Question related to Hydration of Alkynes There will be Markovnikov addition of water and hence —OH will be added to the carbon with triple bond having less number of hydrogen and —H will be added to the one having more number of hydrogen. Then the adduct will tautomerise to form the product corresponding to option A but the answer given is option B. Where am I going wrong?

Due to -I effect from the carbonyl group and the halogen, not far enough for it to be negligible, the stability of the primary vinylic carbocation is counter-intuitively more than that of the secondary vinylic carbocation. Markonikov rule is applicable in absence of permanent electron shift effects, like resonance and inductive effects. The Markonikov rule, under these restrictions, translates into the stability of the carbocation, decided by the positive inductive effect of neighbouring carbons. As a clearer example, consider the hydration of acrolein,

The reaction predominantly yields the primary alcohol, as opposed to the product predicted by the markonikov rule, which again is not applicable in presence of permanent electronic effects. Hope that helps!


Why is the reaction of calcium oxide with sulfur dioxide to form calcium sulfite not a combination redox reaction? I understand that a combination reaction is one where two substances combine to form a third substance, like with
$$\ce{2Na + Cl2 -> 2NaCl}.$$
I do not understand why $$\ce{CaO + SO2 -> CaSO3}$$ is not a combination redox reaction. After all, it is creating a third substance, so why is it not a combination redox reaction?

To determine whether a reaction is a redox reaction, you first have to find the oxidation number of each element in the reactants and products. You do this by either counting the number of electrons in each bond and assigning them to the most electronegative atom in the bond, or by memorizing some rules and applying them. When you do that for this reaction, you should get the following:

\begin{align}
  \ce{   &&CaO + SO2 &-> CaSO3}\\
  \ce{Ca:&& +2 &-> +2}\\
  \ce{O: && -2, -2 &-> -2}\\
  \ce{S: && +4 &-> +4}
\end{align}

Since the oxidation number hasn't changed for any element, no reduction or oxidation has taken place, and this is not a redox reaction.


Commercial chalk for kids composition I have no background in chemistry whatsoever. However, I'm very interested in sidewalk chalk composition. I read that chalk can be both either made out of gypsum or chalk (calcium carbonate). However, I'm confused, since a person that it is connected to a chalk (final product) manufacturer told me that the main ingredient in their product (chalk for kids) is sulfuric acid carbon.

As I said, my knowledge in chemistry is very reduced. I was just hoping that someone that knows well the subject can say if there can be a connection between the sulfuric acid carbon and gypsum.

I am afraid whoever told you "sulfuric acid carbon" does not know the basic terminology (must be from non-chemistry background). There is no such thing as "sulfuric acid carbon" to begin with. Any carbon obtained by charring with sulfuric acid would be jet black. Nobody would like to have a black chalk.

It is very easy to parse $\ce{CaSO4.2H2O}$ incorrectly.

I think he/she misread the calcium $\ce{Ca}$ symbol for carbon which is $\ce{C}$.

$\ce{SO_4^2-}$ can be easily misrepresented to be sulfuric acid.


How to calculate the concentration of lead(II) liberated from lead(IV) oxide under acidic conditions? What is the concentration of $\ce{Pb^2+}$ in a lake that has a pH of $6.0$ and is in equilibrium with $\ce{PbO2(s)}$ and atmospheric oxygen?

\begin{align}
\ce{PbO2(s) + 4H+ + 2e- &-> Pb^2+ + 2H2O} &
\log K &= 49.2
\end{align}

I spent so much time to solve this, but I could not, my final answer does not appear to make sense:
\begin{align}
  \mathrm{pH} &= 6 &
  \implies&& [\ce{H+}] &=10^{-6}\\
  \log K      &= 49.2 &
  \implies&& K &= 10^{49.2}\\
  K &= \frac{[\ce{Pb^2+}]}{[\ce{H+}]^4} &
  \implies&& 10^{49.2} &=\frac{[\ce{Pb^2+}]}{(10^{-6})^4} \\
  &&\implies&& [\ce{Pb^2+}] & =10^{25.2}
\end{align}

What is my mistake?

I think that the problem here is one of notation. I'm pretty sure that the logarithm intended to be used is the natural logarithm, $\log_e(K)$ also represented as $\ln(K)$, and not the base 10 one. Otherwise your method is good.

If you use this, you get a much more reasonable answer (work it out). It also makes sense that the question was worded in this way, as the equations in which one tends to use the equilibrium constant $K$ actually use its natural logarithm, for example the Gibbs's free energy and the Nernst equation.

By the way, this reaction is also part of the reduction process at the cathode of a lead-acid battery.


etymology of supernatant Why do we call a solution that has been centrifuged a supernatant? It seems to me that a "natant" should be something that floats (from Latin "natare" meaning "to swim"), and a "supernatant" should be something which lies above the natant, not below it. Shouldn't the liquid be called a subnatant? Or maybe a superprecipitate?

Edit: fixed Latin (see below)

It means "the liquid that swims above", from the Latin terms super = above and natare = to swim.
So it is the liquid that remains on top.

The subnatant would be the liquid at the bottom, in case you should centrifuge two liquid phases.

The term you name, nato, means son in Latin.


Purification of Ethanol In Vogel's organic book, they have mentioned that in the Lund and Bjerrum method of purification of ethanol, magnesium is added, which forms ethanolate. But they said to activate the magnesium first by iodine.

Why is the activation needed and how exactly does this activation process work in the activation of magnesium?

The refluxing of methanol or ethanol over magnesium is indeed the method of choice to dry these solvents.

As far as the iodine is concerned, this is a classical trick, which is also used to initiate Grignard rections. Is is assumed that magnesium iodide is formed, which is soluble in the alkanols. As a result, a fresh, activated metal surface is left behind.

This might even work, when the magnesium metal is covered with a very thin oxide layer.


I conducted some experiment and I want to know the resulting redish black thing I placed two electrodes in a jar
and poured the sparking water with lemon flavor.
After about 5 hours, the water turned to blue and
the negative probe had the redish black thing.
I'd like to know what it is. Please, reply for me.

Materials : 9V battery, battery holder&wire, jar, copper wire and trevi (sparkling
water)

Connecting battery

After about 5 hours

Result : redish and black thing.

Solvated copper(II) ions for the blue color, copper(II) oxide for the black (as hinted by IvanNeretin). To prove this hypothesis, one could dissolve the copper oxide in sulphuric acid, neutralize the solution and add ammonia. The deep blue color indicates the $\ce{[Cu(NH3)4(H2O)2]^{2+}}$ complex. The latter can also be done for the blue solution currently present.


Finding equilibrium concentrations, trouble simplifying equation For $\ce{CO (g) + H2 (g) <=> CH2O (g)}$

with Kc = 0.068 @ 273K

initial concentrations being $\ce{[CO]}$ = 1.25M, $\ce{[H2]}$ = 2.00M, $\ce{[CH2O]}$ = 1.00M

I did $\frac{1.00}{((1.25)(2.00)} = 0.4$, which is more then Kc, so it's leaning to the left, will produce more reactants, and products will lower. So the I did:

$\frac{1.00 - x}{(1.25 + x)(2.00 + x)} = 0.068$

Multiplied the bottom and got $X^2 + 3.25x + 2.5$

so

$\frac{1.00 - x}{X^2 + 3.25x + 2.5} = 0.068$

I then tried multiplying the denominator by 0.068 and rearranging to solve for $x$ from there, but that's where I got stuck. $X$ is supposed to be 0.65.

Everything you do is correct and it is only the final algebraic transformation missing:
\begin{align}
0.068 &= \frac{1-x}{(1.25+x)(2.00+x)}\\
0.068 &= \frac{1-x}{x^2 + 3.25x + 2.5} &&|\cdot (x^2 + 3.25x + 2.5)\\
1-x   &= 0.068(x^2 + 3.25x + 2.5) &&|-1 +x\\
0.0   &= 0.068x^2 + 1.221x -0.83\\
\end{align}

Now you can solve for $x$ according to
\begin{align}
0&=ax^2+bx+c & x_{1,2}&=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\
\end{align}

And that gives me the values $x_1\approx0.656$ and $x_2\approx-18.6$. The latter one of course makes no sense.


Are heterolytic dissociations barrier-less or do they have an activation energy? When a bond is broken into ionic fragments, and I'm specifically thinking about organometallic or coordination chemistry, the two products carry opposite charges and the potential energy curve should, in my humble opinion, be attractive everywhere (at every point where the distance is larger than its equilibrium value).

Does this mean that the dissociation of a compound like ferrocene into $\ce{FeCp+}$ and $\ce{Cp-}$ (in the gas phase) occurs without a barrier?

In my theoretical research I have often seen that, indeed, there is no barrier for such reactions. In modeling the kinetics you take the bond dissociation energy as the forward barrier and assume a zero barrier going backward. This will also be true for radical recombination reactions.


Why doesn't limestone thermally decompose instead of turning to marble? When limestone is buried and put under intense heat and pressure, why does it turn to marble instead of thermally decomposing into calcium oxide and carbon dioxide?

Short answer:

It's not hot enough, and the pressure is too high.

Long answer:

You say:

  intense heat and pressure

The key word is pressure rather than heat. Let's look at the phase diagram of the CaO and CO2 compositional system:

(Calculated and plotted using Perple_X)

Pure limestone and marble are composed of calcium carbonate: CaCO3. In air, the temperature required for calcium carbonate decomposition is 898 °C (considering simple equilibrium and ignoring any vapour pressure effects). As pressure increases, the temperature requirement for this equilibrium decomposition increases very rapidly. It increases so fast that in fact, at above 80 bar, the calcium carbonate will melt into a liquid (call it calcium carbonate magma or lava) instead of decomposing to lime and carbon dioxide.

While 80 bar may seem a lot for us humans (it is higher than the pressure exerted by a 45 kg woman wearing stiletto heels), it is nothing compared to the pressure below us, deep in the Earth. Marble forms at a wide pressure range, but for this discussion let's say that it's somewhere between 2 and 10 kbar. That's 2000 to 10000 bar! Way off the scale of the phase diagram. Also, marble forms in the temperature range of 300 to 800 °C. Again, it's not even enough to melt the calcium carbonate, let alone decompose it to gas.


Titrating and the addition of acidic distilled water I have read numerous online questions similar to the one I am about to ask. But for most of them the answer has been that the reason one can add water to the unknown concentration solution during a titration is because:

Water has a pH of 7
This is a volumetric analysis, so as long as one record the volume of the titrant and analyte used then there is no problem.

My questions are these:

Due to the effect of $\ce{CO2}$ in the atmosphere mixing with the distilled water, it is obvious that the pH of distilled water is on the acidic side (which we confirmed using a pH sensor). Therefore, why does the addition of distilled water not affect the pH? If it is acidic that should mean there are $\ce{H3O+}$ molecules in the distilled water when we add it into the analyte.

How do we take into account the imbalance of the moles of $\ce{H3O+}$ molecules versus $\ce{OH-}$ in each drop of the acidic distilled water (ex. pH of 5.6)

The only solution I have in mind right now is that all of these are acceptable uncertainties (such as the effect of the addition of an indicator), and that significant figures will wipe out these small errors.

Unfortunately I could not find any sources that specifically address your question so I will be answering using logic. If you notice any fallacies or points you disagree with, please point them out.

$\ce{CO2}$ dissolves in water at such low concentrations normally, the pH would not be greatly affected. Unless you are working in extremely small concentrations that require high accuracy, it would not be a very large problem. If you were, you could try to minimize the time the distilled water is opened, and perhaps work in a vacuum really quickly or find some way to remove that factor.
If we were adding the distilled water to another solution, it is possible that the other solution was already in equilibrium with the partial pressure of $\ce{CO2}$ in the atmosphere and thus the concentration would remain constant ( a very small concentration ).

I do agree that protons may be added, but they would be so few, especially if you are reacting the reaction with acid or some strong base that it would be insignificant. Often times, conditions are not perfect, but good enough to be estimated and be close enough. However, measures can be taken to prevent your worries if the experiment demands it.

Any $\ce{CO2}$ probably already dissociated and is in equilibrium, or comes to it very closely. Opening up a coke and watching it enter equilibrium may serve as a nice visual. Don't worry about it in a titration unless you are trying to do something really really unusual. The $\ce{CO2}$ is insignificant in most titrations.

Note, the solubility is 1.45 g/L at 25 °C, 100 kPa max, so even a saturated solution would be not greatly affected. Furthermore, your solution is probably already in equilibrium so you have already accounted for the carbon dioxide. Think about how when you do a titration, it doesn't matter if you begin at 0 or whatever. As long as you account for it.


Reason for decreased boiling point of hexafluoroisoproponal? Ethanol and isopropyl alcohol have boiling points 78.37 °C and 82.6 °C respectively. The increase in the boiling point is obvious due to increase in carbon chain length which resulted in increase in boiling point of isopropyl alcohol.

But recently, I come across fluorinated ethanol and isopropanol, and the trend of boiling point is the reverse. Trifluoroethanol has boiling point 78 °C whereas hexafluoroisopropanol has boiling point 58.2 °C.

what is the reason for the decrease in boiling point of hexafluoroisopropanol? Is it due to increase in the number of fluorine atom, if so then why this resulted in decrease in the boiling point?

Since fluorine is so much more electronegative than hydrogen, its electrons are considerably less polarizable. This decreases the extent to which induced dipoles are able to form (London dispersion forces), which in turn lowers the energy necessary to break the intramolecular forces. This is what gives fluorocarbons lower boiling points than their hydrocarbon counterparts.


Are these numbers Madelung constants? I have no idea of chemistry, just a little bit what I know from school, but I found through some search some connection between entropy (Shannon entropy) of the divisors of some natural numbers and the Madelung constants.

My question is this:

Are these numbers Madelung constants?

1 -0.000000000000000
2 1.38629436111989 ( http://oeis.org/A016627 )
3 1.38629436111989
4 3.31573052232910
5 1.38629436111989
6 5.54531864369184  ( http://oeis.org/A257872 )
7 1.38629436111989
8 5.57849454296371
9 3.29975689063759
10 5.54521210457048
11 1.38629436111989
12 10.7943529563000
13 1.38629436111989
14 5.54518969515687
15 5.54519368543309
16 8.08924328274917
17 1.38629436111989
18 10.7599543952466
19 1.38629436111989
20 10.7921021686465

Thanks for your help!

Related question:
https://mathoverflow.net/questions/346239/entropy-magnitude-diversity-of-finite-metric-spaces-in-number-theory

Edit:
I think I found an explanation for the numbers ( entropy ). It is approximately:

$\tau(n) \log(\tau(n))$, where $\tau(n)$ is the number of divisors of $n$.

So the $1.386$ and $5.545$ are just coincidences with Madelung constants I think.

I don't know if this would be very helpful or not but as i had some free time on hand I got a list of Madelung Constant Values from the Internet

In the list most of values were reported to 3-4 decimal places so I have Compared only the first decimal places as once the second decimal place was used none of the values matched. Maybe this would be a helpful start.

$$M_a$$ of ZnS (Wurtzite Hexagonal) is $10.7153$ (nearby the 12th and 18th entry in your table)

As Oscar Lanzi has commented $1.386$ is approximately $$2\log(2)$$ which is the Madelung constant for a one-dimensional line of alternating charges

As the question has been edited to add further details maybe the second one is a coincidence.

Only Nearby Matches So Far


What is the driving force for Claisen rearrangements? I came across this mechanism (3,3 sigmatropic rearrangement). Why does the allyl vinyl ether have the ability to rearrange itself giving a new product? What drives the reaction?

Is it the instability of allyl vinyl ether (if yes, state why)?

Thank you in advance!

The formation of the carbonyl group is a gain in thermodynamic stability over the vinyl ether. In your example, the monosubstituted alkene is a "wash". In any case, the alkyl substitution pattern of the non-enol ether double bond is not a problem.
The heat of this Claisen rearrangement, first conducted by Hurd and Pollack1 in 1938, can be calculated by determining the heat of formation (ΔHfo) of allyl vinyl ether (1) and 5-hexenal (2). Heats of formation of allyl vinyl ether (1), hexanal (3), 1-hexene (4) and hexane (5) are known (NIST and Chemeo). These heats of formation are shown in black. However, the ΔHfo of 5-hexenal (2) is not accessible in tables but it can be calculated. The heat of hydrogenation of monoalkyl substituted alkenes is ~125 kJ/mol (~30 kcal/mol). The reduction of 1-hexene (4) to hexane (5) is -124.6 kJ/mol, which is computed from the respective heats of formation of 4 and 5. Applying this heat of hydrogenation to the reaction 2 ---> 3 gives a ΔHfo of -128.2 kJ/mol for 5-hexenal (2). Accordingly, the Claisen rearrangement of allyl vinyl ether (1) is exothermic by -100.6 kJ/mol.

1. C. D. Hurd and M. A. Pollack, J. Am. Chem. Soc., 1938, 60, 1905.


Difference between microcanonical and canonical ensemble I feel a little unclarity about this. Canonical ensemble is pictured as many systems in heat reservoir of infinite capacity having $N$ (number of particles), $V$ (volume) and $T$ (temperature) constant whereas microcanonical ensemble is the analogous system having $E$ (energy) instead of $T$ fixed. From my simple viewpoint of introductory pchem, it seems obvious to me that heat capacity is constant hence how is it possible for canonical ensemble to have different energies while having same temperature. Feeling temporarily frustrated I grab McQuarrie's pchem it reads on the topic:

  "Each system has the same values of N, V, and T but is likely to be in
  a different quantum state, consistent with the values of N and V"

A light shines into my lower mood state.
So we talk about very little changes where the concept of heat capacity must be taken with reserve? That is, it is possible that system sucks heat from this infinite heat reservoir which is used to promote a particle in the system to higher rotational state for instance, which causes the system to acquire a tiny bit of energy while temperature will not change at all? Temperature is understood as describing translational motion of particles, right? Is my idea of the difference between these to ensembles correct?

Typically (at least how I learned this after three years of chemistry) the canonical NVT ensemble is created by considering many microcanonical NVE "ensembles" as microstates in this NVT ensemble. I quote from my lecture script for statistical thermodynamics:

  An ensemble with a constant number $N$ of particles in a constant volume $V$ and at thermal equilibrium with a heat bath at constant temperature $T$ can be considered as an ensemble of microcanonical subensembles with different energies $\epsilon_i$.
  The energy dependence of [the] probability density conforms to the Boltzmann distribution. Such an ensemble is called a canonical ensemble.

So your NVT ensemble is many NVE ensembles at different energies. Their statistical weights (the probability of finding a microstate in that particular NVE state) are Boltzmann distributed. The way this distribution of NVE subensembles looks like then defines the temperature of the overarching NVT ensemble.

To illustrate this I have written a small Python program that calculates this distribution for different temperatures of the NVT ensemble. The probability density as a function of the energy of the 48 possible (this number was arbitrarily defined, in principle it could be infinite) NVE subensembles is given below:

As you can see, the probability of finding NVE microstates with low energies drops drastically as the temperature increases. You can think of this as the thermostat taking a NVE microstate at higher energy, removing some of that energy, and then putting it in a lower energy state. Or the other way around if you want to "heat up" a system.


Is there more than one equation for Delta T(reaction)? So, I have to find the $\Delta T_{\mathrm{reaction}}$ of a solution for my lab report. Earlier in the lab, it gave the equation as

$\Delta T_{\mathrm{reaction}}=T_{\mathrm{mixture}}+\frac{1}{2}(T_{\mathrm{substance~1}}+T_{\mathrm{substance~2}})$

and all temperatures were at the exact time of mixing found from a linear equation of temperature points on a graph. Now, I have another mixture which is water and a substance and I am supposed to find the $\Delta T_{\mathrm{reaction}}$.

Here's the thing, we don't have the temperature of the substance, only the temp of the water and the solution it made, so how could I go about doing it? I thought about using room temperature since it was a solid substance but it doesn't say anything about that.

The substance is solid $\ce{NaOH}$.

If You are stating you have your initial water temperature and the final solution of your temperature, then your change in temperature is just your final temperature minus your initial temperature. No need for linear equations. The change in temperature for the reaction is the temperature change of the water due to the heat given off or absorbed by the system to the surroundings.


Why is BuLi a better nucleophile than base? In Carey, Sundberg Advanced Organic Chemistry it is written:

  The relative slowness of the abstraction of protons from carbon acids by organolithium
  reagents is probably also due to the compact character of the carbon-lithium
  clusters. Since the electrons associated with the carbanion are tightly associated with
  the cluster of lithium cations, some activation energy is required to break the bond
  before the carbanion can act as a base. This kinetic sluggishness of organometallic
  compounds as bases permits important reactions in which the organometallic species
  acts as a nucleophile in preference to functioning as a strong base.

Won't the same reasons apply for nucleophilic displacement? Compactness of the aggregates should affect nucleophilicity, too. So why will BuLi add to a carbonyl instead of deprotonating the alpha carbon atom?

This is a trick question, which explains why the first sentence from the book takes so many precautions. It is very difficult to provide a general explanation because there are many factors influencing the result.

Solvent: the solvent is extremely important in such reactions. It
plays an important role in the order of oligomerization of the
organolithium reagent; monomers, dimers, tetramers or polymers. Of
course, the reactivity is really different.
Co-solvent: a co-solvent like HMPT or TMEDA can drastically change
the course of the reaction by changing the order of oligomerization.
Temperature: it is well known that n-butyllithium is a very strong
base at low temperatures (<-60°C) but will become a good nucleophile
at higher temperatures. It can even open a THF ring.
Thermodynamic or kinetic control: organolithium compounds are
usually used under kinetic control, but...
HSAB: how the oligomerization affects the hardness/softness of the
base and its nucleophilicity?
3D shape of the organolithium compound: it greatly influences the
shape and order of oligomers.
...

So now you feel completely depressed, I can try to answer your question. I think that a key factor is actually that in a cluster (dimer, tetramer and so on), charges are more diluted which enhances the nucleophilic behavior. Also, the geometry of the cluster and of its 3D-complex with the substrate could explain why a reaction is favored over the other.


Why is Gibbs free energy more useful than internal energy? $$\mathrm{d}U<T\,\mathrm{d}S-p\,\mathrm{d}V=
\left(\frac{\partial U}{\partial S}\right)_{\!V}\,\mathrm{d}S
+\left(\frac{\partial U}{\partial V}\right)_{\!S}\,\mathrm{d}V$$

So for a change to occur at a constant entropy and volume, the internal energy must decrease. I get that this statement isn't useful because you can't control entropy in a lab. Furthermore, given that:
$$\mathrm{d}G<V\,\mathrm{d}p-S\,\mathrm{d}T=
\left(\frac{\partial G}{\partial p}\right)_{\!T}\,\mathrm{d}p
+\left(\frac{\partial G}{\partial T}\right)_{\!p}\,\mathrm{d}T$$
this shows that a spontaneous change at a constant pressure and temperature is accompanied by a decrease in Gibbs free energy – far more useful as these variables can be controlled.

However, since, for a spontaneous process, $\mathrm{d}U$ must decrease at a constant entropy and volume which is expressed by $\mathrm dU<T\,\mathrm dS-p\,\mathrm dV$. However, expressions for $\mathrm dS$ and $\mathrm dV$ can be expressed as a function of $p$ and $T$ making this inequality a function of pressure an temperature.
\begin{align}
\mathrm{d}U &<T\,\mathrm{d}S-p\,\mathrm{d}V\\
S&=S(p,T)\\
V&=V(p,T)\\
\end{align}

Sub in the total derivatives and factorize:
$$\mathrm{d}U<\left[T\left(\frac{\partial S}{\partial T}\right)_{\!p}
- p\left(\frac{\partial V}{\partial T}\right)_{\!p}\right]\,\mathrm{d}T
+ \left[T\left(\frac{\partial S}{\partial p}\right)_{\!T}
- p\left(\frac{\partial V}{\partial p}\right)_{\!T}\right]\,\mathrm{d}p$$
Thus, $U=U(p,T)$

This implies that for a spontaneous change at constant pressure and temperature the internal energy must decrease however this is not the case clearly. Does making the substitutions invalidate the inequality? Where is the fault in my reasoning?

So, what's the need for Gibbs free energy. Surely a spontaneous change at constant pressure and temperature is accompanied by a decrease in internal energy? There must be a fault in my reasoning, but where and why?

Thanks for the edit - I see now what you meant.

The short answer for why we "need" Gibbs energy is that the internal energy for a spontaneous process at constant temperature $T$ and pressure $p$ does not necessarily decrease. Internal energy is the thermodynamic potential (wiki link) for constant $S$ and $V$ - this makes sense if you think about it:

If the internal energy decreases, then heat must be released by the system, and/or work must be being done on the surroundings. However, real spontaneous processes happen where heat is absorbed and work is not done on the surroundings - the reason is that if the net entropy increase is greater than the enthalpy decrease, the process will still be spontaneous.

The equation for $\mathrm{d}U$ includes entropy terms, but not in a way that reflects spontaneity - the equation for Gibbs free energy does.

The other reason we "need" Gibbs energy is that it also tells us how much "thermodynamically useful" work we can get out of a process.

Mathematical Explanation:

The two inequalities that you started with only represent spontaneous process under very specific conditions:

$$\mathrm{d}U < 0$$

is true for spontaneous conditions only when $S$ and $V$ are constant. The reason for this is that we define spontaneity as the direction in which total entropy of the universe increases - it is at the root of our concept of "spontaneous" vs. "nonspontaneous."

For example, to arrive at this expression for $\mathrm{d}U$, we start with the Clausius inequality:

$$\mathrm{d}S - \frac{\mathrm{d}q}{T} \geq 0$$

This is just a way of saying that the net change of the entropy of the universe must be greater than or equal to zero for any process, and that the net entropy change of the universe is given by the change for the system ($\mathrm{d}S$) plus the change in surroundings, $-\frac{\mathrm{d}q}{T}$.

To turn this into something useful, we make a restriction: constant volume. This lets us assume no $pV$ work, and we can therefore write:

$$\mathrm{d}S - \frac{\mathrm{d}U}{T} \geq 0$$

and

$$T\,\mathrm{d}S \geq \mathrm{d}U$$

since $\mathrm{d}U = \mathrm{d}q$ at constant $V$.

If we then assume no entropy change for the system, we can also write:

$$\mathrm{d}U \leq 0$$

In other words, the restriction of constant $S$ and $V$ are built-in into the inequality. You can substitute in other thermodynamic variables to get $\mathrm{d}U$ in terms of $T$ and $p$, but it doesn't tell you anything about spontaneity under conditions other than constant $S$ and $V$.

Gibbs energy is useful because the inequality $\mathrm{d}G < 0$ is true for spontaneous processes under constant $T$ and $p$ conditions, which as you mentioned, are easy to achieve in a lab.

The other thermodynamic potentials are derived in similar ways - by holding different variables constant and seeing what happens when change in entropy is greater than zero.


Fluoride concentration in LiF and CdF2 solutions Trying to learn how to use $\ce{K_{\mathrm{sp}}}$ values and not understanding. See question below:

Two saturated aqueous solutions are prepared at $\pu{25^\circ C}$
Solution1 is made by dissolving cadmium fluoride ($\ce{CdF2}$, $\ce{K_{sp}} = \pu{6.44E−3}$) in $\pu{100.0 mL}$ of water until excess solid is present.
Solution 2 is prepared by dissolving lithium fluoride ($\ce{LiF}$, $\ce{K_{sp}} = \pu{1.84E−3}$) in $\pu{200.0 mL}$ of water until excess solid is present.
Which solution has a higher fluoride ion concentration?"

I used the $\ce{K_{sp}}$ given for $\ce{CdF2}$ to get $s = 0.11720$, so $\ce{[F]} = 2s = \pu{0.2344 M}$, is that right? I guess I'm not sure what to do after that to get find the concentration in $\pu{100 mL}$.  Wouldn't it be $\pu{2.344 M}$?  I am told this answer is incorrect.

The problem as stated is not asking you which solution has the higher amount of fluoride but which one has the higher concentration of fluoride, i.e. the higher molarity (as this is usually the concentration unit more often used by chemists).
You need to calculate the fluoride concentration for the $\ce{LiF}$ solution in the same way you did for the $\ce{CdF2}$ solution. My answer comes to 0.04290 M.
Thus, the $\ce{CdF2}$ solution has a fluoride concentration of 0.2344 M while the $\ce{LiF}$ solution has a fluoride concentration of 0.04290 M. Therefore, the answer to the problem is: the $\ce{CdF2}$ has a higher fluoride ion concentration.


If two electrons move up to a higher energy level in an atom, is it considered excited? In Germanium, if two electrons go from 1s go to 4p is it in its excited state?

http://physics.nist.gov/PhysRefData/Handbook/Tables/germaniumtable5.htm
Energy levels of a neutral germanium atom
https://www.webelements.com/germanium/orbital_properties.html
Middle, electron binding energies for germanium

It will certainly be in one of its excited states, yes, probably followed by an Auger cascade.

Label   Orbital   eV
K       1s       11103
L I     2s       1414.6
L II    2p1/2    1248.1
L III   2p3/2    1217
M I     3s       180.1
M II    3p1/2    124.9
M III   3p3/2    120.8
M IV    3d3/2    29.8
M V     3d5/2    29.2


How to calculate the variation of enthalpy in a reaction that does not occur in standard conditions? Standard enthalpy of formation is a value that represents the enthalpy variation in the formation of compounds with the purpose of using this values to know the enthalpy of a reaction where these compounds participate.

Those values are referred to the formation of said compounts at 273 K and 1 atm (known as standard conditions).

My question is:

Does the value change, for distinct pressures or temperatures? If yes, how?

Why am I asking this?

Let's suppose I have a reaction that goes like this:

$A + B \rightarrow C$

At 1 atm and 273 K (standard conditions).

I can say that in this reaction $\Delta H = Hf^0(C) - Hf^0(A) - Hf^0(B)$

Where $Hf^0$ means standard enthalpy of formation.

Because the reaction happens at 1 atm and 273 K.

But, what if the reaction happens at 3 atm and 350 K?

I asked my University teacher and he said: "You don't need that because we will always be working at 1 atm and 273 K".

What tool do I have to know the variation of enthalpy in this case?

A small extra confusion is the fact that in some examples I see standard conditions being 1 atm and 273 K, and some other times it's at 298 K.

For a pure substance you can write generally that the enthalpy depends on pressure as follows:

$$\left(\frac{\partial H}{\partial p}\right)_T = V - T \left(\frac{\partial V}{\partial T}\right)_p \tag{1}$$

If you evaluate this for an ideal gas you find that  $\left(\frac{\partial H}{\partial p}\right)_T = 0$. If the substance is present in a condensed phase with a density (or molar volume) that is weakly dependent on the temperature, then roughly

$$\left(\frac{\partial H}{\partial p}\right)_T = V$$

If you speak of standard enthalpies then by definition there is no pressure dependence, as they are defined at a particular pressure.

If your reaction involves gases which can be assumed ideal, then you can use the standard enthalpy for all reagents and products, since according to my answer the enthalpy of an ideal gas is independent of pressure. On the other hand if it involves condensed phases you need to evaluate the molar volume and perhaps also the thermal expansion coefficient $\alpha$, since Eq. (1) can be rewritten as

$$\left(\frac{\partial H}{\partial p}\right)_T = V (1- T\alpha)$$

For the enthalpy of reaction you might write

$$\left(\frac{\partial \Delta_r H}{\partial p}\right)_T = \sum_i \nu_i V_{m,i} (1- T\alpha_i)$$

where $\nu_i$ are stoichiometric coefficients. In the summation in the rhs account should be taken for sign changes depending on whether species are reactants or products.

If you want to evaluate the enthalpy of reaction at one pressure given the value is known at another pressure, and the molar volumes are known, you can integrate the previous equation, giving (at constant T)

$$\Delta_r H(p)  =  \Delta_r H(p^\circ) + \int_{p^\circ}^{p} \sum \nu_i V_{m ,i}(1- T\alpha_i) dp$$

where $p^\circ$ is the reference pressure at which the enthalpy is already known. This last expression may be simplified in various ways depending on the state of the species and the pressure integration range. If the molar volumes and expansion coefficients are approximately constant in the pressure range then

$$\Delta_r H(p)  =  \Delta_r H(p^\circ) + (p -p^\circ) \sum \nu_i V_{m,i} (1- T\alpha_i)$$

If the species have negligible thermal expansion coefficients then

$$\Delta_r H(p)  =  \Delta_r H(p^\circ) + (p -p^\circ)\sum \nu_i V_{m,i}$$

and so on. For instance, for a reaction converting a solid between two forms $\text{react}$ and $\text{prod}$ having constant molar volume and negligible thermal expansion coefficients in the pressure range $p^\circ$ to $p$,

$$\ce{\text{react}<=>\text{prod}}$$

$$\Delta_r H(p)  =  \Delta_r H(p^\circ) + (p -p^\circ)(V_m(\text{prod})-V_m(\text{react}))$$


Does turning propane or butane gas tank upside down help to use the remaining gas? Some think that when gas (propane-butane mixture for cooking) in a gas tank is almost used (only remainings are left inside and the flame is hardly noticeable) and you turn it upside down it will improve gas burning. Is it true?

In principle it won't make any difference, but in practice it sometimes can.

Since what you want to emerge from the tank is gas, in principle nothing you do will make more gas come out. The liquid inside will vaporise until there is no gas left and the pressure won't change while there is still liquid in the vessel (both gases liquefy easily under pressure and the pressure will be relatively constant while there is still liquid.) All this is true at equilibrium and constant temperature.

However, in practice the assumption that the inside of the vessel is in equilibrium and constant temperature is not always true. If you use a lot of gas, the vessel will cool as the liquid evaporates (and you sometimes see a ring of frost on the outside of the vessel corresponding to the level of the liquid). If this happens then shaking the vessel may have the effect of warming the liquid and increasing the pressure of the gas. So shaking it might help a little.


Can superglue (cyanoacrylite) be thinned with acetone and still harden? I'm looking for something with an extremely low viscosity to use as a binder for cold casting iron oxide with very fine detail (it's to make a slotted winding core for an electric motor with $\pu{1mm}$ slots). My plan is to thin out the superglue with acetone enough to make a slurry for pouring in the mold - possibly also mix in some water just before the pour to speed the setting process.

My question will superglue still harden if it has been thinned with acetone or would this destroy whatever properties that make superglue such a useful substance once it hardens? It's stated in many places on the internet that acetone works as a solvent for CA glue but this is only ever mentioned in the context of dissolving the end result.

As you can see from the reaction of cyanoacrylate from Wikipedia, the carbonyl ($\ce{C=O}$) groups do not react with the cyano acrylate which means that acetone should not react with your super glue.

That does not mean though that any water present in the acetone won't. So you should make sure your acetone is dry. But even if the super glue does cure, it is dissolved in acetone and will still act as a binder for the Iron oxide.


Half-Life of Radioactive Isotopes: Why? How? Why do radioactive isotopes have a half-life? I know that they decay in order to become stable but why would it take out enough subatomic particles to be half? Or am I approaching this question the wrong way: the remaining radioactive substance is going to be half in x amount of time and the other half will be stable atoms.

As commenters have pointed out, there is nothing about half-life that makes it special; we could as easily measure and tabulate one-tenth lives or one-fifth-lives.

Half-life has two marginal advantages: the half that has decomposed and the half that hasn't are equal, so whether we are referring to remaining or already decomposed nuclei is unambiguous, while for a third-life we could be confused as to whether we are talking about the time it takes for a third of the nuclei to decompose or for all but a third of the nuclei to decompose; and experimentally speaking, half is convenient because it's distinctly smaller from the initial amount of nuclei but large enough to be easily measured compared to the initial amount. However, these two advantages are fairly unimportant, and if we had decided to measure and tabulate one-tenth lives for all radioactive isotopes, we would be fine.

What's conceptually more important is that we can use half-lives as a proxy for isotopic disintegration rates, because radioactive decay follows first-order kinetics. Nothing prevents us from using half-life to describe any kinetic process, such as a chemical reaction rate. However, most of the time that's of limited value, because in general, half-lives are not constant, but change depending on the amount of reagents. It is only for first-order kinetics (i.e. when the amount of nuclei decays exponentially with time) that half-lives (or any fractional life) are constant.

For a general kinetic equation with only one reagent $\ce{A}$,

$$ -\frac{\mathrm{d} [\ce{A}]}{\mathrm{d} t} = k [\ce{A}]^n$$

where $n$ is the kinetic order of the process, we can always define the half-life of the process by integrating from an initial amount of $\ce{A}$, $N_\ce{A}$, to half that amount:

$$ \int\limits_{N_\ce{A}}^{N_\ce{A}/2} \frac{\mathrm{d} [\ce{A}]}{[\ce{A}]^n} = \int\limits_{0}^{t_{1/2}} - k\,\mathrm{d} t $$

which means that the half-life can be expressed in terms of the integral on the left side:

$$ t_{1/2} = - \frac{1}{k} \int\limits_{N_\ce{A}}^{N_\ce{A}/2} \frac{\mathrm{d} [\ce{A}]}{[\ce{A}]^n}  $$

This integral will in turn depend on the kinetic order $n$. While in general the half-life will take the form

$$ t_{1/2} = \frac{1}{k} \frac{1}{n-1} \left( (N_\ce{A}/2)^{1-n} - N_\ce{A}^{1-n} \right) $$

and therefore will be dependent on the starting conditions $N_\ce{A}$, and will change as $\ce{A}$ is consumed, for first-order kintetics ($n=1$)

$$ t_{1/2} = - \frac{1}{k} \left( \ln{|N_\ce{A}/2|} - \ln{|N_\ce{A}|} \right) = \frac{\ln{2}}{k} $$

so half-life does not change with the amount of $\ce{A}$.

As long as the decay is first-order (or exponential, which is the same), half-life is such a constant that we can measure and tabulate, as it will be the same under all possible circumstances; no matter if we are talking about tonnes of fissile material or a handful of nuclei, the value for half-life will be the same.

If we start with $N_\ce{A}$ radioactive nuclei, after a time $t_{1/2}$ you'll have half that amount, $N_\ce{A}/2$; since now you have $N_\ce{A}/2$, after another $t_{1/2}$ you'll have $N_\ce{A}/4$; after a third interval equal to $t_{1/2}$ you'll have $N_\ce{A}/8$, and so on.

This makes tabulating half-lives equivalent to tabulating kinetic coefficients $k$, as these can be transformed into each other quickly and unambiguously, while for other kintetic orders we'd need additional informations to perform that transformation.

Note as well that the last step also shows why we could arbitrarily decide to use any other fractional lives in the same way: if, for instance, we used one-tenth lives, we'd get

$$ t_{1/10} = \frac{\ln{10}}{k} $$


Is there an error in a Wikipedia article explaining the influence of oxidation states? Referring to the series of oxoacids of chlorine: $\ce{HClO, HClO2, HClO3},$ and $\ce{HClO4}$, tabulated in the Wikipedia article on electronegativity, the article states, in the section near the end titled "Variation of electronegativity with oxidation number":
“As the oxidation state of the central chlorine atom increases, more electron density is drawn from the oxygen atoms onto the chlorine, reducing the partial negative charge on the oxygen atoms …” and illustrates it with a table. (NB There is an ambiguity in the use of the word "reducing" which is the opposite of "oxidising", but the intended meaning in terms of increasing or decreasing electron density is clear in the Wiki statement.)
But as I understand it (and as also given by the rules for assigning oxidation numbers), insofar as oxidation state represents the number of electrons that an atom can gain, lose, or share when chemically bonding with an atom of another element, the table is correct, but shouldn’t the textual statement be the converse:
“As the oxidation state of the central chlorine atom increases, more electron density is drawn from the central chlorine atom onto the oxygen atoms, increasing the partial negative charge on the oxygen atoms ...”
I reason it thus And just to be clear, let's go back one step to start with the even simpler case of $\ce{HCl}$, where clearly the oxidation states must be ($\ce{H} = +1, \ce{Cl} = -1$, respectively; total 0):
When we move on to $\ce{HClO}$, then it becomes (+1, +1, -2 respectively) as the single oxygen with stronger electronegativity than chlorine is assigned the oxidation state -2, so that balancing the total means the electronegativity of the central chlorine must change from -1 to +1 (in agreement with the table on the Wiki page, but not with the article’s statement about the direction of change of electron density.
And as we move along the series to $\ce{HClO4}$, the central chlorine is having more and more electron density sucked off it by the four more electronegative oxygen atoms, resulting in the final oxidation states (+1, +7, -8), in agreement with the table in the above-cited Wiki article, but not with the article's textual explanation.
So I conclude that the text statement in the Wiki article is currently wrong (and if so I will amend the Wikipedia article) - or else I'm wrong, and would appreciate someone pointing out my own error…?

Please do not edit the Wikipedia article, as your wording is incorrect, and the existing wording, while awkward, is correct.
What the article is trying to say is that the electron density on each oxygen is less as the oxidation state of the Cl increases. We can understand this by imagining progressive oxidation starting with $\ce{HClO}$. Here, the Cl has oxidation state +1 and the O is -2. But of course the oxygen doesn't have a full negative 2 charge. Some of that charge is shared with the chlorine and the hydrogen, neither of which has a full charge of +1.
To get to $\ce{HClO2}$, we can imagine reacting $\ce{HClO}$ with a neutral oxygen atom (oxidation state 0). After the reaction, the oxygen will have oxidation state -2, and the Cl will now be +3. Again, these are not the formal charges, but give us a hint about where charge will be located. The oxygen will have a partial negative charge, which means the chlorine will have to lose some of its electron density, increasing its actual charge. The original oxygen from $\ce{HClO}$ is still there and feels a stronger "pull" on its electron density, so it gives more to the chlorine, and the oxygen's partial negative charge becomes less negative.
Completely making numbers up, we could imagine that in $\ce{HClO}$, the charge on oxygen is -1, on chlorine is +0.25 and on H is +0.75. After adding the second oxygen to make $\ce{HClO2}$, the charge on the new oxygen might be -0.5, on the already attached O might be -0.75, +0.75 on H and +0.5 on Cl. Do you see how the negative charge on each O gets smaller in magnitude, but the total negative charge on oxygen(s) gets larger in order to offset the increased positive charge of the more oxidized chlorine? The Wikipedia statement refers to the individual O's not the sum.


Electrolytic silver deposition A current is applied to two electrolytic cells in a series. In the first, silver is deposited; in the second, a zinc electrode is consumed. How much $\ce{Ag}$ is plated out if $\pu{7.5 g}$ of $\ce{Zn}$ dissolves?

What I ended up doing was

$$\frac{\pu{7.5 g}~\ce{Zn}}{\pu{65.382 g mol}~\ce{Zn}} \times \frac{\pu{2 mol}~\ce{Ag}}{\pu{1 mol}~\ce{Zn}} \times \pu{107 g mol-1}~\ce{Ag}$$

but I'm not even sure I'm doing this right.

I'd suggest to start with writing down the processes occurring during the electrolysis:

\begin{align}
&\text{Cathode (–):} &\ce{2 Ag+ + 2 e- &-> Ag^0} \\
&\text{Anode (+):} &\ce{Zn^0 &-> Zn^2+ + 2e} \\
\hline
&\text{Net} &\ce{2 Ag+ +Zn^0 &-> 2 Ag^0 + Zn^2+}
\end{align}

Applying the unified equation for Faraday's laws of electrolysis:

$$m = \frac{QM}{zF}$$

for each cell, and taking into account that in series circuits current is the same for every element ($I = \text{const}$):

$$\frac{Q}{F} = \frac{z_i m_i}{M_i} = \text{const}$$

$$\frac{z(\ce{Ag}) \cdot m(\ce{Ag})}{M(\ce{Ag})} = \frac{z(\ce{Zn}) \cdot m(\ce{Zn})}{M(\ce{Zn})}$$

The valency numbers of ions are 1 and 2 for silver and zinc, correspondingly. The last thing to do is to express and calculate $m(\ce{Ag})$.

   $$m(\ce{Ag}) = \frac{z(\ce{Zn}) \cdot M(\ce{Ag})}{z(\ce{Ag}) \cdot M(\ce{Zn})} \cdot m(\ce{Zn}) = \frac{2 \cdot \pu{107.87 g mol-1}}{1 \cdot \pu{65.38 g mol-1}} \cdot \pu{7.5 g} = \pu{24.7 g}$$


Why are lead and zinc commonly found together? Copper, silver, and gold are often found in combination. That makes sense to me, because they are members of the same chemical family. Likewise, I would expect nickel palladium, and platinum to be found together.

But lead and zinc are often found together, even though they are not of the same chemical family. At least, these metals seem to appear "jointly" more often than lead and tin, even though these are members of the same family.

Why would that be, or am I missing something?

Lead and Zinc have in common two electrons "s" in their outer shell. That is why they may sometimes have a similar behavior. You would object that Lead has also two electrons "p" in its outer shell, which are missing in Zinc. It is true. But these p electrons are "modified" and can be "forgotten" by an unusual phenomena in chemistry, namely Einstein's relativity.

When the number of protons in the nucleus is as high as about $80$, and it is $82$ for $\ce{Pb}$,  the nucleus charge is so high that the electron has to move at a relativistic speed to stay rotating around the nucleus, according to Bohr's model. Of course Bohr's model is wrong and the electron is not rotating and has no speed. But the calculations made with it are still valid, as was shown by Pekka Pyykko, Relativity and the Periodic Table, Accounts of Chemical Research, Vol. 12, No. 8 (1979) p. 276 - 281.

At relativistic speeds, the dimensions of all moving objects decrease. And Pekka Pyykko has shown that this effect is proportional to the sum of the two quantum numbers n and l. Lead has the following electronic configuration : $\ce{[Xe] 4f^{14} 5d^{10} 6s^2 6p^2}$. The sum n+l is equal to $8$ for $\ce{4f, 5d, 6p}$. It is $7$ for $\ce{6s}$. So the electrons $\ce{4f, 5d, 6p}$ are so contracted that they are mixed with inner electrons and "disappear" from the periphery. Only the two electrons $\ce{6s}$ , with n+l = 7, remain in the outer shell of the atom. That is why Lead $\ce{Pb}$ often behaves like an atom of the $2$nd column : The most usual oxidation number of Pb is $2$ (like in $\ce{Pb^{2+}}$), and not $4$ (like the $4$th column), which exists but is not frequent. Also $\ce{PbSO_4}$ is insoluble in water, like $\ce{BaSO_4}$. So $\ce{Ba}$ is made of $\ce{[Xe] 6s^2}$, and Pb may be considered as a sort of [pseudo-$\ce{Xe] 6s^2}$...

To go back to $\ce{Zn}$ and $\ce{Pb}$ configuration, relativity explains why $\ce{Pb}$ looks like a noble gas (Xenon) plus $2$ electrons "s" in its outer shell. By comparaison, Zinc is $\ce{[Ar] 3d^{10} 4s^2}$. Like $\ce{Pb}$, it also looks like a noble gas (Argon) plus $2$ electrons "s" in its outer shell, if the filled shell $\ce{4d^{10}}$ is "forgotten". Mother Earth is probably not able to recognize the existence of this $\ce{4d}$ shell and has often mixed $\ce{Zn}$ and $\ce{Pb}$ minerals.


What makes selectivity other than activation energy? Selectivity and kinetics of reactions are somehow linked to the activation energies of reactions. However, there's something not totally clear in my mind about the selectivity. I will get a general example as many systems can fit to this.
Let's consider a reaction between A and B. The reaction between A and B can proceed through different pathways (1 & 2). A selectivity of 100% is observed for pathway 1 and it is demonstrated that the activation energy for this pathway is slightly lower than pathway 2.
Let's consider than a system of C and D. The reaction between C and D can also proceed through 2 pathways, let's name them 3 & 4. In this scenario, a selectivity of 75% for example is observed for pathway 3. The activation energy for pathway 3 is lower than pathway 4 in a more marked way than when we considered the reaction of A and B.
My question is then. Why are sometimes observed a total selectivity when the difference in activation is low, while sometimes the products result from different selectivities when the difference in activation energies between the pathways is rather high.

It's not just the activation energy that determines the selectivity of a reaction. There is a thermodynamic factor too. This is the classic kinetic-vs-thermodynamic control that you will find in most chemistry textbooks at a degree level.

Reactions under thermodynamic control have outcomes that depend on the position of the equilibrium and therefore the relative stability of the possible products
Reactions under kinetic control have outcomes that depend on the rate at which the reaction processds, and therefore on the relative energies of the transition states leading to the alternative products.

[Organic Chemistry, 2nd ed., J. Clayden, N. Greeves, S. Warren, OUP] page 264

You are considering only activation energies ($\Delta G^\ddagger$) i.e. the difference in energy between the transition states and the reactants.

If your reaction has two pathways, then the pathway with the lower activation energy is faster. This means that the product of that pathway is formed faster. This is kinetic control.
However, the reaction can be reversible. The activation energy for the reverse reaction is $\Delta G^\ddagger + \Delta_r G$. Therefore, if the reaction energy (i.e. $\Delta_r G$) is not too large, then there would be backward reaction.
In this condition, the products and reactants are in equilibrium, so the side with the lowest Gibbs free energy is favoured. If the product from one pathway is more stable (i.e. more negative $\Delta_r G$) than the other product, then the stable product will the be the one with higher yield when the equilibrium is reached.
This is thermodynamic control.
In the above diagram for example, pathway 1 is faster as TS1 is lower in energy. So product 1 is the kinetic product. Pathway 2 is slower, but the product is lower in energy, so product 2 is the thermodynamic product.
Whether a reaction is under thermodynamic or kinetic control would depend on the reaction conditions, the specific identity of the reactants, and how much time is allowed for the reaction.


Pourbaix diagram and the order of the species Figure 1. Pourbaix diagram of iron (Source)

The line between $\ce{Fe}$ and $\ce{Fe^2+}$ is there because the potential $\ce{Fe/Fe^2+}$ is $\pu{-0.44 V}$. The same applies for $\ce{Fe^2+/Fe^3+}$ with $\pu{0.77 V}$. But $\ce{Fe/Fe^3+}$ is $\pu{-0.03 V}$, but it's not on the graph as a line.

How do I know which species have a line, and which don't? And if I'm drawing a graph, how do I know that I also shouldn't include this line?

Pourbaix Diagram In Environmental Chemistry
Pourbaix diagrams are very useful to identify the nature chemical species in different pH environment.
The figure contains additional information on the stability of water. The lines at 1.23 V and 0.00 V represent the redox chemistry of water. At 1.23 V water is being oxidised to oxygen. At 0.00 V the water is reduced to hydrogen. As labelled in Fig.1, there are 3 regions where behaviour of water is represented. The middle region is where the water in "stable state".
Note: These voltages are obtained at 298 K. In order to derive these lines, you can use Nernst Equation (where gradient is: -RT/nF, as the both reactions involve 2 electrons the gradient is the same in both cases.)

Figure 1: Stability of water.
Now moving on to iron. Overlapping water behaviour and iron as in Fig.2. (The dashed lines - redox chemistry of water). In the case of water, knowing the pH given you can predict the behaviour of iron. For the sake of understanding how to utilise the Pourbaix diagrams.
Assuming you are analysing alkaline lake (high pH: 10-12) and assume there is not much oxygen content (to utilise also the y-axis, low oxygen level - E decreases). In Fig.2, the region of your interest is pH 10-12 and since low oxygen, and region around 0.00 V. The Pourbaix indidicates that at these conditions you can have some amounts of Fe(OH)2, Fe(OH)3.

Figure 2: Analysis of Iron species in 1 M of water (298 K).
To see different behaviour of other species:
http://www.crct.polymtl.ca/ephweb.php


How does a "sea of electrons" hold metal atom together? How do delocalized electrons hold metal atoms together? What are the forces of attraction between the atoms and the electrons called?

The metal atoms lose electrons to form positive metal ions and a sea of delocalised electrons. The positive ions are attracted to the oppositely charged electrons and form an electrostatic attraction.


The effect of structure on the boiling points of organic molecules Why do molecules like methoxyethane, ethoxyethane and so on have lower boiling points than isomeric alcohols?

I think it has something to do with the structure, but I don't exactly know what it is.

Yes, structure is important in this respect. Most importantly, the alcohols are able to form hydrogen bonds among each other, which is not possible for the ethers. Since evaporation always means breaking intermolecular interactions in the liquid phase, any additional stabilizing interaction raises the boiling point.


Why does Co2+ have 7 electrons in the 3d orbital, and not 5 like Mn? I'm taking general chemistry 2 this half of the summer. We are currently going over coordination complexes, ligands and transition metal ions. In particular, this question is in regards to the ligand field stabilization energy.

Why does $\mathrm{Co}^{2+}$ have 7 valence electrons in the $3d$ orbital? Doesn't the superscript $^{2+}$ denote that it is missing 2 electrons, and therefore isoelectric with $\mathrm{Mn}$?

Well, you can find some kind of explanation why the electronic configuration of $\ce{Co^2+}$ is $[\ce{Ar}] \, 3d^7$ here and there, but at the end of the day, you just need to * memorize it*.

  When d-block elements form ions, the $4s$ electrons are lost first. Source

The $n+l$ rule tells you the order in which atomic orbitals are filled, and according to the rule the $4s$ orbital is occupied before the $3d$ orbital because it has lower energy.
Thus, the electron configuration of $\ce{Mn}$ is $[\ce{Ar}] \, 3d^5 4s^2$ while that of $\ce{Co}$ is $[\ce{Ar}] \, 3d^7 4s^2$.
But, the $n+l$ rule, as many other rules of old quantum theory, is not 100% working, and thus, sometimes gives wrong electron configuration.
For instance, in general, it is applicable only for neutral atoms in their ground state, and thus, if you apply it to cation $\ce{Co}^{2+}$ you will get the wrong electron configuration [\ce{Ar}] $3d^5 4s^2$.

As it is sometimes explained, the statement that $4s$ orbital is lower in energy than $3d$ orbital is true only when the orbitals are unoccupied. But while you fill $3d$ orbital with electrons it becomes lower and lower in energy and eventually ends up lower in energy than the $4s$ orbital. Thus, when electrons are lost from $\ce{Co}$ atom, they are lost from the 4$s$ orbital first because it is actually higher in energy when both $3d$ and $4s$ are filled with electrons.

P.S. And if I remember correctly this has nothing to do with stabilization in ligand fields.


Can baking soda or citric acid help clean a pot with burned-in milk rice? The other day I was cooking milk rice and forgot about it so it burned a little on the bottom. This lower layer was very hard to clean. A lot of websites recommend boiling water in the pot with baking soda (Sodium bicarbonate) or citric acid to dissolve the burned residue and make it easier to clean the pot. I tried this but it did not help. I doubt there is even a chemical reaction between these reagents that would help in any way.

Is this another myth or is there scientific basis for given recommendation?

The hotter the water, the more likely it will be to remove burnt-on starch. Not only does $\ce{NaHCO3}$ raise the boiling point of the water, but it also helps saponify fats (actually, "washing soda", $\ce{Na2CO3}$ would be better).

However, you still need to do a lot of scrubbing afterwards to remove the carbonized layer.


Why is potassium permanganate used to kill mollusks? I know that potassium permanganate is used as a disinfectant for bacteria. Permanganate acts as a strong oxidizing agent that can rip the electrons off of bacterial cell membranes, resulting in a disruption that can lead to death. I've also read that potassium permanganate can be used as an oxidizing agent to kill mollusks (source: http://www.chemistry.wustl.edu/~edudev/LabTutorials/Water/PublicWaterSupply/PublicWaterSupply.html). I've tried looking into it and I can't find anything helpful. Does permanganate kill mollusks in the same way? If that is the case, will the disruption of the outer cells of the mollusks really kill the whole organism? Any insights would be helpful, thanks in advance!

The provided link (Lab Tutorial at Washington University in St. Louis)  states the following:

Another problem for water-treatment plants in the Great Lakes and Mississippi Valley is zebra mussels, which can "congregate" at the water intake and clog the screen, eventually sealing it off. Potassium permanganate, KMnO4, can be used to kill these mollusks.

Zebra mussels are a problematic invasive species in the midwestern USA. Permanganate has been found to inhibit their proliferation. The main effect is apparently oxidation. From a report released by the US military (ref [1.]) on use of oxidizing molluscicides including permanganates:

In general they have similar modes of action based on the oxidation of organic matter,which leads to toxic and lethal effects. They are suitable for use in preventative treatment, where they are added to a system throughout the breeding season at from 0.1 to 0.5 mg/L (ppm) total residual oxidant (TRO) to prevent settling. For reactive treatments,continuous application of 0.5 to 1.0 mg/L TRO for 2 to 4 weeks can eliminate established adult colonies, but concentration and contact time required depends on temperature, water chemistry, and physiological state of the zebra mussels. Mussels do detect oxidants, and shell closure for up to 2 weeks may reduce efficacy in adults(Claudi and Mackie1994)

See also ref [2] cited in ref [1.].
Ref [3] compares the toxicity of various molluscicides, noting that

An  LC50 value  could  not  be  estimated  for antimycin,  Bulab  6002,  Bultab  6009.  Calgon DM -DACC,  KML  V2.  KML  V54.  potassium  chloride, and  potassium  permanganate  due  to  insufficient mortality  of  the  test  organisms.

In addition, potassium can also have an effect on mollusks. For instance, from a thesis (Ref [4]):

Potassium can reach increased levels that are toxic to aquatic organisms, such as mollusks, in the environment (Romano and Zeng 2007). For example, Fisher et al. (1991) reported that increased potassium concentrations were toxic to Dreissena polymorpha, causing asphyxiation by damaging gill epithelium. In addition, Imlay (1973) suggested that high potassium concentrations could be the cause of absent mussel populations or lower diversity within Upper Midwest mussel assemblages. Any river or stream with a potassium concentration of 7 mg/L or higher lacked mussels, but mussels could be found in rivers with a concentration lower than 4 mg/L.

Note responses vary across organisms. Discussion of that is bound to take us increasingly away from chemistry.
References

Zebra mussel chemical control guide (ERDC/ELTR-00-1) by Susan L Sprecher, Kurt D. Getsinger; prepared for U.S. Army Corps of Engineers (2000).

Klerks,P.L., Fraleigh, P.C,and Stevenson, R.C.(1993)."Controlling zebra mussel {Dreissena polymorpha) veligers with three oxidizing chemicals: Chlorine, permanganate, and peroxide+iron."Zebra mussels: Biology, impacts,  and control. T.F.Nalepa and D.W.Schloesser, eds., Lewis Publishers, Boca Raton,FL,621-641.

Diane L. Waller, Jeffrey J. Rach, W. Gregory Cope, and Leif L. Marking, Susan W. Fisher, Henrycka Dabrowska. Toxicity of Candidate Molluscicides to Zebra Mussels (Dreissena polymorpha) and Selected Nontarget Organisms. J.  Great  Lakes  Res.  19(4):695-702 Internat.  Assoc.  Great  Lakes  Res.,  1993.

Gibson, Kesley. (2015). ACUTE TOXICITY TESTING ON FRESHWATER MUSSELS (BIVALVIA: UNIONIDAE) AND FRESHWATER SNAILS (GASTROPODA: CAENOGASTROPODA).


Activation energy in physical adsorption Physical adsorption of molecules on surface requires activation energy, because bonds of the adsorbed molecules are broken.

Is this a valid statement?

I have been under the assumption that physical adsorption does not involve any energy of activation as there is no chemical bond formation between the adsorbed molecules and the surface (and hence there is no necessity for bonds of adsorbed molecules to be broken).

The definition given above would more closely describe chemisorption. At large separation from the surface the interaction is zero. In physisorption the interaction becomes attractive and forms a potential well as the surface is approached. There is no activation energy needed to do this. At shorter separations than the potential minimum, the potential becomes repulsive. The minimum is usually at larger separation than the minimum for equivalent species exhibiting chemisorption. Species may physisorp before chemisorption occurs

Comment by porphyrin converted to answer


Is there any abbreviation for hydrates? I'm getting tired of writing out the full formulas for hydrates, adding $\ce{.$x$\,H2O}$ to all formulas. I'm wondering if there's standard shorthand for these.

There is, as far as I know, no shorthand for that, but you might be able to omit it completely in a first order approximation. The closest might be adding $\ce{(aq)}$ when working in aqueous environment.

Adding $\ce{.$x$\,H2O}$ to the formula usually indicates, that it is a solid and that there are $x$ water molecules enclosed in the crystal structure. It depends on what you are looking at, which kind of reaction, what you might be able to omit.

For example, when you are describing a drying process, it is not possible:
$$\ce{\overset{blue}{CuSO4.5H2O} ->[\Delta] \overset{white}{CuSO4} + 5H2O}$$

When you are working in aqueous solution it is usually not important, as there is much more water around it, hence
$$\ce{\overset{blue}{CuSO4.5H2O} + Na2S -> \overset{black}{CuS} v + 2Na+ + SO4^{2-} + 5H2O}$$
is essentially the same as
$$\ce{CuSO4~(aq)  + Na2S~(aq) ->CuS~(s) v + 2Na+ ~(aq) + SO4^{2-}~(s)}$$

If you are describing a redox reaction, it might be the same, hence the following should be fine
$$\ce{CuSO4~(aq) + Zn~(s) -> Cu~(s) v + Zn^{2+}~(aq) + SO4^{2-}~(aq)}$$
Often you can omit the $\ce{(aq)}$ or $\ce{(s)}$ all together.

So depending on what kind of chemistry you are talking about, some information can be implied and therefore be omitted. If you add an example of what kind of reactions/ substances you are working with, I can expand my answer to take that into account.


Why don't we consider VΔP when we define Q? We only define $Q = \Delta U + W_\text{exp}$ (expansion work = $-P\Delta V$).
If heat can cause $\Delta U$ and work, why work is defined only as expansion work in the first place where there are other forms of works, such as isochoric work ($=V\Delta P$)?

Let's say, there is a container with ideal gas $T, P, V$ and $V$ can't be changed by this container. I give it heat therefore $T, P$ changes to $2T$, $2P$. In this constant volume process, expansion work, Wexp is 0. That's why $Q_v = \Delta U$. In the mean time (in this constant volume process), $\Delta H = Q_v + V\Delta P$ and $V\Delta P$ is work done in constant volume process. My question is that why can't we just say this heat $Q = \Delta + V\Delta P$? Do we just ignore the non-expansion work? or is it included in $\Delta U$? or because the definition of $Q$ is $\Delta U+P\Delta V$ originally?

One more thing. I googled it all day and someone wrote that $V\Delta P$ is work in flow process. I can't imagine how matter can flow in the constant volume container, and even though it can flow, what is the relationship between $V\Delta P$? They didn't explain it why so my question get bigger and bigger.

Let's start by the fundamental ideas. Assume, you have a container with volume $V$ with a piston on it of  cross sectional area $A$ and container is filled with gas.
Suppose, you push the container by $\mathrm dx$ amount. So, the gas is compressed and work is done on the gas.
By the definition of work classically,
$$\mathrm dW=F\,\mathrm dx$$
Now if we use the fact that $P=\frac FA$, and $\mathrm dV=A\,\mathrm dx$, we will have $$\mathrm dW=\frac FAA\,\mathrm dx= P\,\mathrm dV$$ So, the classical definition of work only tells us the that work done on the gas should be $$\int_{V_1}^{V_2}P\,\mathrm dV=P\,\Delta V$$ if $P$ is constant.
So, the definition tells us there is no $V\,\Delta P$ term in the expresion for work. And, therfore, in the first law of thermodynamics, i.e. $\Delta Q = \Delta U - W_\text{on the system}$, also, that expression doesn't occur from work, and it is not also included in $\Delta U$ either.
But if you consider enthalpy change for a system, there $V\,\mathrm dP$ term will occur as $$\mathrm dH=\mathrm dU+\mathrm d(PV)=\mathrm dU+P\,\mathrm dV+V\,\mathrm dP$$


If we use Magnesium instead of the sodium with alkyl halide in dry ether then will a reaction similar to Wurtz reaction will occur I know that a grignard reagent will formed will be formed but if you look at his question then the answer of this question is b. so I just want to know if something like wurtz reaction will take place if there is some another explanation for this answer.

In 1930, Palmer and Kipping1 reported the formation of cyclohexyl magnesium bromide in ether in 50-60% yield. They did not make an effort to detect the coupling product, "bicyclohexyl" [choice (b)] in the Grignard reagent itself. Subsequent reaction of the Grignard reagent with SiCl4 led to the isolation of bicyclohexyl and some cyclohexene. Although the authors did not speculate on the source of these products, it is likely that they are part of the 40-50% of the "missing" Grignard reagent formed by Wurtz-like coupling and elimination, respectively.

The following year, Gilman and Zoeller2 reported their results on the relative merits of forming cyclohexyl magnesium bromide and chloride. Cyclohexyl chloride gave generally higher yields than the corresponding bromide, and under their conditions, higher yields than what Palmer and Kipping had obtained.

To answer your question, it appears that both bicyclohexyl and cyclohexene may be by-products during the formation of cyclohexyl Grignard reagents.
1) K. W. Palmer and F. C. Kipping, J. Chem. Soc., 1930, 1020.
2) H. Gilman and E. A. Zoellner, J. Am. Chem. Soc., 53, 1931, 1945.


Why might the O—H stretch in the IR of MY triphenylmethanol product be less intense than expected? Why might the O—H stretch in the IR of MY triphenylmethanol product be less intense than expected when compared to the expected IR for triphenylmethanol?

I'm interpreting the IR I took of my product, triphenylmethanol, to the expected IR I obtained from SDBS.

My IR:

SDBS:

We performed a Grignard Reaction in lab using bromobenzene and methyl benzoate in order to synthesize triphenylmethanol.

ANY clues why my triphenylmethanol (O—H stretch) is not as intense? I really don't know.

A possible explanation would be the concentration of sample you used. Of course, if you are confident that you had utilized sufficient concentration then there may be another reason for the decreased intensity. However, some of the other peaks are also somewhat decreased in intensity so it is a possibility.

This point is noted here (http://chemwiki.ucdavis.edu/Physical_Chemistry/Spectroscopy/Vibrational_Spectroscopy/Infrared_Spectroscopy/Infrared%3A_Interpretation#The_Origin_of_Peak_Intensities)


Why cobalt reacts favorably with nitrogen? I used also iron, nickel, zinc, copper. However, cobalt gives the easiest and straight forward yield of nitrides (XPS and SAED verified).
It is so easy to get cobalt nitride that I don't even have to use high purity nitrogen as I need with $\ce{Fe},$ $\ce{Ni},$ $\ce{Zn},$ $\ce{Cu}.$

Is there any reason for that? Is it because of back-bonding?

According to the recent first principles ab initio DFT calculations done with VASP of structural, electronic and mechanical properties of 3d transition metals mononitrides $\ce{MN}$ carried out by Liu et al. [1], there is a local maximum of cohesion energy $E_\mathrm{coh}$ at cobalt for all possible/discovered structure types:

Ultra-soft Vanderbilt pseudo-potentials (US-PP) … as supplied by Kresse and Hafner …, with local density approximation (LDA) … and PW91 general gradient approximation (GGA) … were both used. It is known that LDA underestimates the lattice constant and overestimates the elastic constants and cohesive energy, while PW91 GGA does the opposite for the lattice constant and elastic constants by a smaller amount and yields more accurate values of cohesive energy.

Figure 1.
Cohesive energy per atom of the nitrides $(\ce{MN})$ versus their corresponding 3d transition metals $(\ce{M}).$
The three structures are zincblende (zb, black square), rocksalt (rs, red circle) and
cesium chloride (cc, blue triangle).
Experimental values for the rocksalt structure (green ×) … are also presented.

Since we are discussing synthesis from the elements, it's especially handy to use cohesion energy since for solids it signifies the energy required to disassemble the structure into an array of neutral free atoms.
As a rule of thumb, the higher the atomic cohesion energy is, the more stable the solid is.
Numerical values of $E_\mathrm{coh}^\mathrm{GGA}$ for cobalt nitrides are always exceeding the ones for iron, nickel, copper and zinc, and also $\ce{CoN}$ demonstrates mechanical stability across all structure types:

There is still a lot of controversy between the studies of preferred phase and magnetism [2]; I'm not sure though it is currently possible to interpret/rationalize or let alone predict this behavior by other, simplified means.
References

Liu, Z. T. Y.; Zhou, X.; Khare, S. V.; Gall, D. Structural, Mechanical and Electronic Properties of 3d Transition Metal Nitrides in Cubic Zincblende, Rocksalt and Cesium Chloride Structures: A First-Principles Investigation. J. Phys.: Condens. Matter 2013, 26 (2), 025404. DOI: 10.1088/0953-8984/26/2/025404.
Soni, H. R.; Mankad, V.; Gupta, S. K.; Jha, P. K. A First Principles Calculations of Structural, Electronic, Magnetic and Dynamical Properties of Mononitrides $\ce{FeN}$ and $\ce{CoN}.$ Journal of Alloys and Compounds 2012, 522, 106–113. DOI: 10.1016/j.jallcom.2012.01.100.


Redox titration or iron(II) The concentration of $\ce{Fe^2+(aq)}$ can be determined by a redox titration using

A. $\ce{KBr}$
B. $\ce{SnCl2}$
C. $\ce{KMnO4}$ (basic)
D. $\ce{KBrO3}$ (acidic)

Can anyone please help me on this question? I know the answer is between C and D, since those two are the only redox reactions. However, I don't know how to determine between C and D, as they will both act as an oxidizing agent and Fe will be reduced in both cases (the correct answer is D).

Can you please elaborate on that part? I know that the E value of
  Mno4- is +1.51, and the one for BrO3- is +1.48. What does that
  suggest?? that Mno4- will more likely to be reduced by Fe+ than
  Bro3-?? please explain

The first and foremost hint, even before you calculate the electrode potentials is the solubility rule taught in general chemistry. Hydroxides of most metals are insoluble. Write the equations : Fe(II) + hydroxide ions = iron (II) hydroxide. This is why alkaline permanganate is not a good choice. However, this is not the final answer because you would like to further confirm whether C or D is the correct choice.

Next determine the E_cell= E(cathode)-E(anode) > 0 V or not.

Now you need to oxidize Fe(II) to Fe(III), so the iron half cell will be the anode. As a mnemonic, anode and oxidation both start with vowels; cathode and reduction don't. You would like to reduce the permanganate in an alkaline medium, this half cell is the cathode.

Repeat the same for bromate half cell in acidic medium and Fe(II). Both (C) and (D) would turn out to be positive E_cells. However, the (C) is not an answer because you don't want iron to precipitate out as iron(II) hydroxide.

In real life analysis, Fe(II) can be routinely titrated with potassium permanganate but in highly acidic medium. For practice, check the E_cell after writing a balanced reaction.


Reaction between lead dioxide and nitric acid A wikipedia article suggests that lead dioxide is produced by this reaction:
$$\ce{Pb3O4 + 4 HNO3 → PbO2 + 2 Pb(NO3)2 + 2 H2O}$$

The same article also suggests that lead dioxide reacts with nitric acid in the following way:
$$\ce{2 PbO2 + 4 HNO3 → 2 Pb(NO3)2 + 2 H2O + O2}$$

I am unable to understand why then, in the first reaction, $\ce{PbO2}$ doesn't further react with the residual nitric acid. Is it because the quantity of the acid is regulated? Or is there some error in the reactions?

You need to consider the reaction conditions.

The Wikipedia Lead Dioxide article is saying that in dilute nitric acid the first reaction occurs, while in warm nitric acid the second reaction occurs.

So the first reaction would only proceed to the second reaction under harsher conditions, heat and more concentrated acid.


Cyanide hydrolysis; could it yield an oxime? The mechanisms for both acid and base catalyzed hydrolysis of nitriles are clear;

(With the subsequent deprotonation from the hydroxyl) and

(Source)
In both cases, the hydroxyl adds to the carbon while the proton adds to the nitrogen.
I thought up an alternate mechanism, with the proton attaching to the carbon, and the hydroxyl going to the thereby positively-charged nitrogen, yielding an oxime.
Why doesn't this occur? I thought of the fact that the nitrogen atom, with a lone pair, would cause any nearby protons to come straight to it, but I'd like clarification on that.

Yes, since nitrogen is more electronegative than carbon, and also since it has the lone pair available for taking, the proton is likelier to go to it.


Why do highly branched alkanes have higher octane numbers than their corresponding linear isomer? The octane number of fuels are based on a set of primary reference fuels, which are mixtures of iso-octane and n-heptane. Now iso-octane being branched alkane has very less tendency to react and given ON=100 and n-heptane reacts easily and given ON=0. However n-octane that is the corresponding linear isomer of iso-octane has an ON<0. Now why does branching in a alkanes reduce its reactivity?

The reaction desired in an automobile piston is a rapid burning, not an explosion (very very rapid reaction). If the air + gasoline vapor burns much faster than the camshaft is turning, the explosion causes a ping, or high pressure rise in a smaller volume (because the piston has not moved far enough). This can damage the piston, or at least add an unusual stress.

Burning is a radical reaction and the radicals produced from linear alkanes are quite reactive (i.e., fast reactors). Branched alkanes might lose a hydrogen, even break apart, but the free electron is more stable, and the reaction rate is slower.

There are compounds with a higher octane number than iso-octane. E.g., toluene has an octane number ~120, and was highly sought after for airplane fuel during WW II. The benzylic radical which is the likely first product of a radical reaction is quite stable (but not inert).


Stability of geometrical isomers in cycloalkanes Among the following, which should be the most stable compound?
1)Cis-cyclohexane-1,2-diol
2)Trans-cyclohexane-1,2-diol
3)Cis-cyclohexane-1,3-diol
4)Trans-cyclohexane-1,3-diol

My thought process is-cis isomers with adjacent OH groups lead to Hydrogen bonding so 1) should be the most stable, I'm not sure if repulsion will be the dominating factor and change the answer- what is the correct way to think about this?

To answer this question, you needs to know  what are the values for each acting forces on the cyclohexane frame work. Without that information, nobody would know if the repulsion forces (steric) would or would not be the dominating factor over hyrdogen bonding (electronic). Once we know the answer for that, you'd be able to calculate and find out the most stable structure. For your convenience, I attached a diagram with the chair conformations of all four structures and their Newman projections to make you understand the hydrogen bonding capabilities:

As you see, if the substitutions are not capable of hydrogen bonding, it is obvious that 1,3-cis structure is the most stable considering only stecic factors (eq-eq conformer). 1,3-trans structure (eq-eq conformer) is also stable relative to others but has an additional gauche interaction of vicinal bulky groups (1,2-substitutions) compared to its 1,3-counterpart.

In your case, however, all four molecules contain hydroxyl groups, so that three of them could be able to have structure stabilizing hydrogen bonding (the exception is trans-cyclohexane-1,3-diol; see Newman projection of IV-eq-ax). Both cis- and trans-cyclohexane-1,2-diols are structurally similar to 1,2-ethanediol, which favors a gauche conformation instead of an anti conformation, because only the gauche conformation allows hydrogen bonding between the two vicinal hydroxyl groups. Thus, both cis- and trans-cyclohexane-1,2-diols would adopt gauche conformations (see Newman projections of I-eq-ax and II-eq-eq) Ref.1. Keep in mind that, for cis-cyclohexane-1,2-diol (I), the anti conformation is impossible because of the restriction of ring structure. In trans-1,2-cyclohexanediol (II), however, the anti conformation is physically possible, but it would require the hydroxyl groups to be in diaxial positions, which is much less less stable than diequatorial due to 1,3-interactions imposed by axial 3,5-hydrogens (ax-ax chair confirmation).  According to the literature (Ref.1), cis-cyclohexane-1,2-diol is more acidic than trans-cyclohexane-1,2-diol by about $\pu{1.5 kcal/mol}$, hence 1,2-cis-compound would impose stronger hydrogen bonding than its trans-counterpart.

According to the same study (Ref.1), cis-cyclohexane-1,3-diol is $\pu{10.9 kcal/mol}$ more acidic than trans-cyclohexane-1,3-diol, thus cis-compound is highly stable than its trans-counterpart (also remember, trans-cyclohexane-1,3-diol cannot have hydrogen bonding (vide spupra)). However, between conformers, the diaxial and diequatorial conformations in cis-cyclohexane-1,3-diol are relatively close in energy, so it is not immediately clear which conformation is dominant (Ref.1). Yet, a solution NMR study of this molecule showed that the diequatorial conformer (III-eq-eq) is more stable than diaxial conformar (III-eq-eq) Ref.2, although  III-eq-eq is impossible to make hydrogen bonding (see Newman projections of III-eq-eq and III-ax-ax). This shows that hydrogen bonding itself cannot overcome the steric repulsions induced by the 1,3-ax-ax-interactions. My conclusion is without knowing values of these attraction and repulsion interactions, it is impossible to say which of the three structures, I, II, and III), is the most stable.

References

X. Chen, D. A. Walthall, and J. I. Brauman, Acidities in Cyclohexanediols Enhanced by Intramolecular Hydrogen Bonds, J. Am. Chem. Soc., 2004, 126(39), pp. 12614–12620.
R. J. Abraham, E. J. Chambers, W. A. Thomas, Conformational analysis. Part 21. Conformational isomerism in cis-cyclohexane-1,3-diol, J. Chem. Soc., Perkin Trans. 2, 1993, (6), pp. 1061–1066.


Does energy come from fuel or oxygen? (surprise?) Question: Is it oxygen or fuel that releases energy during burning?

The answer from this article:

Why Combustions Are Always Exothermic, Yielding About 418 kJ per Mole of O2 (Journal of Chemical Education, 2015)

  Our analysis provides simple answers to two related, important
  questions: “What makes fire hot?” and “How much heat is produced?” in
  the combustion of a certain fuel. One can now explain that the
  double bond in O2 is unusually weak, and therefore the formation of
  the stronger bonds in CO2 and H2O results in the release of heat or
  an increase in thermal motion.

  Generally, our analysis indicates that it is not the organic fuel
  but rather O2 that is “energy-rich”.

It is usually said that fuel contains energy and that oxygen only enables the release of energy in the sense like enzymes enable reactions. Does anyone want to comment this? Because, from this article it seems that it is just the opposite: fuel enables the release of energy from oxygen.

(This question was initially nutrition-related: Does energy come from nutrients or oxygen?)

[OP:] It is usually said that fuel contains energy and that oxygen only enables the release of energy in the sense like enzymes enable reactions.

An enzyme is a catalyst, so it does not change the enthalpy of a reaction. That part is correct. "Oxygen only enables the release of energy" is incorrect. Oxygen is one of the reactants, and the oxygen atoms also appear as products.

[Dr. Schmidt-Rohr in JChemEd paper] Generally, our analysis indicates that it is not the organic fuel but rather O2 that is “energy-rich”.

This might be true in the context the combustion reaction with the set of organic molecules he chose to include. The organic fuels exclusively had C-H, C-C, C=C, O-H, C-O and C=O bonds, if I understand correctly from the abstract and the figure. He is saying that the bond energy of most bonds in reactants and products are similar (e.g. C-H vs O-H, C=C vs C=O, 2 C-C vs C=O), and that the O=O "double bond" is an outlier.
I think it is worthwhile to state the amazing fact that we have a biradical (see https://chemistry.stackexchange.com/a/15061) in our atmosphere and are breathing in a very reactive substance. The only reason that our red blood cells aren't constantly undergoing mutations from oxidative damage is that they lack DNA once they are released into the bloodstream.
If I were to study the combustion reactions with elemental metals, I might come to a different conclusion than the paper. Try comparing $\ce{Na + O2}$ with $\ce{Hg + O2}$.
Final remark (quoting myself):

When a raw egg rolls off a countertop and falls on the floor, does it crash because the countertop is so high or because the floor is so low?


Will the ions in salt bridge be used up? Ions in salt bridge move to the electrolytes of two half cells to balance the charges.

But will the ions in salt bridge be used up?

If it will, is the electron flow then stop?

The ions in the salt bridge are meant as charge carriers, and using a suitable diaphragm intends to keep the mass transport pretty low.

Nethertheless, some of the electrolyte (like $\ce{KCl}$, $\ce{KNO3}$, $\ce{Na2SO4}$ or $\ce{Bu4N+ ClO4-}$ to mention a few) does enter the other half cell.  This is why, as one criterion, the electrolyte in the salt bridge is choosen to not interfer with the electrochemical reaction you want to characterize (analysis like cyclic voltammetry) or execute (electrochemical synthesis).  Likewise for solvents used in cyclic voltammetry, there is an electrochemical window, the intervall of electrode potentials where it is safe to use them so that they are not oxidized, or reduced, respectively.

And if the electrolyte is consumed and charge carriers are absent, provided the electrical potential applied is not altered, yes, the electrical current drops to 0.


Why are the t2 orbitals above the e orbitals in a tetrahedral complex using the crystal field theory? In case of octahedral coordination entity formation, the $\mathrm{e_g}$ is above in energy level diagram than the $\mathrm{t_{2g}}$ due to more electrostatic force acting when the orbitals are along the axes in case of $\mathrm d_{z^2}$ etc. But in case of tetrahedral, why is the energy diagram inverted and becomes somewhat like this? I can't think of any possible reason. Can anyone please help?

Take into account that $\mathrm{e_g}$ and $\mathrm{t_{2g}}$ (or $\mathrm{e}$ and $\mathrm{t_2}$) are symmetry terms, and they make sense in the context of a certain symmetry. In other words, they do not describe something that is implicit to the orbitals, but, rather, how are orbitals affected by an environment with a certain symmetry.

In crystal field theory, we look only at the electrostatic effect of a certain distribution of charges around the nucleus on the orbitals; the discussion is primarily qualitative and geometric in nature.

In an environment with spheric symmetry (such as an isolated atom), all 5 d-orbitals are degenerate; since electrostatics are the same in any radial direction, and the set of 5 d-orbitals are only distinguished by their angular part (how are they distributed in the solid angle around the nucleus), they are equivalent. [Incidentally, this would be true not only for an isolated atom but also for an atom surrounded by a spherically symmetric charge distribution, such as a uniform charge distribution a certain distance away from the nucleus; however, I can't think of a real system that would have exact spherical symmetry.]

As soon as you introduce point charges around the nucleus, however, symmetry will break down. All 5 d-orbitals will no longer interact equally with the point charge; some orbitals will be closer to the point charge, some will be further away; and, most importantly from the point of view of symmetry, perhaps several orbitals will interact equally with the point charges.

This may be easier to visualize with the three $\mathrm p_x$, $\mathrm p_y$, $\mathrm p_z$ orbitals: if there are two point charges along the $z$ axis, it is clear that the $\mathrm p_z$ will interact more strongly with them than the $\mathrm p_x$ and $\mathrm p_y$ orbitals, but, crucially, $\mathrm p_x$ and $\mathrm p_y$ will interact exactly the same way with the charges; their interactions with the point charges are symmetry-equivalent.

In the extreme case where the distribution of point charges around the atom has no symmetry, all five d-orbitals would in general have different energies, because their interactions with the charges would be different; but if our distribution of charges is symmetric (as happens in both the octahedral and the tetrahedral case), we need to look at how the orbitals behave under the symmetry.

If you do that (and Jan's excellent answer explains what happens with the Cartesian axes under both symmetries), you will see that in the octahedral case two orbitals behave the same way (and are destabilised to the same extent), while another set of three orbitals behave equally (and are stabilised to the same extent). Looking further at their symmetry properties, these sets of two and three orbitals behave in the ways we describe with $\mathrm{e_g}$ and $\mathrm{t_{2g}}$, respectively. In the tetrahedral case, per contrast, we find that two orbitals are equally stabilised while three orbitals are equally destabilised; the symmetry label that fits their behaviour is $\mathrm e$ and $\mathrm t_2$. [In both cases, $\mathrm e$/$\mathrm{e_g}$ is a symmetry label that corresponds to two elements, and $\mathrm t_2$/$\mathrm{t_{2g}}$ corresponds to three.] These "blocs" of orbitals are degenerate (i.e. these two or three orbitals all have the same energy because they interact similarly with the point charges in the environment), while the energy of these blocs is not the same - that's what we call "splitting".

Note that the temptation to think that there are two "$\mathrm e$" d-orbitals and three "$\mathrm t_2$" d-orbitals in an atom is to be avoided: there is nothing intrinsic in the orbitals that produces this particular splitting into two blocs; rather, it's a product of the symmetry of the environment. The same 5 orbitals, if put in a square planar environment, do not form blocs of 2 and 3 orbitals, but 2, 1, 1 and 1; in a trigonal bipyramidal environment, they split into 2, 2, and 1; and so on.

So it just happens that two orbitals are stabilised in a tetrahedral environment (and thus are labelled $\mathrm e$) and three are destabilised (thus labelled $\mathrm t_2$), while it's the opposite in an octahedral environment.

[As a thought exercise: in crystal field theory, we assume that we are dealing with a central cation surrounded by anions, because that's the problem for which CFT was developed. But, what would happen if the point charges around the atoms were positive instead of negative? You'll find that you end up with stabilised $\mathrm{e_g}$ (octahedral) and $\mathrm t_2$ (tetrahedral) and destabilised $\mathrm{t_{2g}}$ (octahedral) and $\mathrm e$ (tetrahedral), exactly the opposite splitting but the same symmetry behaviour.]

There are rigorous mathematical ways to deduce the symmetry labels of d-orbitals (or any orbital, for that matter) under an environment of a certain symmetry, as provided by point group theory, which involve symmetry tables and so on. Most likely, your textbook will include a discussion of the derivation for at least the octahedral case.


Why does sodium react with water? The reaction of sodium with water is quite infamous.

$\ce{2Na + 2H_2O \rightarrow 2NaOH + H_2}$

But exactly why does $\ce{Na}$ displace $\ce H$ in $\ce{H_2O}$?

Alkali metals in general are extremely reactive metals. They are highly reducing in nature. Hence, they react with water forming their corresponding hydroxides evolving dihydrogen gas.

In case of sodium - as cited in your question - its reduction potential is $\pu{-2.7109 V}$ hence it easily reduces hydrogen.

Also, as the density of sodium is quite low it floats on the water. It immediately reacts with water to form a white trail of $\ce{NaOH}$ which further gets dissolved to give a colourless solution.

This reaction is highly exothermic evolving a lot of heat which is sufficient enough to melt sodium due to its low melting point.

Sodium moves in water as the dihydrogen gas formed below the waterline pushes sodium. If sodium is trapped in the container the increasing temperature could induce dihydrogen to catch fire.


How do I distinguish between internal energy and enthalpy? I am finding it hard to distinguish between the two. Can someone explain the two terms and the difference between them?

I tried learning from Wikipedia, but it said for both the above terms that it is the total energy contained by a thermodynamic system.

Internal energy, $U$, is the total energy contained in a thermodynamic system. However, absolute internal energy is hard to determine, and even relative internal energy and changes to internal energy are hard to determine. Here's why:

Changes to internal energy usually occur through heat transfer $q$ or work done $w$. There are other ways to transfer energy also (like radiation), but these two are common. Thus,

$$dU = q + w$$

The heat transfer can be determined easily by calorimetry, where $q=cdT$, where $c$ is the heat capacity of the system and $T$ is the temperature of the system. Alternatively, it may be easier to measure the temperature change of surroundings with known heat capacity, since $q_{sys}=-q_{surr}$

Work is a little more challenging to determine, since work is defined as the change in volume at constant pressure. $w=-PdV$ (the minus sign is there to show that when the volume of the system increases, the system does work on the surroundings). However, unless you have a closed system, volume changes are difficult to measure. Additionally, pressure may not be constant.

Thus,

$$dU=q+w=cdT-PdV$$

Internal energy changes can be determined easily at constant volume $dV=0$ so $w=0$ so $dU=cdT$. Internal energy changes can also be determined in closed systems at constant pressure (so $PdV$ is evaluatable).

However, in open systems (like a beaker), $dV$ is hard to determine, making $w$ hard to determine. A new thermodynamic state function was devised to remedy this problem.

Enthalpy is a measure of the total energy of a system. More precisely, it is a thermodynamic potential. Enthalpy is defined as $H=U+PV$. $PV$ is a state function, even though $P$, $V$, $PdV$, and $VdP$ are not state functions. Thus the change in enthalpy is:

$$dH=dU+d(PV)=cdT-PdV+PdV+VdP=cdT+VdP$$

Enthalpy changes are easy to determine at constant pressure (most open systems are constant pressure systems, which is why enthalpy is so useful to most chemists), since $dP=0$ and $dH=cdT$. Enthalpy changes can also be determined in closed systems of constant volume where the pressure change can be determined.

To summarize, internal energy and enthalpy are used to estimate the thermodynamic potential of the system. There are other such estimates, like the Gibbs free energy $G$. Which one you choose is determined by the conditions and how easy it is to determine pressure and volume changes.

At constant volume: $dU=cdT$ and $dH=cdT+VdP$
At constant pressure: $dU = cdT-PdV$ and $dH=cdT$
At constant volume and pressure: $dU=dH=cdT$
At variable volume and pressure, $dU$ and $dH$ are both hard to
determine (but $dH$ may be easier since you may be able to numerically estimate $d(PV)$ from pressure-volume data).


Why can't electrolysis utilize specific ions for reduction/oxidation In the electrolysis of water, reduction half equation is $$\ce{2H2O + 2e- -> H2 + 2OH-}$$ with a standard $E$ value of $-0.83\ \mathrm V$.

If $\ce{H+}$ is present in water, why can $$\ce{2H+ + 2e- -> H2}$$ not occur, with potential $0\ \mathrm V$, which is more favourable?

See, this is pretty much the same reaction. The potential is not always exactly what they say in the reference book. It depends on concentrations. The potential of $\ce{2H+ + 2e- -> H2}$ is $0 V$ only in standard conditions, that is, when the concentration of $\ce{H+}$ is 1 mol/L. When it is lower, the potential is different (see Nernst equation). Indeed, $\ce{H+}$ is present in pure water, but the concentration is quite low ($10^{-7}$ mol/L). If you plug that concentration into the Nernst equation, you'll get the same value that is listed for $\ce{2H2O + 2e- -> H2 + 2OH-}$.