other side of the fence, ketones, such as acetophenone or benzophenone, hardly show fluorescence. For these compounds, $S_0\rightarrow S_1$ excitation of the $\ce{C=O}$ group is a $(n,\pi^*)$ transition. From a $^1(n,\pi^*)$ state, spin-orbit coupling can both flip the spin and move one electron to a more distant and geometrically different orbital. $S_1 \rightarrow T_1$ thus is in fact a $^1(n,\pi^*) \rightarrow {}^3(\pi,\pi^*)$ transition.

For compounds like rhodamin B in their $^1(\pi,\pi^*)$ states, the  corresponding transition to a $^3(\pi,\pi^*)$ state is largely spin-forbidden ans does not occur. Provided that no external effects come into play, these molecules exhibit strong fluorescence.


Reaction between Sudan IV and lipids The Sudan IV is a molecule used to detect lipids in a substance, especially the triglycerides. The Sudan IV molecule formula is  $\ce{C24H20N4O}$ and a triglyceride molecule could have this molecule formula $\ce{C55H98O6}$.

When the two solutions are mixed, there is a color change. What causes this? I assume that it is because a new compound is produced and this compound is red colored.

Is it possible to find the chemical equation of the reaction between Sudan IV and triglyceride?

I don't think that's how Sudan IV works. Sudan IV is a highly conjugated aromatic diazo compound, so it is already highly colored. In the lipid, the Sudan IV is basically hydrophobic so it will stain any lipid droplets suspended in water. The color is due purely to the Sudan IV, not a reaction.


hydroxychloroquine (HCQ) Synthesis: Which mechanism? Draw the product of the following reaction and provide an explanation for this reaction knowing that $\mathrm{ROH}$ or $\mathrm{RNH_2}$ +  $\mathrm{PhCl}$ do not react at all.

I have no doubt that the answer is hydroxychloroquine (HCQ), but how would it be formed if the reaction:
$\mathrm{RNH_2}$ +  $\mathrm{PhCl}$
does not happen? Or, is the question hinting that the reaction takes the $\mathrm{S_N1}$ pathway where the phenylcation gets formed? If so, why does the Cl atom para to the nitrogen leave more readily than the other Chloride?

The 2 and 4 positions of quinolines are similar to the 2 and 4 positions of pyridine in reactivity. 2 and 4 haloquinolines/pyridines are highly reactive to nucleophiles by SNAr. Indeed 4-Cl-pyridine has to be handled as the hydrochloride because of its propensity to self condense.

The 7 position, by contrast, is unactivated and similar in reactivity to benzenes, so the 7-Cl is not going to undergo SNAr without specialist catalysis.


Why does chlorine oxidise iron to iron(III), not iron(II)? I was asked to predict the product of the reaction between iron metal and chlorine gas:

$$\ce{Fe +Cl2->} ?$$

The product here is supposed to be $\ce{FeCl3}$. But how would we know if the product is $\ce{FeCl2}$ or $\ce{FeCl3}$? Why should it be $\ce{FeCl3}$ and not $\ce{FeCl2}$?

The enthalpy of formation ($\Delta H_\mathrm{f}^\circ$) of $\ce{FeCl3}$ is $\pu{-399.49 kJ mol-1}$, while the $\Delta H_\mathrm{f}^\circ$ of $\ce{FeCl2}$ is $\pu{-341.79 kJ mol-1}$. This means that $\ce{FeCl3}$ is $\pu{57.7 kJ mol-1}$ more stable than $\ce{FeCl2}$, a considerable amount. This means that it is more thermodynamically favorable for $\ce{FeCl3}$ to form than $\ce{FeCl2}$, likely due to the larger lattice energy.

Furthermore, in the +2 oxidation state, one electron remains paired in the $\mathrm{3d}$ orbital. When $\ce{Fe}$ is in the +3 oxidation state, however, it has a half filled $\mathrm{3d}$ orbital, a state which is known to be particularly stable, which you can read about further here.


What will be the name of CH2=CH-O-CO-CH3? We have this question in the textbook but there is no answer given and I have no idea how to name this compound.

Edit: I tried naming the compound.
Correct me if I am wrong here
1-Methylpropenoate

Most people would call it vinyl acetate. If you want the systematic name, then it is ethenyl ethanoate.

It is formally derived from "vinyl alcohol" and acetic acid, from which the name follows easily. See here (https://en.wikipedia.org/wiki/Ester#IUPAC_nomenclature), for example.

The compound is https://en.wikipedia.org/wiki/Vinyl_acetate.


What quantity of dessicant agent It is known that dessicant agents such as magnesium sulfate can be used in order to trap water. But I was not able to find what quantity has to be added ideally compared to the amount of water that would be trapped. I know you can put it in excess but it could be a waste sometimes.

You add dissicant until it suspends easily in the solvent when you swirl the flask. Some people say: "Until it snows".

When you add a bit of magnesium sulfate or other fine powdered dissicant, it forms a lump and goes to the bottom.
Many times it just sticks to the bottom when you swirl the flask.
You add more solid and swirl until it suspends and then slowly falls to the bottom, like it is snowing. At that point it is taking no more water and it has no sense to add more.


For Hess' law type questions, when do you take the inverse of the constant and when do you switch the sign? For example, when you're given the enthalpy $\Delta H$ of a  reaction and you want to switch that reaction around in order to add it to something else, you switch the sign. Furthermore, when adding two reactions, you add their $\Delta H$ values.

But when you're working with equilibium problems and you have the $K$ of a reaction, if you want to switch the reaction you take the inverse of $K$ and when you add reactions you multiply $K$ values.

When working with cell potentials $E_{cell}$ of a redox reaction, you switch the sign if you switch the equation around, but when you're adding two equations you do not multiply $E_{cell}$ by whatever you multiply the coeffecients on the species by.

Are there any types of patterns other than the three I've described here? Any help is greatly appreciated.

Equilibrium constant

When a reaction is at equilibrium, the species will have concentrations called the equilibrium concentrations. They do not depend on whether the reaction is written in one direction or in the reverse direction, and whether all the stoichiometric coefficients are multiplied by the same value. However, your equilibrium expression will be different (switching concentrations from numerator to denominator if reversing the chemical equation, or changing exponents when changing stoichiometric coefficients). As a result, the value of the equilibrium constant will change while the equilibrium concentrations themselves will remain the same.

Enthalpy of reaction

For a given reaction, the enthalpy change associated with making - say - 1 mol of species A is a given. If you use 1 mol of species A in the reaction instead, the enthalpy change will be of the same magnitude but opposite sign. If I make a different amount of species A, the enthalpy change will change in a manner proportional to that amount. In a similar manner, if I change the chemical equation for $\ce{A -> B}$ to $\ce{2A -> 2B}$, the molar enthalpy of reaction will double as well. The same goes for Gibbs energy or for entropy of reaction.

Cell potential

For a given reaction with certain concentrations of species in the two half cells of a voltaic cell, I will measure a certain cell potential. If I switch the role of reactants and products (in the equation), I will still measure the same potential (the direction of the actual reaction does not change). If I multiply all coefficients in the chemical equation by the same number, it still correctly describes the reaction, but this will not change the potential.

Mathematical relationships

$$\Delta G^\circ = - R T \ln(K)$$

The way logarithms work, if you multiply $\Delta G^\circ$ by a number, you have to raise K to the power of that number. If you add Gibbs energies, you have to multiply equilibrium constants.

$$\Delta G^\circ = - z F E^\circ $$

If you multiply $\Delta G^\circ$ by a number, z (the number of electrons transferred) will also increase by that number. Therefore, $E^\circ $ does not change.

Multiplying all coefficient in a chemical equation by (-1) changes the role of reactants and products. Thus, once you know the rules for changing coefficients by a common factor, you don't need to memorize another rule for reversing the equation.


Nomenclature of simple molecules I'm from Slovakia, 16, and I'm really confused about the English simple molecule (oxides, hydroxides, salts, acids) nomenclature. I guess we do it differently. To sum it up: we have suffixes from 1-8 to communicate the oxidation number of an element in a molecule. We put those suffixes at the end of the elements stem, combine it with the word for what kind of molecule it's forming (oxide, hydroxide, ...), and then we can determine from that name and the oxidation number of the other elements (if it's an oxide then oxygen if it's a hydroxide then the hydroxide anion and so on) what it's chemical formula is. Is it done the same way in English speaking countries? Do you just have to search for the oxidation number in a periodic table? If not then how?

Thanks!

Edit: we use these suffixes for oxidation numbers:

-ný/á
-natý/á
-itý/á
-ičitý/á
-ičný/ečný/á
-ový/á
-istý/á
-íčelý/á

(ý or á is based on gender of the word...)

They all represent the positive oxidation number of a part of a simple molecule, for example, Al2O3 would be "oxid hlin-ITÝ" because aluminum is hliník in Slovak and -itý means that the oxidation number of the aluminum is +3.
How would you write down $\ce{Al2O3}$ in English? And also how do you name salts, bases, and acids?

EDIT2(for Karsten Theis)

\begin{array}{|c|c|c|}
\hline
\text{Formula} &  \text{English name} & \text{Slovak name}  &\text{Oxidation #}  \\
\hline
\ce{FeCl2 }& \text{ferrous chloride, iron(II) chloride} &  \text{chlorid želez-natý}  & +2\\
\hline
\ce{FeCl3 }& \text{ferric chloride, iron (III) chloride} & \text{chlorid želez-itý} & +3\\
\hline
\ce{FeO }& \text{iron(II) oxide} &  \text{oxid želez-natý}  & +2\\
\hline
\ce{Fe2O3 }& \text{iron (III) oxide} & \text{oxid želez-itý} & +3\\
\hline
\ce{N2O}& \text{nitrous oxide} & \text{oxid dus-ný} &+1 \\
\hline
\ce{NO}& \text{nitric oxide} & \text{oxid dus-natý} &+2\\
\hline
\ce{HNO2}& \text{nitrous acid} & \text{kyselina dus-itá} &+3\\
\hline
\ce{HNO3}& \text{nitric acid} & \text{kyselina dus-ičná} &+5\\
\hline
\ce{NO2-}& \text{nitrite} & \text{dus-ita-n} &+3\\
\hline
\ce{NO3-}& \text{nitrate} & \text{dus-ičita-n} &+5\\
\hline
\ce{N^3-}& \text{nitride} & \text{nitrid} &-3\\
\hline
\ce{CrO}& \text{chromous oxide} & \text{oxid chróm-natý} &+2\\
\hline
\ce{CrO3}& \text{chromic oxide, chromium trioxide} & \text{oxid chróm-ový} &+6\\
\hline
\ce{Cr2O3}& \text{chromium (III) oxide} & \text{oxid chróm-itý} &+3\\
\hline
\ce{H2CrO4}& \text{chromic acid} & \text{kyselina chróm-ová} &+6\\
\hline
\ce{HCl}& \text{hydrochloric acid} & \text{kyselina chlóro-vodíková *} &-1\\
\hline
\ce{HClO}& \text{hypochlorous acid} & \text{kyselina chlór-na} &+1\\
\hline
\ce{HClO2}& \text{chlorous acid} & \text{kyselina chlor-itá} &+3\\
\hline
\ce{HClO3}& \text{chloric acid} & \text{kyselina chlor-ečná} &+5\\
\hline
\ce{HClO4}& \text{perchloric acid} & \text{kyselina chlor-istá} &+7\\
\hline
\ce{Al2O3}& \text{aluminum oxide} & \text{oxid hlin-itý} &+3\\
\hline
\ce{CaO}& \text{calcium oxide} & \text{oxid vápen-natý} &+2\\
\hline
\ce{Li2O}& \text{lithium oxide} & \text{oxid lít-ný} &+1\\
\hline
\end{array}

*(vodíková means that that acid only has hydrogen and not oxygen)

In English, French and German, the names are less systematic. Here are some examples from English:

\begin{array}{|c|c|c|}
\hline
\text{Formula} &  \text{Name} & \text{Oxidation #}  \\
\hline
\ce{FeCl2 }& \text{ferrous chloride, iron(II) chloride} &  +2\\
\hline
\ce{FeCl3 }& \text{ferric chloride, iron (III) chloride} &  +3\\
\hline
\ce{N2O}& \text{nitrous oxide} &  +1 \\
\hline
\ce{NO}& \text{nitric oxide} &  +2\\
\hline
\ce{HNO2}& \text{nitrous acid} &  +3\\
\hline
\ce{HNO3}& \text{nitric acid} &  +5\\
\hline
\ce{NO2-}& \text{nitrite} &  +3\\
\hline
\ce{NO3-}& \text{nitrate} &  +5\\
\hline
\ce{N^3-}& \text{nitride} &  -3\\
\hline
\ce{CrO}& \text{chromous oxide} &  +2\\
\hline
\ce{CrO3}& \text{chromic oxide, chromium trioxide} &  +6\\
\hline
\ce{Cr2O3}& \text{chromium (III) oxide} &  +3\\
\hline
\ce{H2CrO4}& \text{chromic acid} &  +6\\
\hline
\ce{HCl}& \text{hydrochloric acid} &  -1\\
\hline
\ce{HClO}& \text{hypochlorous acid} &  +1\\
\hline
\ce{HClO2}& \text{chlorous acid} &  +3\\
\hline
\ce{HClO3}& \text{chloric acid} &  +5\\
\hline
\ce{HClO4}& \text{perchloric acid} &  +7\\
\hline
\ce{Al2O3}& \text{aluminum oxide} &  +3\\
\hline
\ce{CaO}& \text{calcium oxide} &  +2\\
\hline
\ce{Li2O}& \text{lithium oxide} &  +1\\
\hline
\end{array}

I think this is what happens if names of substances were chosen before the concept of oxidation state (and electrons etc) existed. It would be nice to have something more systematic, but that would mess up all the older literature.

It would be fascinating (to me, at least) to see the Slovak names in another column of this table. Also, I am glad that chemical formulae exist that are universal in any language.


Why is the chemical formula for ethanoic acid CH3COOH and not C2H4O2? Why does the hydrogen ion move to the front of the formula and not to the subscript of the H already in the formula?

The first one is preferred since its makes it easy to see that it has a carboxyl group and a methyl group, so its easier to draw the Lewis structure and figure out the molecule's properties. For example, from the first formula one can see that it is an acid. However, its more difficult to figure that out from the second one.


What is the relationship between equilibrium concentrations in two similar experiments? Two experiments were performed involving the following equilibrium. The temperature was the same in both experiments.
  $$\ce{H2 (g) + I2 (g) <--> 2HI (g)}$$

  In experiment A, $\pu{1.0 M}~\ce{H2}$ and $\pu{1.0 M}~\ce{I2}$ were initially added to a flask and equilibrium was established. In experiment B, $\pu{2.0 M}~\ce{HI}$ was initially added to a second flask and equilibrium was established. Which of the following statements is always true about the equilibrium concentrations?

  $\ce{[H2]}$ equals $\ce{[HI]}$ in experiment A.
  $\ce{[HI]}$ equals $\ce{2[H2]}$ in experiment A.
  $\ce{[HI]}$ in experiment A equals $\ce{[HI]}$ in experiment B.
  $\ce{[HI]}$ in experiment A equals 2 $\ce{[I2]}$ in experiment B.

Can someone explain why the answer is 3. Doesn't that mean 2 and 4 are true as well?

Note that you are not given the equilibrium constant so you do not know the absolute equilibrium concentration of any species at equilibrium.  This eliminates 2 and 4.

Since all conditions and total concentration of reactants + products are the same between the two experiments, you do know that the concentration of any species in experiment A is equal to that species in experiment B.  Again, only at equilibrium, and you don't know to what degree that equilibrium lies in terms of absolute concentrations.


major products of this reaction? My attempt:
polar protic solvent, low temp, tertiary carbon, and a weak (?) nucleophile and a weak base.

It seems like this is a pretty standard SN1 reaction so "I and II" would be the correct answer?

I think your answer is correct as long as you can justify it. (IV) is clearly absurd but why do you eliminate (III) ? However in my opinion at $\ce{20 ^{\circ} C}$ there would be no reaction from a practical point of view, i.e. one with reasonable yield and reaction time.


Why is phenyl carbanion aromatic? In phenyl carbanion, one carbon atom has a lone pair on it:

This answer claims that the anion is aromatic, because "the lone pair on the benzene is in the plane of the ring and hence orthogonal to the conjugated $\pi$-system so that it cannot interact with it."
Why is the lone pair perpendicular to the plane of the molecule? It seems clear to me that the carbon in question has three $\mathrm{sp^2}$ orbitals, perpendicular to which is a pure $\mathrm{p}$-orbital containing two electrons. Assigning one extra electron (i.e., a negative charge) to an uncharged carbon in a benzene ring shouldn't rotate its $\mathrm{p}$-orbital by $90^\circ$.
The $\mathrm{p}$-orbitals in carbon are, of course, always perpendicular to each other. The $\mathrm{p}$-orbital containing the lone pair should, therefore, be perpendicular to the hybridized orbitals.

The lone pair of electrons will be in a $\mathrm{sp^2}$ hybridized orbital and not in  a pure p-orbital. Thus, it wil be in the same plane as the molecule. Only the lone pairs which are in an orbital perpendicular to the molecular plane can take part in resonance, so those $\mathrm{sp^2}$ electrons will not be counted in Hückel's rule.


Equation for the reaction of iodide to iodine using hydrogen peroxide under acidic conditions? I'm building a home lab around All Lab, No Lecture Illustrated Guide to Home Chemistry Experiments and I'd like to understand the chemistry that I'm doing a bit better. For example, here's an iodine experiment.

  Weigh out 2.0 g of potassium iodide and transfer it to a test tube.
  Add about 1.5 mL of distilled water to the test tube and swirl to dissolve the potassium iodide.
  Add 1.5 mL of concentrated hydrochloric acid (or about 1.8 mL of hardware store muriatic acid) to the test tube and swirl to mix the
  solutions.
  Add about 10 mL of drugstore 3% hydrogen peroxide. The solution immediately turns dark brown as the iodide ions are oxidized to
  elemental iodine, which precipitates out.
  Swirl the test tube to suspend the iodine and pour the liquid through a funnel with a folded piece of filter paper to capture the
  iodine crystals.
  Rinse the iodine crystals on the filter paper several times with a few mL of distilled water. The rinse solution appears brown from
  dissolved iodine, but iodine is not very soluble in water, so you're
  not losing much of your yield.
  Spread out the filter paper on a watch glass or saucer and allow the crystals to dry thoroughly. Iodine gradually sublimates (passes
  directly from solid to gaseous form) at room temperature, so don't
  leave the crystals exposed to air any longer than necessary to dry
  them.
  Once the crystals are dry, transfer them to a sealed storage bottle or vial.

It's easy enough to follow and I know I have $$\ce{KI, H2O, HCl, H2O2}$$ with a result of $$\ce{I2\, +\, ???}$$ but then what?

I've had HS Chemistry, skipped Chem in college :(, so I know how to balance equations, convert from g or ml to moles, etc, but where can I learn to translate an experiment into equations without having to ask others (Textbooks or online resources are ideal, going and taking chem at a local uni. is a distant second, since I work fulltime)?

Currently I'm sticking to inorganic chem, so that should simplify things a bit.

I'm currently studying EdX's Solid State Chemistry, so I hope that'll help some.

We know that this is a redox situation, where iodide ion is oxidized to iodine, and where hydrogen peroxide is reduced to water.

The oxidation equation is as such: $\ce{2I- -> I2 + 2e-}$

The reduction equation is as such: $\ce{2H+ + H2O2 + 2e- -> 2H2O}$

Therefore, the overal equation is as such: $\ce{2H+ + H2O2 + 2I- -> 2H2O + I2}$

You can read more about redox reactions here.


Understanding the increase in pH of a buffer solution upon incremental additions of NaOH analytically So let's say we have $200\ \mathrm{mL}$ of $1\ \mathrm M$ $\ce{CH3COOH}$ solution. In this solution we have the equilibrium $\ce{CH3COOH <=> CH3COO- + H+}$. To that we add $100\ \mathrm{mL}$ of $1\ \mathrm M$ $\ce{NaOH}$ solution. Then after reacting we get a buffer solution where $[\ce{CH3COOH}] = [\ce{CH3COO-}] = \frac{1}{3}\ \mathrm{mol\ dm^{-3}}$. In moles, we have $0.1\ \mathrm{mol}$ of both $\ce{CH3COOH}$ and $\ce{CH3COO-}$.

Then suppose we add $x\ \mathrm{mL}$ of $0.1\ \mathrm M$ $\ce{NaOH}$. This is, in effect, reacting $0.1\ \mathrm{mol}$ of $\ce{CH3COOH}$ with $0.0001 x\ \mathrm{mol}$ of $\ce{NaOH}$, the result of which is that now $n(\ce{CH3COOH}) = (0.1 - 0.0001 x)\ \mathrm{mol}$ and $n(\ce{CH3COO-}) = (0.1 + 0.0001 x)\ \mathrm{mol}$.

Could we not then model, by the Henderson-Hasselbalch equation, $\mathrm{pH} = \mathrm pK_\mathrm a + \log\left(\dfrac{0.1 + 0.0001 x}{0.1 - 0.0001 x}\right) = 4.76 + \log\left(\dfrac{0.1 + 0.0001 x}{0.1 - 0.0001 x}\right)$?

But this would mean that for the pH of the buffer solution to rise by $1$, we would need $818.82\ \mathrm{cm^3}$ of $0.1\ \mathrm M$ $\ce{NaOH}$ solution, which sounds absurd. Moreover, for different acids, like propanoic acid and butanoic acid, the same line of logic could be used to deduce that the volume required to increase the $\mathrm{pH}$ by $1$ would be the same for all, but they have different buffering capacities. So what's wrong with the logic?

I just answered what I thought was this question but realized that it was essentially the same question asked 5 years ago. So here's that answer modified for the slight difference in the way it is asked here:

We need to know the the pH of the buffer i.e. what does it measure before any NaOH is added. Later we will assume that it is 4.76, the pK of acetic acid.
$[\ce{HAc}] = [\ce{Ac^-}]$ ([·] symbolizes the molar concentration of ·). In order to solve the general problem we need to be able to calculate the fraction of the total Ac that is dissociated and the fraction that isn't.

The fraction that is dissociated comes right out of the Henderson - Hasselbalch equation and is $$f_1 = 1/(1 + r_1)$$ where $$r_1 = 10^{(pH - pK_1)}= [\ce{Ac^-}]/[\ce{HAc}]$$ as is clear from inspection of the Henderson - Hasselbalch equation. The subscript 1 indicates that $r_1$ is the ratio of the number of acid ions that have lost 1 proton to the number that have lost $ 1 - 1 = 0$. In $f_1$ the subscript is indicative that $f_1$ is the fraction of the total Ac molecules that has become singly charged by loss of a single proton. When dealing with monoprotic acetic acid the subscripts aren't that important as there is only one proton to loose. But if the acid is polyprotic we have unionized, once ionized, twice ionized etc. ions to consider. The fractions of those ions are $f_0$, $f_1$, $f_2$... and we have other ratios as well

$$r_j = 10^{(pH - pK_j)} = [H_{n-j-1}Ac^{−j}]/[H_{n-j}Ac^{-(j-1}]$$ Here "Ac" stands for the acid anion and n is the number of protons it can yield when fully dissociated. For acetic acid. $\ce{CH_3COOH}$, and Ac is $\ce{CH_3COO}$, n = 1 and j only has values of 0 or 1. With phosphoric acid, $\ce{H_3(PO_4)}$, Ac is $\ce{(PO_4)}$, n = 3 and j = 0,1,2 or 3.

So suppose now we have a polyprotic acid with values for $r_1$, $r_2$, $r_3$... and that there are x moles of the acid in a solution. Then there would be $xr_1$ moles of the singly deprotonated species, $xr_1r_2$ moles of the doubly deprotonated, $xr_1r_2r_3$ moles of the triply deprotonated and so on.Then the total number of moles of Ac would be the sum of the number of moles of each$$C_{Ac} = x + xr_1 +xr_1r_2 +xr_1r_2r_3...$$ The fraction of the total that is undissociated is $$f_0 = x/(x + xr_1 +xr_1r_2 +xr_1r_2r_3...) = 1/(1 + r_1 +r_1r_2 + r_1r_2r_3...)$$ The fraction that is singly dissociated is $r_1$ times this $$f_1 = r_1f_0$$ and the fraction that is doubly dissociated is $r_2$ times that $$f_2 = r_2f_1$$ and, in general $$f_j = r_jf_{(j-1)}$$

At this point let's remember that $f_j$ is a function of the solution pH and all the pK's of the acid in question and that it is the fraction of the anions of that acid that carry charge -j. Thus we can write an expression for the total charge on all species of Ac at a given pH. This is $$Q_{Ac} = -C_{Ac}(0f_0 + 1f_1 + 2f_2 + ...)$$

Before going on to show you how Q solves buffering problems let me stop to suggest that the simplest way to work with it is to make an Excel (or other) spread sheet. Designate a column for pKs and a cell into which the pH goes. For example, put pH into cell A1 and start the pKs list in A2, Then in B2 put =10^($A$1 - A2).  Using $A$1 lets you copy and paste B2 into as many cells as you have pKs. The B column now contains the r corresponding to the pK in the cell to the left of it. Now enter the formula for $f_0$ in a cell and make another column with the $f_j = r_jf_{(j-1)}$ formula in it. So how many pK's. I say make the spread sheet for 5 or 6. Why? Well lets go back to $\ce{CH_3COOH}$ for a minute. It doesn't have 1 proton to give, as we have been assuming. It actually has 4. Are the other three ever coming off? Not with any base in my lab but we can model those other protons simply by assigning pKs that are so high (say 50) that the f values for anything other than $f_0$ or $f_1$ are 0. The point being that if you are going to go to the trouble to make the spreadsheet you might as well make it big enough to handle any acid you may ever encounter as it easily handles anything up to its maximum size using this trick.

Now how to use $Q(pH)$. If there are a total of $C_{Ac}$ moles of Ac in a solution the negative charge on them at $pH_0$ is $C_{Ac}Q(pH_0)$. At $pH_1$ it is $C_{Ac}Q(pH_1)$. Thus to move the solution from $pH_0$ to $pH_1$ you must supply or remove charge of $$\Delta Q_{Ac}(pH_0\ce{->}pH_1) = C_{Ac}Q(pH_1) - C_{Ac}Q(pH_0)$$ by adding or absorbing protons. If $ pH_0  > pH_1$ then $Q(pH_1) > Q(pH_0)$ (less negative) and so the difference will be positive indicating that protons (acid) will need to be added to effect this pH shift.

The acid species in the solution are not the only thing that emits or absorbs protons when pH changes. The solvent does too.

$$\Delta Q_{W}(pH_0\ce{->}pH_1) = 10^{-pH_1} - 10^{-pH_0} + (10^{(pH_0 - pK_w)} - 10^{(pH_1 - pK_w)})$$ represents the number of protons that must be supplied (or absorbed) to change the pH of water from $pH_0$ to $pH_1$.

Now let's use this to solve the original question. We have 300 mL of 4.76 buffer with $C_{Ac}$ = 0.2 mol ( 0.1 mol of Ac and 0.1 mol of $Ac^{-1}$. This implies that $f_0 = f_1 = 0.5$ Also
$$\Delta Q_{Ac}(pH_0\ce{->}pH_1) = 0.02(Q(pH_1) - Q(4.76))$$
$$\Delta Q_{W}(pH_0\ce{->}pH_1) = 10^{-pH_1} - 10^{-4.76} + (10^{(4.76 - pK_w)} - 10^{(pH_1 - pK_w)})$$

Aparently we want to know how much NaOH we would need to raise the pH to 5.76.  With our formulas in a simple Excel spread sheet and using 5.76 for $pH_1$ we quickly find that $\Delta Q_{Ac}(4.76\ce{->}5.76) = -0.0818 mole. IOW protons must be absorbed. At the same time we find $\Delta Q_{W}(4.76\ce{->}5.76) = -1.56E-5 mole per liter. We only have 0.3 L so the protons to be removed to adjust the water are insignificant. We apparently need 81.8 mL of 1 M NaOH. Note that your digits are pretty similar so one of us is off by a power of 10. As 0.8 mol OH- to shift an 0.2 mol buffer is a lot, I think it's you. As we both used Henderson Hasselbalch, we had better get the same answer.

So why would you do what I'm suggesting instead of what you are doing (assuming you find the factor of 10). If you do a lot of problems like this it makes cranking them out a snap.

Now this post has gone on long enough so I won't have space to tell you about extension of this technique to much more complicated systems. You can use it to find the pH of complicated mixes of weak and strong acids and bases or any materials which have buffering capacity. I ginned it up (this is just using the 'proton condition' to solve a complex system - I didn't invent it) to predict the pH of brewer's mash which is a mixture of weak acid (bicarbonate) and several malts (each of which has its own buffering properties).


Acidity of substituted phenols Phenol has a $\mathrm pK_\mathrm a$ approximately equal to $9.9$.

When one studies the acidity of chlorophenols, one notices the following:

First of all, chlorophenols are more acidic than phenol, due the negative inductive effect (−I) of chlorine, that reduces the negative charge, located on the oxygen of the phenolate anion.

Among the different chlorophenols, the observed trend is explained by the fact that the stabilizing −I effect of chlorine decreases with the distance from oxygen.

Now focusing on another class of substituted phenols, such as methylphenols (= cresols), the $\mathrm pK_\mathrm a$ varies as follows:

In this case, the methyl group yields a positive inductive effect (+I), thus increasing the negative charge on phenolate oxygen. That's why cresols are less acidic than phenol. One can further realize that the meta compound is more acidic, and this is due to the resonance structures which show a negative charge in ortho and para, but not in meta.

I would appreciate to learn whether such a reasoning, for these two type of substituted phenols, is plausible or whether further more solid arguments should be invoked.

In particular, I would appreciate to understand why the difference in acidity between 2- and 4-chlorophenols is much more pronounced than the same difference between o- and p-cresols.

I would appreciate learning whether such reasoning, for these two
  type of substituted phenols is plausible or whether furthermore
  solid arguments should be invoked.

Your reasoning seems generally well thought out.  You clearly have a good understanding of inductive effects and you mention resonance effects in the cresol series.  Structural effects don't play a role with these compounds, but let me mention them for completeness.  If we have a compound such as I with a carbonyl ortho to the hydroxyl substituent, then in addition to inductive (stabilizes the anion) and resonance (stabilizes the anion) effects, we should also consider hydrogen bonding; which, in this case, would stabilize the phenol making it less acidic.  On the other hand, in compound 2, the bulky t-butyl group would force the ester carbonyl to twist out of the plane of the benzene ring.  In II, the inductive effect of the carbonyl would still operate, but the resonance and hydrogen bonding effects on acidity would be diminished.

  I would appreciate learning why the difference in acidity between
  2- and 4-chlorophenols is much more pronounced than the same
  difference between o- and p-cresols

This link discusses electrophilic aromatic substitution, where again both inductive and resonance effects need to be considered.  About two-thirds of the way down the page we see that chlorobenzene undergoes electrophilic nitration about 30 times slower than benzene, but is still ortho-para directing.  We also see that toluene reacts about 25 times faster than benzene and is also ortho-para directing.  From these and other examples, we learn that chlorine donates electrons weakly by resonance and withdraws electrons strongly through its inductive effect; and that a methyl group is a mild electron donor through resonance (hyperconjugation), while its inductive effect is also electron-donating, but is small.

Let's apply these concepts from electrophilic aromatic substitution to your two phenol series.  In the chlorophenols, we would expect the chloro substituent to influence the series primarily through a strong electron-withdrawing inductive effect.  Indeed, as the chlorine moves closer to the hydroxyl substituent, the inductive electron withdrawal increases in power and the $\mathrm pK_\mathrm a$ increases dramatically.  On the other hand, in the methyl series we expect mild resonance (hyperconjugation) effects to dominate (even though the inductive effect is in the same direction), therefore in the ortho- and para-cresols we expect the anion to be weakly destabilized and consequently a bit less acidic than phenol itself; while the meta isomer, with minimal resonance and inductive effects affecting the anion, should have an acidity close to phenol.


Sodium Perborate as a detergent ingredient Sodium Perborate is a well known bleaching agent in detergents. What makes Sodium Perborate an effective bleaching agent and how is this related to its oxidation chemistry?

Connecting with water, sodium perborate releases hydrogen peroxide.
$$\ce{NaBO2*H2O2*3H2O->Na+ + BO2- + H2O2 + 3H2O}$$
$$\ce{H2O2->H+ + HOO-}$$
$$\ce{HOO- ->HO- + O}$$

Dissolution of sodium percarbonate in water at hand washing and the washing machine washing, leads to the formation of the perhydroxy anion. This anion is an effective bleach, but it is formed in sufficient quantities only at temperatures above 60 ° C.
From the perhydroxy anion obtain atomic oxygen, which has a strong oxidizing capacity.


Metal rod and NMR machine I have a metal rod in my leg from previous surgery. I believe it is titanium. Would this significantly affect/be affected by an nmr machine? I noticed that a sign by it said not to bring metal objects in the room and I wanted to be sure.

A little bit about me - I manage a busy research NMR Facility at a leading international tertiary institution. I deal with questions like this not infrequently, and there are some very important things you need to understand about implants and magnetic fields. I have all users of our Facility sign a waiver prior to access to indicate whether they have medical or surgical implants, and for those that do - there is probably 3-4 a year - we make a specific case assessment to determine whether it is safe for them to use the instruments. For some people, we simply arrange an alternative pathway to have samples run. This may be having others run them altogether, or load up samples into an autosampler for them.

A bit about you - that you 'think' it is a titanium implant is not sufficient to allow a proper assessment in your case. Unless you can get this is writing from your doctor, then I would have to assume a surgical grade of stainless steel. And it needs to be made perfectly clear that most surgeons don't know this information. There is no such thing as completely non-magnetic stainless steels - even the highest grade austenitic steels contain minute amounts of ferrite materials. I have had many examples over the years of people believing one thing about their implant and then discovering the truth later. In the most interesting case, I had a user who had surgery and believed they had a titanium plate in their skull. Their doctor put this in writing. Only after a follow-up MRI scan some years later did they discover that they did NOT have any implant. Nothing at all. The doctor checked his records.....ah yes - they didn't need to use one after all! Some implants have additional attachments and electronics that meed to be considered above and beyond the basic 'rod' and these can both be affected by, and affect magnetic fields.

Typically, implants will rarely affect the magnet of the spectrometer to the point of creating problems. In any case, problems only arise through moving through the field - a static object is generally no problem. With MRI, this 'movement' comes from the application of field gradients. Just as passing a current through a coil generates a magnetic field, moving a metal conductor through a magnetic field creates an electric current.

Even surgical grade metals are affected by the stray fields of magnets, and depending on the age, design and size of magnet, this may be of concern to you as a spectrometer operator. Movement of metals through the field will create local eddy currents which can result in localised heating. In MRIs, a common incident is serious burns from the embedded metal in bra straps. For biomedical implants, this risk is still considerable, and consideration should be given to:

how long ago the surgery was done. Recent surgery may not have allowed adequate time for bone growth to sufficiently embed the implant, and heating can create swelling and cause problems that may need to rectified. I am aware of one such case with an implant in the wrist that required follow-up surgery to rectify intense swelling that resulted from exposure to a strong magnetic field.
where the implant is. Some areas will be more susceptible to problems, or least symptoms. Not everyone that has an implant can detect any changes. Some people do. I'd say that most people who say they experience a symptom have implants associated with joints - elbows and wrists are the most common that come to mind. Symptoms have ranged from tingling in the joint to feeling warm. How much of this is genuine, and how much psychosomatic, I can't say. I once had an older tradesman working within the stray field for a day who commented that the tattoos on his forearm were itchy by the end of the day.
what the implant is. Some implants are just not worth taking a risk with. Aneurysm clips, for instance, are simply not worth taking a risk with. Finding an alternative way for a user to access the spectrometer is much easier than cleaning up an exploding head (yes, an exaggeration, I know) and doing the paperwork. Some pacemakers are MRI approved, and most are actually perfectly ok within the small stray fields of a spectrometer. But some aren't ok, so I wouldn't take this risk without plenty of written evidence from doctors and medical suppliers.
the extent of stray field form the magnet. All magnets are different through age, design and magnitude. An older unshielded 600 poses a very high dose of exposure; a newer shielded 300 has the 5G line contained within the magnet case, and has an almost neglible exposure. However, magnetic fields are invisible, and unless you know the details about the specific magnet you are using, it is best to assume the worst case. We base all of our stray field limits on our unshielded 400, which has a much greater stray field than our 500 for instance.

Regardless of the final assessment of your condition, I would still advise against undertaking high field procedures, such as changing probes, or doing low temperature connections etc.

Ferromagnetic materials can obviously be a problem because of the large attractive forces created. Most incidents with these arise from large external metal objects being brought within the stray field; gas cylinders and trolleys etc. MRIs are usually more problematic with stray fields due to the bore size, but small objects can also pose a serious concern in strong fields. These include handheld tools, but for NMR spectrometers the biggest problems actually come from things like staples and paperclips and even iron filings that migrate across the floor over time and can cause serious problems if they get near the magnet bore.

The take-home message for you is not to take advice from a bunch of stackexchange users beyond understanding that you need to discuss your individual case with your NMR Facility Manager, your doctor, and quite possibly your supervisor if alternative arrangements need to be considered for your access. A great resource for you to be aware of is Frank Shellock's MRI Safety site.


Why are alkali metal compounds like sodium hydride or sodium amide strong bases but weak nucleophiles? I understand why compounds such as $\ce{NaH}$ or $\ce{NaNH2}$ are weak nucleophiles: as they aren't very soluble in organic solvents, they react only on the clusters' surface. But why are they strong bases?

I could understand why they are in water, as they are soluble in water, but in my organic chemistry class it was said that they accelerate E2 reactions. So how can they be strong bases in organic solvents if they aren't soluble in organic solvents?

You're right they're difficult to use in organic reactions because of the solubility problem you mention. However, by using crown ethers like dibenzo-18-crown-6 they can be captured and the crown ethers are subsequently dissolved so that the bases are "brought into solution" and react.


Is there a site where I can input RGB values and the site will respond with a pH? I have a tool where I point it at something, and it gives me an RGB value. I want to add universal indicator to a liquid, measure the RGB value and input it into the site to find the pH. Does anyone have any ideas?

Your best bet would be to prepare 15 solutions of buffers at pH 0, 1,.. 14 and calibrate your instrument. Dip three different test papers in each solution, get average numbers. Normalize colors and see the r/g, r/b, and g/b ratios for each pH. Now you can just make a long if statement in excel and plug your measured numbers there.

Fill excel with 15 statements like
=if(r/g > 5 and r/g <8 and r/b>3 and r/b<3.5; "pH=3"; "")

Also, many pH papers come with a map of colors for each pH. you can use that for calibration.

The thing is, chemists use pH meters when they need accurate readings. accuracy down to mili-pH is achievable. pH meters are build on very different principles and are reasonably priced.


Is it still possible for an amateur, a hobbyist, or a science buff to make relevant discoveries in chemistry today? Consider astronomy. Hobbyists and amateurs are today still able to make small to moderate scientific contributions to their field. Is the same true in chemistry?

I think there might be at least two restricting factors:

The equipment needed to make scientific discoveries is far too expensive (e.g. hundreds of thousands of dollars/euros/whatever) for an amateur to obtain
Not having a bachelor's + master's degree (so being ready for a PhD) is too limiting ("too much to study alone").

If there is no "correct" answer for the field of chemistry in general, please post an answer per bigger sub-field (e.g. organic chemistry).

It's conceivable that somewhere out there, someone with a completely fresh point of view could have an annus mirabilis in an area like theoretical chemistry, working with just a paper and pencil (or a laptop) and home-brewed equipment and resources.  Yes, for most of us, all of the low-hanging fruit is gone---but for a real giant, the whole tree might be full of low-hanging fruit.

Most of us aren't giants, but as Newton said, we can stand on the shoulders those who are. Doing that requires some ability to search the extant literature, even for "small to moderate contributions".

In addition to the obstacles you've listed above, there is a third impediment: paywalls.

There was a time when you could walk into a university library and sit down with the Science Citation Index and do a good job of finding research related to yours. Nowadays, you need a faculty or student university ID to do such a search and retrieve the papers you need. Without an ID at a good university, our amateur, hobbyist, or science buff will  hit paywall after paywall. Amateurs might have to pay half a day's wages or more for the pdf of a single paper!

Paywalls are not only a major obstacle for amateurs---they discourage professionals in the third world, and they shut out young people who are beginning their careers outside of university.

Consider Jack Andraka's story, here: http://blogs.plos.org/thestudentblog/2013/02/18/why-science-journal-paywalls-have-to-go/


Why is the empirical formula of sodium chloride NaCl? Since we know in sodium chloride crystal sodium ions are surrounded by $6$ chloride ions and each chloride ions are again bonded to $6$ sodium ions. What I want to know is that on what basis empirical formula of sodium chloride was assigned to be NaCl? does that formula mean sodium ion should atleast bonded to one chloride ion?

The empirical formula of, say, $\ce{P4O10}$ is $\ce{P2O5}$ but that doesn't mean that each phosphorus is bonded to only 2.5 oxygens. It's perfectly possible (and in fact it is pretty much a requirement) that if Na has 6 nearest-neighbour Cl and Cl has 6 nearest-neighbour Na, then the overall stoichiometry is NaCl.

Let's say, for example, that there are $n$ sodium atoms in a sample of NaCl. Then, if each $\ce{Na+}$ atom has 6 nearest-neighbour $\ce{Cl-}$ ions, then at first glance we might say there are $6n$ chloride ions.

However, since each $\ce{Cl-}$ ion has 6 nearest-neighbour $\ce{Na+}$ ions, this means that in the earlier calculation we counted each $\ce{Cl-}$ ion 6 times (once per $\ce{Na+}$ neighbour). So in fact there are only $6n/6 = n$ chloride ions, leading to a stoichiometry of $1:1$.

Maybe if this is too complicated for you, you can reduce it to one dimension. Consider the following (infinite) arrangement of red and green trees along a road:

Surely you'd agree that each red tree has two green neighbours, and each green tree has two red neighbours. Yet the "stoichiometry" is obviously $1:1$.

Now if you tile that in three dimensions, you get the $\ce{NaCl}$ structure.


What is the oxidation mechanism of gunpowder? I've been recently looking into fireworks and the chemistry of explosives, and I found out that, amongst others, one of the most important reactions in gunpowder ignition is the oxidation of charcoal and sulfur by $\mathrm{KNO_3(s)}$. I am curious as to whether it is known if this reaction occurs between charcoal/sulfur and solid state potassium nitrate or if the potassium nitrate decomposes first and the reaction occurs via gaseous oxygen as an intermediate.

Is there a general mechanism of oxidation between solid oxidizing salts and organic matter/sulfur? Does the same thing happen with $\mathrm{KClO_3(s)}$ or $\mathrm{KMnO_4(s)}$ for example?

$\mathrm{KNO_3(s)}$ can oxidize charcoal without sulfur. It needs a "hot" primer through to get the reaction started.  Sulfur's role is to lessen the ignition temperature.

See section in Wikipedia article.

Since the reaction isn't complete and has a number of possible products bthere isn't an exact chemical formula for black gunpowder burning. The Wikipedia article lists several approximations to black gunpowder burning. What is supposedly the best follows.

    Although charcoal's chemical formula varies, it can be best summed up by its empirical formula: $\ce{C7H4O}$. Therefore, an even more accurate equation of the decomposition of regular black powder with the use of sulfur can be described as:
    $\ce{KNO3 + C7H4O + 2S → K2CO3 + K2SO4 + K2S + 4CO2 + 2CO + 2H2O + 3N2}$


Exchange-correlation hole I did a google and chemistry.exchange search and found several technical definitions of the exchange correlation hole.

The questions that I want to proposes are:

What the exchange-correlation hole means in common words?
How do you explain it to a non-literate person?

What the exchange-correlation hole means in common words?

Assuming that the probability is the common word, one could say that the exchange-correlation hole, is a region of space around an electron in which the probability of finding another electron is close to zero due to electron correlation.

  How do you explain it to a non-literate person?

Long story short is that mathematically electron correlation arises as a consequence of the instantaneous Coulomb repulsive forces between electrons and also as a consequence of the fundamental principle of antisymmetry of an electronic (fermionic in general) wave function. Note that the first kind of electron correlation which is due to Coulomb repulsions is known as the Coulomb correlation while the second one which is due to antisymmetry of a wave function is knows as the exchange correlation. However, since the root of all ab-initio wave function methods, the HF method, already accounts for exchange correlation, the Coulomb correlation is usually also referred to as just the correlation.

With respect to these two different kinds of electron correlation note that exchange-correlation hole does not refer specifically to just the exchange correlation, rather it is the hole due to both types of the electron correlation mentioned. I think that one would better call it, say, exchange-Coulomb correlation hole to avoid an ambiguity, but I'm afraid we're stuck with the exchange-correlation hole term.

Long story

Consider the simplest many-electron system, a two-electron one, and let us examine the following two important events

$\vec{r}_{1}$ - the event of finding electron-one at a point $\vec{r}_{1}$;
$\vec{r}_{2}$ - the event of finding electron-two at a point $\vec{r}_{2}$.

Probability theory reminds us that in the general case the so-called joint probability of two events, say, $\vec{r}_{1}$ and $\vec{r}_{2}$, i.e. the probability of finding electron-one at point $\vec{r}_{1}$ and at the same time electron-two at point $\vec{r}_{2}$, is given by
\begin{equation*}
    \Pr(\vec{r}_{1} \cap \vec{r}_{2})
    =
    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) \Pr(\vec{r}_{2})
    =
    \Pr(\vec{r}_{1}) \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) \, ,
\end{equation*}
where

$\Pr(\vec{r}_{1})$ is the probability of finding electron-one at point $\vec{r}_{1}$ irrespective of the position of electron-two;
$\Pr(\vec{r}_{2})$ is the probability of finding electron-two at point $\vec{r}_{2}$ irrespective of the position of electron-one;
$\Pr(\vec{r}_{1}\,|\,\vec{r}_{2})$ is the probability of finding electron-one at point $\vec{r}_{1}$, given that electron-two is at $\vec{r}_{2}$;
$\Pr(\vec{r}_{2}\,|\,\vec{r}_{1})$ is the probability of finding electron-two at point $\vec{r}_{2}$, given that electron-one is at $\vec{r}_{1}$.

The first two of the above probabilities are referred to as unconditional probabilities, while the last two are referred to as conditional probabilities, and in general $\Pr(A\,|\,B) \neq \Pr(A)$, unless events $A$ and $B$ are independent of each other.

If the above mentioned events $\vec{r}_{1}$ and $\vec{r}_{2}$ were independent, then the conditional probabilities would be equal to their unconditional counterparts
\begin{equation*}
    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) = \Pr(\vec{r}_{1}) \, , \quad
    \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) = \Pr(\vec{r}_{2}) \, ,
\end{equation*}
and the joint probability of $\vec{r}_{1}$ and $\vec{r}_{2}$ would simply be equal to the product of unconditional probabilities,
\begin{equation*}
    \Pr(\vec{r}_{1} \cap \vec{r}_{2}) = \Pr(\vec{r}_{1}) \Pr(\vec{r}_{2}) \, .
\end{equation*}
Note that this is the picture in which electrons are sort of billiard balls, i.e. neutral macroscopic particles.

In reality, electrons are, of course, not like billiard balls; they are charged microscopic particles, and consequently,
\begin{equation*}
    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) \neq \Pr(\vec{r}_{1}) \, ,
    \quad
    \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) \neq \Pr(\vec{r}_{2}) \, ,
\end{equation*}
and
\begin{equation}
    \Pr(\vec{r}_{1} \cap \vec{r}_{2}) \neq \Pr(\vec{r}_{1}) \Pr(\vec{r}_{2}) \, .
\end{equation}
And since the spatial probabilities of finding electrons are inevitably related to their states, the inequality above means that the state of electron-one is not independent of the state of electron-two and vice versa.
This interdependency of the states of electrons is referred to as the electron correlation.

The inequalities above, which are the mathematical essence of electron correlation, hold true for two reasons. First, electrons repel each other by Coulomb forces, and, as a consequence, at small distances between the electrons the conditional probability is less than half of the corresponding unconditional one
$$
    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) < \Pr(\vec{r}_{1}) \, ,
    \quad
    \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) < \Pr(\vec{r}_{2}) \, ,
$$
while at large distances between the electrons the conditional probability is greater than half of the corresponding unconditional one
$$
    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) > \Pr(\vec{r}_{1}) \, ,
    \quad
    \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) > \Pr(\vec{r}_{2}) \, ,
$$
In the extreme case when $\vec{r}_{1} = \vec{r}_{2}$, the Coulomb repulsion between the electrons becomes infinite, and thus, $\Pr(\vec{r}_{1}\,|\,\vec{r}_{1}) = 0$ and consequently $\Pr(\vec{r}_{1} \cap \vec{r}_{1}) = 0$, i.e. the probability of finding two electrons at the same point in space is zero.
The correlation due to Coulomb repulsion is termed as Coulomb correlation and one sometimes speaks about the presence of the so-called Coulomb hole around each electron - a region of space around it in which the probability of finding another electron is close to zero due to Coulomb repulsion.

Secondly, as a consequence of the Pauli exclusion principle, electrons in the same spin state can not be found at the same location in space, so that for electrons in the same spin state there is an additional contribution into the inequalities between conditional and unconditional probabilities above.
This effect is relatively localized as compared to one due to Coulomb repulsion, but bearing in mind the relationship between probabilities and wave functions and taking into account the continuity of the later, it is still noticeable when electrons are close to each other and not just at the same location in space.
This type of electron correlation, which is a consequence of the Pauli exclusion principle, is referred to as Fermi correlation, or exchange correlation, and one sometimes speaks about the presence of the so-called Fermi hole around each electron - a region of space around it in which the probability of finding another electron in the same spin state is close to zero.


What is the thermal state of a reaction at equilibrium? I was asked this question at class,

  What is the thermal state of a reaction at equilibrium, the forward reaction of which, is exothermic?

I conjectured that since the forward (exothermic) reaction and the backward (endothermic) reaction occur at equal rates, the reaction at equilibrium would be athermic since there doesn't seem to be any net heat evolution. However, my teacher said this was not true.

Could someone tell me where I'm going wrong with my reasoning?

You're right, as far as I can see. If we look at a particular transformation $A \leftrightharpoons B$ there is a forward reaction and a backwards reaction. They are associated with enthalpy change of different signs and the same magnitude. At equilibrium the rates are the same, so the forwards and backwards reactions release and absorb the same amount of heat and it appears as if nothing is happening.

There is no separate "reaction at equilibrium" with its own enthalpy change. It's not really clear to me what "thermal state of a reaction" means, as systems have states and reactions are pathways between systems.


The effect of temperature on the caffeine intensity Does the brewing temperature have an effect on the caffeine level?
For instance, coffee with 100*C have higher caffeine level compared to coffee with 50*C?

When coffee is being brewed in a moka pot, the water keeps evaporating until it reaches the coffee powder, then it condenses and falls down again, bringing some extracts from the coffee powder: it's just like a small reflux condenser.

When the pressure is high enough, the extract reaches the valve, and it escapes, giving the brewed coffee.

When temperature is too high, the pressure raises too quickly, and water fails to make a good extraction (just think about a reflux condenser which is kept running only for a few cycles).

If the temperature is appropriate, the water performs many evaporization/condensation cycles before it comes out of the valve, hence performing a better extraction (which, I guess, implies an higher caffeine level).

That's why usually a low temperature and a slow brewing are recommended.


Restricted synthesis; formation of nucleophile and electrophile from one carbon source I am tasked by my textbook to synthesize 2-hexyne from 1-propanol. All carbon atoms must come from the starting material, 1-propanol.

My solutions manual (an older version) does not have this additional problem, so I am forced to ask here.

I understand that in this problem there is a double in carbon count in the product, so we are expected to somehow "fuse" the carbons from the starting material. I also understand that this means making an electrophilic carbon and a nucleophilic carbon from the starting materials. The electrophilic carbons that I know about are alkyl halides, alkyl tosylates, epoxides, and acid chlorides.

I'll choose to make an alkyl halide from the starting material through reaction of the primary alcohol (1-propanol) with thionyl bromide and pyridine as a base. Now I have my electrophilic carbon.

My nucleophilic carbon will be the acetylide anion. I could use a cyanide ion but that only gives me a one carbon increase. Plus that's not an allowed carbon source. So I dehydrate my primary alcohol (is this possible?) and creates 1-propene. I can brominate this in an inert solvent (i.e. methylene chloride). From here with the vicinal dihalide I can form the alkyne and then methyl acetylide. I can then react this with the alkyl halide I made previously.

How does this look?

Your proposed synthesis looks reasonable. Dehydration of the primary alcohol is indeed possible, it proceeds via the E2 mechanism (although primary alcohols are less reactive than secondary or tertiary ones). Chlorinated solvents, like methylene chloride and chloroform, are commonly used for the bromination of alkenes, for the reason you have provided (inertness).


Which among phenol and 1,2-dihydroxybenzene has the higher boiling point? My attempt

Based on symmetry: I think that looking for symmetry phenol has a higher boiling point than 1,2-dihydroxybenzene because the two -OH groups projecting out from the benzene ring of 1,2-dihydroxybenzene will not allow close packing of them which results in a lower boiling point.

Based on hydrogen bonding: In this sense phenol would have a lower boiling point because 1,2-dihydroxybenzene would have more inter-molecular hydrogen bonding than phenol.

How do I know which factor wins?

Your structures (grabbed from Wikimedia Commons)

Phenol:

Catechol (1,2-dihydroxybenzene)

The three main influences on boiling point are:

Shape - molecules with a more spherical shape have lower boiling points than molecules that are more cylindrical, all else as equal as possible. Spheres have the lowest surface area to volume ratio of any of the solids. This is where symmetry comes in. More symmetric molecules are more spherical. The shape/symmetry argument does not work here since the two compounds are not isomers. The issue of close packing is relevant to the melting points of solids, since in a solid, most particles do not have translational motion - they are locked in a crystal lattice. In a liquid, such random translational motion is common. In the liquid state, molecules of most compounds are so disorderly arranged that it does not matter how well they pack as a solid. In your case, both molecules are rather like flat disks with a little bit of polarness sticking off one side. Their shapes are not so different as far as surface area to volume goes. Boiling point will be dominated by mass and polarity.
Molecular mass - bigger molecules require more energy to vaporize because they have more mass, and $KE=\frac{1}{2}mv^2$, all else as equal as possible. Mass difference applies in your case because the two molecules are similar enough in shape and polarity, except one has an extra oxygen in its formula.
Polarity - polarity controls the types and number of intermolecular forces, including hydrogen bonding. The more and stronger intermolecular forces between two molecules, the higher the boiling point, all else as equal as possible. Since catechol has two hydroxy groups and phenol has one, you have your strong argument for increased intermolecular forces.

The other easy thing to do is use the internet to look up the boiling points, and then explain the difference using the three factors. Once you know what the boiling points are, it is easy to choose the structural features that support the difference.

Phenol - $181.7 \ ^\circ\text{C}$
Catechol - $245.5\ ^\circ\text{C}$


How to evaluate the heat of formation with computational chemistry? I'm sorry if this is a stupid question, but since I'm a beginner with computational chemistry, please, be patient.

Let's consider the reaction of formation of water:

$$\ce{1/2 O2(g) + H2(g) -> H2O(g)}$$

This reaction shoud give a $\Delta_\mathrm fH^\circ = -241.83~\mathrm{kJ/mol}$. The reaction is intentionally simple to let me run fast calculations and tests with single point energies.

First, I've performed a geometry optimization on $\ce{H2O}$, $\ce{O2}$ and $\ce{H2}$ (HF/6-31G*). Then, I've used M06/Def2-QZVPP for the single points:

$E_\mathrm{SP}(\ce{H2O}) = -76.431519932791~E_\mathrm h$
$E_\mathrm{SP}(\ce{O2}) = -150.262365177069~E_\mathrm h$
$E_\mathrm{SP}(\ce{H2}) = -1.171819656537~E_\mathrm h$

In this case, doing $$E_\mathrm{SP}(\ce{H2O}) - \left(\frac12 E_\mathrm{SP}(\ce{O2}) + E_\mathrm{SP}(\ce{H2})\right) = \Delta E$$ I'm getting $\Delta E = -0.1285176877195084~E_\mathrm h$ that is equal to $-337.42312874~\mathrm{kJ/mol}$, almost $96.6~\mathrm{kJ/mol}$ higher than the reference of $-241.83~\mathrm{kJ/mol}$.
Is this the correct approach for a case like this? Or is this discrepancy normal?

Your approach is quite correct, but as Jan already pointed out, it is incomplete. What you calculated is the difference in the electronic energy of the reaction $$\ce{H2 (g) + 1/2 O2 (g,\,{}^1\Delta_{g}) -> H2O (g)}.$$

Let's turn this into some kind of a tutorial and a little exercise, that you can try to reproduce at home. For a more detailed description on how it works in Gaussian, you can read the paper by Joseph W. Ochterski. I will use Gaussian09 rev. D for the calculations involved.

The enthalpy of formation of water from the standard state of the elements involved, is basically the reaction enthalpy of the combustion of hydrogen and oxygen. You are in luck, because most of the quantum chemical programs shall provide you with the necessary data for reactions straight away. What you do need to do is calculate the thermal corrections to the reaction. Essentially this is done by calculating the Hessian, i.e. a frequency calculation. The second derivatives only have a interpretable meaning if they are performed at a stationary point. Hence they must be performed with the same method, that you used to optimise the geometry.
Jan already pointed out, that HF/6-31G(d) is probably not a well enough fit for geometries and frequencies. I'd always consider some level of theory with at least some coverage of correlation energy. I often prefer BP86 or M06L and the def2-SVP basis set as a first anchor point. The smaller the system, the more accurate the methods you can use (or afford).
Jan also pointed out, that especially for oxygen, the spin state is important. (I did a quick check and am almost 100% certain you calculated it in the singlet state; see above equation.)

For this exercise I have matched your suggestion with using M06/def2-QVZPP as it is still feasible. Keep in mind, that most of the times density functional approximations have to be calibrated. It's all about what you can or cannot afford. You should perform calculations to the best of your abilities.

Okay, let's dive into some mathematics. First, we consider the reaction equation
$$\ce{H2 (g) + 1/2 O2 (g,\,{}^3\Sigma_{g}^{-}) -> H2O (g)}.$$

And we would like to know the enthalpy of formation for water at standard state (which we will use as $298.15~\mathrm{K}$ and $1~\mathrm{atm}$)
$$\begin{align}
\Delta_{f}H^\circ(\ce{H2O}) &= H^\circ(\ce{H2O})
                               - (H^\circ(\ce{H2})
                               +\frac12 H^\circ(\ce{O2}))\\
\Delta_{f}H^\circ(\ce{H2O}) &=
   [E_\mathrm{el}(\ce{H2O}) + H_\mathrm{corr}(\ce{H2O})]\\
&\qquad
 - \left([E_\mathrm{el}(\ce{H2}) + H_\mathrm{corr}(\ce{H2})]
 +\frac12 [E_\mathrm{el}(\ce{O2}) + H_\mathrm{corr}(\ce{O2})]\right)\\
\end{align}$$

The data in the expanded form we can directly take from the output of the frequency calculation:

Oxygen  mo6qzvpp.freq.log/.com
 SCF Done:  E(UM06) =  -150.322374247     A.U. after    1 cycles
 Zero-point correction=                           0.003897 (Hartree/Particle)
 Thermal correction to Enthalpy=                  0.007203
 Thermal correction to Gibbs Free Energy=        -0.016049

Hydrogen mo6qzvpp.freq.log/.com
 SCF Done:  E(RM06) =  -1.17221845130     A.U. after    1 cycles
 Zero-point correction=                           0.009902 (Hartree/Particle)
 Thermal correction to Enthalpy=                  0.013206
 Thermal correction to Gibbs Free Energy=        -0.001590

Water mo6qzvpp.freq.log/.com
 SCF Done:  E(RM06) =  -76.4321360023     A.U. after    1 cycles
 Zero-point correction=                           0.021635 (Hartree/Particle)
 Thermal correction to Enthalpy=                  0.025415
 Thermal correction to Gibbs Free Energy=         0.004010

Therefore we will obtain
$$\Delta_{f}H^\circ(\ce{H2O}) = -236.6~\mathrm{kJ/mol}$$
which I think is reasonably close to the experimental value.

Appendix
The following input files for the computation of above values.

%chk=mo6qzvpp.chk
#p M06/def2QZVPP
opt
int(ultrafinegrid)

oxygen

0 3
O
O 1 1.2
 (blank)

%chk=mo6qzvpp.freq.chk
%oldchk=mo6qzvpp.chk
#p M06/def2QZVPP
freq
geom=check guess=read
int(ultrafinegrid)

oxygen freq

0 3
 (blank)

%chk=mo6qzvpp.chk
#p M06/def2QZVPP
opt
int(ultrafinegrid)

hydrogen

0 1
H
H 1 1.2
 (blank)

%chk=mo6qzvpp.freq.chk
%oldchk=mo6qzvpp.chk
#p M06/def2QZVPP
freq
geom=check guess=read
int(ultrafinegrid)

hydrogen freq

0 1
 (blank)

%chk=mo6qzvpp.chk
#p M06/def2QZVPP
opt
int(ultrafinegrid)

water

0 1
O
H 1 1.2
H 1 1.2 2 109.0
 (blank)

%chk=mo6qzvpp.freq.chk
%oldchk=mo6qzvpp.chk
#p M06/def2QZVPP
freq
geom=check guess=read
int(ultrafinegrid)

water freq

0 1
 (blank)

Have a lot of fun.


Determine reaction concentration at chemical equilibrium with multiple reactions I have a problem understanding how to calculate the change/dependence of the concentrations at the chemical equilibrium using multiple reactions.
The concentration of my reactants and products at chemical equilibrium are given.

E.g.: For a simple reaction like $\ce{C + O <=> CO}$ at chemical equilibrium with concentrations $\ce{10C}$, $\ce{5O}$ and $\ce{2CO}$, I calculate the equilibrium constant $K_{eq} = 2/(5*10) = 0.04$. Then I can use the ICE table and calculate $x$ to derive the concentration of $\ce{CO}$ at chemical equilibrium at various input concentrations for the reactants:
e.g., for an input concentration of $\ce{22C}$ and $\ce{7O}$ (and $\ce{0CO}$) I get a concentration of about $\ce{3CO}$.

BUT how do I calculate the change of my products, if I have to consider multiple reactions with the same reactants involved?
E.g., $\ce{C + O <=> CO}$ and $\ce{C + N <=> CN}$ again with given concentrations at chemical equilibrium of $\ce{10C}$, $\ce{5O}$, $\ce{30N}$, $\ce{2CO}$, and $\ce{8CN}$.

These two reactions are obviously coupled, so solving them independently is not possible (without proper decoupling). I'm also not sure if the equilibrium constant for the reactions can be determined. If I just 'ignore' one of the reactions at a time and calculate the equilibrium constant using the remaining concentrations, I would neglect the coupling which can have a strong effect, e.g., an increase in the input concentration of $\ce{C}$ may influence one reaction by ~90% and have almost no influence at the other (depending on a saturation effect).

I'm also not sure if it is NOT allowed to throw all products and reactants together and just consider them as a single reaction like $\ce{10C + 5O + 30N <=> 2CO + 8CN}$, calculate the equilibrium constant and apply my variation of input concentration like in the example above. How would I account for the case N=0 and CO != 0?!

I think that the only reliable approach would be an iterative calculation of the concentration until convergence is achieved. But I don't have a clue how to do this.

EDIT: I have found part of the solution myself. As I mentioned above, I have to use a system of coupled equation and solve it. So here are my thoughts about it. There has to be conservation of mass (or elements), so the following equations have to hold:

$\ce{C + CN + CO -} nC = 0$

$\ce{N + CN -} nN = 0$

$\ce{O + CO -} nO = 0$

where $nX$ represents the concentration of a certain species. Second, the equilibrium constants also need to be specified like the following:

$\ce{CO - K_1*C*O = 0}$

$\ce{CN - K_2*C*N = 0}$

Solving these coupled equations numerically is no problem. I did some testing and can confirm that at least the uncoupled case ($nN = 0$) leads to exact the same results as the method above. Extending this case to include multiple and more complicated reactions should not be a problem. The only remaining question is how to determine the equilibrium constants for these coupled equations?!

EDIT: I used the same approach to solve the following system of equations for the equilibrium constants:

$\ce{K_2*K_1 - \frac{CN}{(C*N)}*\frac{CO}{(C*O)} = 0}$

$\ce{K_2/K_1 - \frac{CN}{CO}*\frac{O}{C}} = 0$

I used the solution for these equilibrium constants, plugged them into the equations above and calculated the concentrations at equilibrium. The solutions are consistent. (Note: Got myself confused and interchanged $K_1$ with $K_2$ within my computations.)

Conclusion To calculate variations of the input concentrations at chemical equilibrium 2 major steps are needed.

1) Calculate the equilibrium constants $K_x$ for every reaction $x$ at given concentrations. Solve that system of $x$ equations (details in 2nd edit) with a numerical solver of your choice.

2) Adjust/variate the input abundance and use another system of $z$ equations (conversation of mass/element and reaction equations - details in 1st edit) to calculate the new concentrations at equilibrium.

Note: Make sure to always have $n$ equations for $n$ variables.

Answer already found by myself. Short conclusion:

To calculate variations of the input concentrations at chemical equilibrium 2 major steps are needed.

1) Calculate the equilibrium constants $K_x$
for every reaction $x$ at given concentrations. Solve that system of x equations (details in 2nd edit) with a numerical solver of your choice.

EDIT: No numerical solver is needed, since the equations simplify to:

$\ce{K_1 - \frac{CN}{C*N} = 0}$

$\ce{K_2 - \frac{CO}{C*O} = 0}$

which is very interesting, since this simplifies the whole problem a lot. It also means that at chemical equilibrium a reaction does not care if a reactant is depleted via another reaction e.g., is bound within another product, or if the (initial) concentration of that reactant is just lowered to achieve the same amount in equilibrium. No reaction describes the connection between $\ce{CO}$ and $\ce{CN}$ e.g., $\ce{CO + N <=> CN + O}$, so basically the concentration of $\ce{CO}$ and $\ce{CN}$ ARE DECOUPLED (looking at the decoupled equilibrium constants), although they seem to be coupled via their common reactant ($\ce{C}$) e.g., changing $\ce{N}$ (or $\ce{O}$), changes the amount of produced $\ce{CN}$ (or $\ce{CO}$), changes the amount of $\ce{C}$ and further the amount of the other product $\ce{CO}$ (or $\ce{CN}$) .

This gives rise to a new question: How would a 'forced coupling' between $\ce{CO}$ and $\ce{CN}$ via $\ce{CO + N <=> CN + O}$ influence the concentration of both products?

Since we are at chemical equilibrium and have already taken 'indirect coupling via $\ce{C}$' into account, I guess that introducing a triangular-like reaction scheme would only introduce a redundant equation and won't give new information. Maybe I will test for that and share my results.

EDIT: Testing showed no difference in using a triangular-like reaction scheme (including the reaction above). As stated above, the 'new' reaction/equation does not contain new information. The 'new' equation can be written as a linear combination of the other 2 reaction equations. Therefore, one of these 3 reaction equations can be canceled/eliminated since 1) we need $n$ equations for $n$ variables and 2) an equation consisting of a linear combination of other equations cancels out (leads to a zero result) when solving the system of equations.

This means we only need to consider (in this case) the 3 equations for mass/element conservation and 2 reaction equations for the 5 variables (concentration of 3 reactants and 2 products).

2) Adjust/variate the input abundance and use another system of $z$
equations (conversation of mass/element and reaction equations - details in the edit of the question) to calculate the new concentrations at equilibrium.

Note: Make sure to always have $n$ equations for $n$ variables and don't mess them up.


Amount of enantiomers I got a question about this molecule, and I'm not sure how to answer it. I know that this one has 4 chiral C-atoms, but I can't seem to get the number of enantiomers right. I know that the formula for the amount of enantiomers is $2^{n-1}$, with $n$ being the amount of chiral centers, which is $4$. So that makes an amount of $8$ enantiomers. But that's not what it says in the answer.

Am I doing something wrong here?

The compound in question is called D-galactose in carbohydrate chemistry. It has 4 chiral centers (C2, C3, C4, & C5): Thus, stereochemistry of them are $(2R,3S,4S,5R)$. The rule is if a compound has $n$ chiral centers, then it has maximum total of $2^n$ possible stereoisomers, if it does not have any symmetry. D-galactose does not have any symmetry within the molecule (e.g., It would have symmetry if aldehyde group, $\ce{-CHO}$ has been reduced to $\ce{-CH2OH}$). Thus, four chiral centers of D-galactose are accounted for a total of 16 stereoisomers ($2^4$). If you draw them all, you would find they are collection of 8 pairs of enantiomers as:

{$(2R,3S,4S,5R)$, $(2S,3R,4R,5S)$}
{$(2R,3R,4S,5R)$, $(2S,3S,4R,5S)$}
{$(2R,3S,4R,5R)$, $(2S,3R,4S,5S)$}
{$(2R,3S,4S,5S)$, $(2S,3R,4R,5R)$}
{$(2S,3S,4S,5R)$, $(2R,3R,4R,5S)$}
{$(2R,3R,4R,5R)$, $(2S,3S,4S,5S)$}
{$(2S,3R,4S,5R)$, $(2R,3S,4R,5S)$}
{$(2R,3R,4S,5S)$, $(2S,3S,4R,5R)$}.

That make maximum total of 8 pairs of enantiomers, meaning $2^3 = 2^{4-1} = 2^{n-1}$, where $n$ is number of chiral centers. Therefore, your formula should have been for pair of enanthiomers.


Is there a haloalkane for astatine? I'm studying organic chemistry, and I noticed that there are fluoro-, chloro-, bromo-, and iodoalkanes. However, there are none for Astatine. Could someone shed some light on this, and what it would be called?

Part of the reason why there has been very little research done on organoastatine chemistry is the extreme radioactivity of astatine; the half-life of $\ce{^210At}$, the longest-lived isotope, is about 8 hours. However, the element can be incorporated into organic compounds (e.g., via a destannylation reaction).1

In organic chemistry, the prefix is "astato-", as in astatobenzene or 3-astatotamoxifen, both of which have been synthesized.1,2 This is analogous to "fluoro-", "chloro-", and so on. As a general term, astatine compounds can be referred to as astatides (e.g. aryl astatides, analogous to aryl fluorides, chlorides, etc.).1

The aforementioned synthesis of 3-astatotamoxifen is shown below.

As for inorganic nomenclature, astatine can be denoted with the prefix "astatido-"; this follows from the IUPAC Red Book for nomenclature of inorganic compounds.

Milius, R. A., W. H. McLaughlin, R. M. Lambrecht, A. P. Wolf, J. J. Carroll, S. J. Adelstein, and W. D. Bloomer. Appl. Radiat. Isot. 1986, 37 (8), 799-802.
Samson, G., and A. H. W. Aten. Radiochimica Acta 1970, 13 (4), 220-221.


Minimising the formation of manganese dioxide in the titration of potassium permanganate For a titration of potassium permanganate into a standard acidified solution of hydrated iron (II) ammonium sulfate, the solution seems to end up brown (due to the formation of manganese dioxide) during the titration unless the titration is carried out extremely slowly. How can I minimise this? Adding sulfuric acid to the solution seemed to help, but is there a physical way to prevent manganese dioxide forming when the titrant is added too rapidly?

Adding a large excess of acid helps greatly as the reduction to $\ce{MnO2}$ only takes place under neutral conditions. $$\ce{2H2O + MnO4- + 3e- -> MnO2 + 4OH-}$$ Under acidic conditions the $\ce{MnO4-}$ is reduced all the way to $\ce{Mn^{2+}}$.$$\ce{8H+
 + MnO4- + 5e- -> Mn^{2+} + 4H2O}$$ Also swirling or stirring the solution as you are titrating makes a big difference as it stops the area where the permanganate is being added from becoming temporarily depleted of acid.Even if you do get some $\ce{MnO2}$ forming then as long as you have enough acid in the flask a few swirls should be enough to reduce the $\ce{MnO2}$ down to $\ce{Mn^{2+}}$.


Which materials/compounds expand in response to cold temperatures? Which materials or compounds expand in response to cold temperatures?
This material or compound should expand at a slow (but constant rate), from about 0–20 months when in the following temperature range −25 °C to −10 °C.

Also, the compound should not react with plastic or be toxic to humans or the environment.

Please explain why the compound does this as well.

Perhaps one of the most studied materials to exhibit negative thermal expansion is cubic zirconium tungstate $(\ce{ZrW2O8}).$


ClO2 chlorine dioxide how to know if it is still active or not ? by color ? by smell ? by reaction? my wife have ClO2 tablet form. she dilutes it to be put inside mister. each time she make the dilution:
package instruction:
1 tablet to 1 liter water => 100ppm ClO2
*we name this 100ppm as "original soln"
she takes 100ml from the "Original soln" to make 1 liter of solution to do household cleaning.
may i know how much ppm is the final solution that she used to do cleaning ?
i think should be 10ppm solution , right ?  10% mix.
thx

If you start with $\pu{100 mL}$ stock solution at $\pu{100 ppm}$ concentration, and dilute this by a factor of 10 to yield $\pu{1 L}$, then you are correct, the later one has a concentration of $\pu{10 ppm}$.  Keep both in a cool place.
As to check if the solution still is active, or not, and both stock and diluted solution at the expected concentration (which will decrease already if the solution is exposed to hot summer temperatures), you may use test strips you immerse dip into the solution, dry at room temperature, and compare against a scale.  An example were the following

(source)
Depending on the manufacturer, the scale of concentration to test may vary.