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What side does heat get added on in an equilibrium expression?
I know this is probably such a basic question with fundamental explanations and I can't help from overthinking.
So if we have an equilibrium would the addition of heat as in an increase in temperature be added to the reaction side or product side. How does this apply to the addition of other chemicals into the equation? Will everything always be added to the left side since it is the reactants that form the product?
(reactants).  (product)
$$\ce{[CuCl4]^2- + 4H2O <=> [Cu(H2O)4]^2+ + 4 Cl-}$$

You better add, on next line after the equation, a sort of explanation like : Delta H = - xxx kJ/mol (or  + xxx kJ/mol)

Energy, usually energy involving the kinetic and quantum mechanical [rotational, vibrational, electronic] energy but that can also be electrical or photonic, is involved in chemical reactions in 2 ways. The first is activation energy, there must be enough energy available for the reaction to have an effective rate. This rate could be fractional seconds or eons. Adding heat energy increases the rate of both forward and reverse reactions and changes the equilibrium position.
The second is the change in chemical potential energy in the breaking and forming of chemical bonds. The reactants with the stronger overall bonds release heat to the system; removing heat [or work] forces the reaction in that direction. If heat is added the reaction is forced in the reverse direction [provided equilibrium can be obtained otherwise an alternative mechanism must be found].
Writing a reaction in the exothermic direction makes heat a product and it can be treated thusly. [or in reverse heat is a reactant]. Whether the reaction is reversible depends on complexity or entropy change the; simpler the more possibility it is reversible.


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Thermoplastics crystallization during nonisothermal cooling cycles
I have tested a PAEK family polymer using differential scanning calorimetry (DSC) with heating-cooling-heating cycles under different cooling rates (heating rates remained unchanged). PAEK are semi-crystalline polymers.
In theory, the crystallization enthalpy during the cooling cycle is similar to the melt enthalpy during the second heating cycle since the crystals formed during the cooling cycle are later melted.
Based on my results, the crystallization enthalpy decreases as the cooling rate increases (expected behavior), however, the subsequent melt enthalpy remains pretty much the same as the crystallization enthalpy value at the lowest cooling rate. The experiments were conducted with a few repeats and the results seem consistent.
What could be the cause(s) of this behavior?

See PAEK as polyaryletherketone.

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Justification for this theoretical experiment to measure temperature change during Joule expansion?
I am reading the book Heat and Thermodynamics 7th Edition by Zemansky and Dittman.
There is a section of a chapter that I would like to understand (Ch. 5.3).
I'm going to go through the reasoning, paraphrasing some parts and quoting others.

Imagine a thermally insulated vessel with rigid walls, divided into
two compartments by a partition. Suppose that there is a gas in one
compartment and that the other contains a vacuum. If the partition is
removed, the gas will undergo what is known as an adiabatic free
expansion in which no work is done and no heat is transferred. From
the first law, since both $Q$ and $W$ are zero, it follows that the
internal energy remains unchanged during a free expansion.
The question of whether or not the temperature of a gas changes during
a free expansion and, if it does, of the magnitude of the temperature
change has engaged the attention of scientists for about a hundred
years. Starting with Joule in 1843, many attempts have been made to
measure either the quantity $(\partial T/\partial V)_U$, which is
called the Joule coefficient, or related quantities that are all a
measure, in one way or another, of the effect of an adiabatic free
expansion, or as it is often called, Joule expansion.

Experiments that try to measure a temperature change during a free expansion are difficult in practice.

Instead of measuring a temperature change during free expansion for
which the internal energy is constant, consider measuring a change of
internal temperature for constant temperature.

The internal energy function of any gas is a function of any two of the coordinates $P,V$, and $T$. We can write
$$dU=\left ( \frac{\partial U}{\partial T} \right )_V dT +\left ( \frac{\partial U}{\partial V} \right )_T dV\tag{1}$$
and if no temperature change ($dT=0$) occurs in a free expansion ($dU=0$) then
$$\left (\frac{\partial U}{\partial V}\right )_T = 0\tag{2}$$
which means that $U$ does not depend on $V$
Similarly,
$$dU=\left ( \frac{\partial U}{\partial T} \right )_P dT +\left ( \frac{\partial U}{\partial P} \right )_T dP\tag{3}$$
and once more we have that if no temperature change ($dT=0$) occurs in a free expansion ($dU=0$) then
$$\left (\frac{\partial U}{\partial P}\right )_T = 0\tag{4}$$
ie, $U$ does not depend on $P$.

Then, it is apparent that, if no temperature change takes place in a
free expansion of a gas, $U$ is independent of $V$ and $P$, and,
therefore, $U$ is a function of $T$ only.
Thus, to determine if the internal energy is a function of
temperature, one must perform an experiment where the temperature is
constant and measure whether either $\left (\frac{\partial U}{\partial
 V}\right )_T$ or $\left (\frac{\partial U}{\partial P}\right )_T = 0$
is zero.

My question
The reasoning here seems to be that if we can determine that $U$ does not depend on $V$ or $P$ in general (actually, not sure about the use of the term "in general" here; I think actually we would be determining this for a particular level of $T$ and $V$ or $T$ and $P$), then this must mean that in the specific case of an adiabatic free expansion (ie, a Joule expansion) we have from (1) and (3) that
$$0=\left ( \frac{\partial U}{\partial T} \right )_V dT +0\cdot dV\tag{1a}$$
$$0=\left ( \frac{\partial U}{\partial T} \right )_P dT +0\cdot dP\tag{3a}$$
Now, assuming this is correct, I am not sure about the justification for $\left ( \frac{\partial U}{\partial T}\right )_V \neq 0$ or $\left ( \frac{\partial U}{\partial T}\right )_P \neq 0$, but if we assume this then it means that $dT=0$ in both equations.
Is this reasoning correct?
Note that I have a follow-up question concerning an actual experiment based on the theoretical analysis above.

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Understanding Rossini and Frandsen experiment for measuring $(\partial u/\partial P)_T$?
In another question, I asked about the theoretical justification for the experiment that I will describe in what follows.
The experiment was done by Rossini and Frandsen in 1932 at the National Bureau of Standards.
Here is the apparatus used

What we have here is $n$ moles of gas in a container $B$ at pressure $P$.
When the valve is open, gas escapes to the atmosphere through the long coil surrounding the container $B$.
There is a water bath around the container whose temperature can be maintained constant at exactly the same value as that of the surrounding atmosphere.
The experiment is as follows

The valve is opened slightly and gas flows slowly through the long coil and out into the air.

At the same time, the temperature of the gas, the container, the coil, and the water is maintained constant by an electric heating coil immersed in the water.

The electrical energy supplied to the water is, therefore, the heat $Q$ absorbed by the gas during the expansion.

Thus, we have a gas expanding in volume at constant temperature.
The work done by the gas is
$$W=-P_0(nv_0-V)\tag{1}$$
where $P_0$ is atmospheric pressure, $v_0$ is the molar volume at atmospheric temperature and pressure, $V$ is the volume of the container $B$, and $nv_0$ is larger than $V$.

If $u(P,T)$ is the molar internal-energy at pressure $P$ and
temperature $T$ and if $u(P_0,T)$ is the molar internal energy at
atmospheric pressure and the same tempeature, then, from the first
law, the change of molar internal-energy can be expressed in terms of
the measured quantities $Q$ and $W$ as
$$u(P,T)-u(P_0,T) = \frac{Q+W}{n}\tag{2}$$
provided that corrections have been made to take account of the energy
changes due to the contraction of the walls of the container.

My first question is about (2). Shouldn't it be $u(P_0,T)-u(P,T)$?

In this way, the change of molar internal-energy $\Delta u$ was
measured for various values of the initial pressure $P$ at constant
temperature $T$. The values of $\Delta u$ were plotted against the
corresponding pressure $P$, as shown in Fig. 5-3.

My main question is about the interpretation of these results contained in the following snippet

Since $u(P_0,T)$ is constant, the slope of the resulting curve is
equal to $(\partial u,\partial P)_T$ at any value of $P$. Within the
pressure range of 1 to 40 standard atmospheres, the experimental
points fall on a straight line, meaning that $(\partial u/\partial
> P)_T$ has the same value at every pressure; that is $(\partial
 u/\partial P)_T$ is independent of the pressure, depending only on
the temperature. Thus,
$$\left ( \frac{\partial u}{\partial P}\right )_T = f(T)\tag{3}$$
Rossini and Frandsen's experiments with air, oxygen, and mixtures o
oxygen and carbon dioxide led to the conclusion that the internal
energy of a real gas is a function of both temperature and pressure.
They found no pressure or temperature range in which the quantity
$(\partial u/\partial P)_T$ was equal to zero. In other words, their
real gases did not reach the low-pressure limit of the ideal gas.

Questions
1)

Since $u(P_0,T)$ is constant, the slope of the resulting curve is
equal to $(\partial u,\partial P)_T$ at any value of $P$.

I think this claim comes from
$$dU=\left (\frac{\partial u}{\partial T}\right )_P dT+\left (\frac{\partial u}{\partial P}\right )_T dP$$
But since $dT=0$ we have
$$dU=\left (\frac{\partial u}{\partial P}\right )_T dP$$
However, if we are always measuring $u(P,T)-u(P_0,T)$, (and actually as I said before the way I understand it the final pressure is $P_0$ and so the difference should be the other way around) then aren't we measuring some kind of average slope instead of the actual slope at at a point?
2.
My second question has to do I believe with this other question which is about why we would want to do this experiment in the first place and the theoretical basis for it.

Rossini and Frandsen's experiments with air, oxygen, and mixtures o
oxygen and carbon dioxide led to the conclusion that the internal
energy of a real gas is a function of both temperature and pressure.

This seems to imply that since they found $(\partial u/\partial
 P)_T\neq 0$ at all pressures, then it is not possible to have an adiabatic free expansion in which the temperature stays constant.
Is this the correct interpretation?
3.

In other words, their
real gases did not reach the low-pressure limit of the ideal gas.

What does this means exactly?

One form of the general virial equation of state for a real gas is $$\frac{Pv}{RT}=1+B^*(T)P+C^*(T)P^2+...$$where v is the molar volume.  At low pressures, this approaches $$\frac{Pv}{RT}=1+B^*(T)P$$
For a real gas, $$\left(\frac{\partial u}{\partial P}\right)_T=\left[-P+T\left(\frac{\partial P}{\partial T}\right)_v    \right]\left(\frac{\partial v}{\partial P}\right)_T$$At low pressure, this approaches $$\left(\frac{\partial u}{\partial P}\right)_T=-RT^2\frac{dB^*}{dT}$$which is a function only of temperature.

Where does the third equation come from?

It come from $du=Tds-Pdv$, $ds=\frac{C_v}{T}dT+\left(\frac{\partial s}{\partial v}\right)_Tdv$, and $\left(\frac{\partial s}{\partial v}\right)_T=\left(\frac{\partial P}{\partial T}\right)_v$

I really don't follow. We have $dU=(\partial U/\partial T)_V dT+(\partial U/\partial V)_T dV=T\frac{C_V}{T} dT +(\partial U/\partial V)_T dV=T\frac{C_V}{T}dT+T(\partial P/\partial T)_V dV +(\partial U/\partial V)_T dV - T(\partial P/\partial T)_V dV=Tds+dV((\partial U/\partial V)_T - T(\partial P/\partial T)_V)$. According to what you wrote we should have then $(\partial U/\partial V)_T - T(\partial P/\partial T)_V = -P$. Not sure where that comes from.

The derivation is in every thermo book.  Have you not learned about entropy yet?

No, entropy is a few chapters ahead still.

I've never seen the derivation done without employing the relationships involving entropy.


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For CVD diamond in a reaction chamber why does a lower temperature lead to a greater height value/thickness
I've learned that for CVD (chemical vapor deposition ) crystals in a reaction chamber, the greater the temperature the smaller the growth rate so the crystals have a smaller height. I know at lower temperatures the growth rate is faster and this is something to do with the balance between etching and growth rate but I'm not sure why this happens. Also, we have to compare the height at different times and why there is a fluctuation as you leave it in the reaction chamber up to an hour (3 min, 10 mins, 20 mins, 40 mins and 60 mins)
Also we have learned that at larger temperatures it is a slower reaction in the reaction chamber, however I am not sure why it is slower so why the grain size is larger.

Huh, homework about making of diamonds? Some people are more lucky ;)


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What's the practical use-cases of Raoult's Law?
So, I am taking my first chemistry class and I started learning about Raoult's Law and it states that, when a non volatile solute is added to a solvent, the vapor pressure of the solvent decreases in proportion to the concentration of the solute. One consequence of this is an increase of boiling point.
So, my question is, what are the applications of this in a real-world lab setting? Genuinely asking, why do we care about decreasing vapor pressure and increasing the boiling point in a lab environment?

Review the guide How to ask and Asking FAQs to prevent clarification requests, objections,  downvoting or closure.

I can think of one right away, that is estimating molecular masses of solutes. Further, making fractional distillation requires mole fractions in the gas phase to be different those in the solution phase, otherwise the solution forms an azeotrope which cannot be separated via fractional distillation. Moreover, Raoult law helps in understanding the bonding between solute and solvent. Deviation from the law also help in understanding our system well.
For large scale applications, and accurate lab experiments, $\Delta_\text{mix}H$ and $\Delta_\text{mix}V$ can have serious implications. Did you know that immiscibility of liquids can be explained based on large deviations from the law?

For further reading, you may refer to Atkin's Physical Chemistry


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Fractional Distillation Question
We're covering fractional distillation in my physical chemistry class. I'm confused regarding this diagram:

From your starting point, you heat the binary mixture (in this case benzene and toluene) to reach point a. Is the change in benzene mole fraction, point a to point b, then spontaneous? If so, why? If not, what triggers this rightward shift in mole fraction?

It should not be surprising that different substances have different volatility, leading to different compositions of liquid and gaseous phase.

Here is an interpretation of the diagram, implemented with two distillation setups (blue and orange):

In the first setup (blue), you have a large amount of your initial mixture, which you have to heat to about 370 K to boil (a). The vapor that boils off is enriched in benzene. You cool down the vapor to under about 355 K (c) to get liquid again.
In the second setup (orange), you have to heat this second mixture a bit again to get it to boil (c). The initial mixture boiling off is almost entirely benzene (d), and will liquify at about 350 K (e).
What is not shown is that as the distillation proceeds, the composition of the liquid changes, the boiling temperature will increase, and you will have more and more toluene in your distillate.

Is the change in benzene mole fraction, point a to point b, then spontaneous? If so, why?

The overall mole fraction can't change. You enrich the liquid in toluene while getting vapor that is enriched in benzene. If you turn the entire sample into vapor, there is no enrichment, of course.

If not, what triggers this rightward shift in mole fraction?

The same as the difference in boiling points of the two pure substances: It requires less energy to get a benzene molecule to boil or evaporate than a toluene molecule (in this particular system).

And we do this enrichment many times (using a fractionating column) to get a pure product (for example a Vigreux column).

Yes, but that's not shown in the picture. Here is a short video showing how the composition and temperature change in a simple distillation (one theoretical stage):

The use of an arrow in the diagram is probably misleading you. This diagram, for the most part, does not show how a system evolves over time.
This graph with the two red curves describes the equilibrium condition, at 1 atm, in a system containing only toluene and benzene. The top curve shows the equilibrium composition of the gas. The bottom curve shows the equilibrium composition of the liquid.
There is no movement from a to b. These two points represent the states achieved by the two phases in a single system at a single point in time. Point a represents the state of the liquid. Point b represents the state of the gas. The system is assumed to be at thermal equilibrium, so both points have the same temperature, but they have different compositions.
Similarly, points c and d both represent a single system at a single point in time, just at a different temperature from the a/b system.
In a fractional distillation, the idea is to create several "mini-systems" in your apparatus. The bottom flask is at 370 K, and conditions in that region of the apparatus are described by points a and b. Somewhere in the middle of the column, the temperature is 355 K, and the conditions there are described by points c and d. The whole a->d path exists in the system at a single time.
To make this more confusing, the diagram includes a point marked "Start." This point is on neither curve, and does not describe a benzene/toluene equilibrium at 1 atmosphere. This could indicate a system that is not at equilibrium, but in practice, for a fractional distillation, it means there is a third substance -- air -- in the system. As you heat the bottom flask to 370 to start the distillation, the air is driven out of the system by the vapor, and the bottom flask reaches the a/b state.


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why are there no pictures or experiments on solid ozone?
wikipedia states that solid ozone is violet black in colur and so is my textbook, but when you search up for solid ozones, there are no pictures or experiments done in youtube. Was the colour or the compound actually made or was it theoritically proven somehow that the colour was violet-black.
Was just curious to know

They cannot physically capture images of solid ozone due to difficulty working with it and not to mention supercold temperature. But studies has been made. See: https://pubs.acs.org/doi/10.1021/j100339a004; https://pubmed.ncbi.nlm.nih.gov/17718620/. Regarding ozonides, see: https://www.youtube.com/watch?v=WhcfBLop5gM

Solid ozone was certainly made, so researchers looked at it. I seriously doubt any dudes specialising in putting videos on YT did it, though. It's not easy and solid ozone is explosive.

Liquid O3 is already highly explosive and even gas could explode if given the chance - it's not something to be trifled with.

https://docslib.org/doc/7696119/the-history-of-ozone-iv-the-isolation-of-pure-ozone-and-determination-of-its-physical-properties-1

This text^ says solid O3 was photographed, but a photo in a paper and one googlable are two different things.

@Mithoron when I bring up your last reference they do not show the photograph (which I would have screenshit onto my device and then uploaded with appropriate sourcing). But photographs of the liquud are available and well document a violet coloration.

@OscarLanzi It's only a mention that one team made a color photo.

The color of liquid ozone is documented in sciencemadness.org:

Liquid diatomic oxygen, for comparison:

The color difference is similar to what we see between aquo and ammine complexes of copper, and for fundamentally the same reason: the deeper-colored form requires more energy to photoexcite an electron into an antibonding orbital. In the case of ozone versus diatomic oxygen, the increased energy comes from the delocalized pi bonding in the triatomic molecule, which raises the energy of the electron-accepting antibonding orbital.

Well, the question asked for solid ozone and not liquid ozone :)

@NilayGhosh But ozone can explode on fast cooling, so it's always something ;)


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Is the half-wave potential concentration dependent?
As I was refreshing my knowledge on cyclic voltammetry, I came across the following doubt.
Is the half-wave potential concentration dependent or is it not?

If we look at this graph, let's assume we have a reversible redox couple. The half wave potential $E_{1/2}$ is defined to be $\frac{E_{pc}+E_{pa}}{2}$, and is equal to the standard formal potential $E^{0}$. For example, if we did the ferrocene/ferrocenium CV experiment in 10 mM, 20 mM or 30 mM concentrations, would the resulting half wave potential $E_{1/2}$ always be the same or would it deviate, why or why not?
Edit: What is the physical explanation why the half wave potential is always around the formal potential of the redox couple, regardless of the concentrations?

So as you said, the halfway potential is equal to the formal potential for a reversible redox couple. However, is the formal potential dependent on concentration? The answer is yes, the activity of either species is dependent on the electrolyte specifications. Which is already one reason why the halfwave potential will shift with changing concentrations of reactants.

Okay you are right. But what is the physical explanation why the half wave potential is always around the formal potential of the redox couple, regardless of the concentrations?

What would be the physical explanation for the half wave potential not being around the formal potential of the redox couple?

@Poutnik yeah I don’t know why the half wave potential is around the formal potential, it was always a given in my electrochemistry course.


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Critical point from ab initio calculations
I was wondering if it is possible to calculate the critical point (e.g., temperature, pressure, and density) other than via experiments or equations of state like ab initio calculations.
Would anyone with more experience have any ideas?

It's more like estimation than calculation, I think, but check here for example: https://agupubs.onlinelibrary.wiley.com/doi/10.1029/2020GL089599

Are you asking for answer "in principle" or "in practice". Given enough computational resource and computing time, any properties of a chemical system can be calculated using ab initio calculation. In practice? I'm not sure. But @Mithoron provide a pretty good resource.


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How to Remove Copper Sulphate from a Leather Belt
I was careless when we were doing a routine footbath of our sheep in copper sulphate solution, and it splashed onto my leather belt. There are now little dark spots all over it, and I was wondering how to get the belt to go back to it's normal colour? It's been a while since it got stained, so does that make any difference?

Unfortunately, money is often the best solvemt for things like this.

Since it's a good preservative, I'd dip the whole belt in sheep dip, then wash and let dry at room temperature. Oil as needed, or ever 5,00 km.

Copper sulfate is water soluble, so water is your best bet. However, if it's bonded to or reacted with anything in the leather, you're probably out of luck.
Detergents and alternative solvents are not likely to help. Bleaches and other stain removers cannot destroy copper, and anything alkaline could make the situation worse by converting it to a less soluble copper hydroxide or carbonate.
A comment by DrMoishe Pippik suggests, possibly ironically, staining the whole thing with copper sulfate, and honestly that might work. You can find articles on dyeing and preserving leather with copper sulfate. But I would be a little worried what would happen if a child or pet chewed on it. Copper is not that toxic (unless you have Wilson's disease) but a new belt isn't that expensive either.


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Specific temperature ranges of kinetic combustion mechanisms
Do kinetic combustion mechanisms typically operate within specific temperature ranges? I tested some mechanisms from the scientific literature, and apparently they were only working at temperatures greater than 1000 K. Is this common to happen or am I probably making an error in my tests?

related https://chemistry.stackexchange.com/questions/16664/what-is-the-lower-bound-to-the-temperature-at-which-a-fire-can-burn


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How can I model the melting of ice as a function of time?
How much time it will take for a ball of ice at 0 °C and radius 3 cm submerged in 1 l of water at 100 °C to completely melt? (Assume the ice melts completely uniformly and treat it as a shrinking sphere.)
What type of theoretical calculations would I use to do this? I was thinking about using Newton's Law of Cooling and maybe related rates but I don't know where to start.

There is too many variables.

Newton's law would only apply  if the water were flowing past the ice. A full calculation would involve using Ficks law of diffusion for ice and water, so not easy as you will need to know about Fourier methods to solve the equations or do so numerically (and allow for enthalpy changes). You could start with a thin rod melting at the end with ice held at 0 degrees C. There are some examples in chapter 10 of https://chemistry-maths-book.com

@porphyrin It would need to solve water convection as well, that would be changing in time and the ice could get irregular shape. It would be complex system with non trivial simulation.

@Poutnik, I agree, its very very complicated but appears not to be. The temperature of the ice is unknown anyway no matter any other complications. This simple question illustrates why we invariably have to build rather simple models to explain things by ignoring this and that to try to reach an answer.

This seems like a rather standard engineering problem you can approach with semi-empirical formulas and in a similar fashion as you would derive Planck's equation of freezing, however the obtained differential equation will become incredibly complex as all these formulas (find the heat transfer coefficient from the Nusselt number, using free convection relations which will need Rayleigh and Prandtl numbers...) have a dependence on the characteristic dimension which is changing over time. This is of course neglecting the bulk temperature changes significantly which can be checked as well.

As a very, very crude, 'ball-park' or 'back of envelope' etc. estimate we can use the thermal diffusion coefficients of ice and water. The values for ice and water are $\approx 1$ and $0.14$ mm$^2$/s respectively (Wikipedia, Thermal diffusivity). If we assume that heat transfer is limited by thermal diffusion in water into the ice (as its so much faster when in the ice) and that the temperature at edge of ice is always the same (100 deg) it takes $L=\sqrt{6Dt}$ to diffuse average length $L$ in time $t$, thus taking $L=15$ mm, the time is $t\approx 4.5$ minutes. This should be the shortest time, as the ice cannot melt faster than the thermal energy can reach it.

Yet another aspect, aside of conduction and convection, is heat transfer caused by thermal radiation of condensed phases. As it is proportional to the difference of the 4th power of temperatures, it is progressively raising. I would not , of course, dare to try to involve it in any simple model.

The simplest model I know is to define a diffusion layer (or boundary layer) around the ice sphere. The temperature of the ice is set to 0 °C, the temperature of the bulk water is set to 100 °C, and there is linear diffusion in the boundary layer to transfer the heat from water to ice.
A similar approach is used for the dissolution of crystals in water. Depending how vigorous you stir, the boundary layer is larger or smaller (we did an experiment on the dissolution of gipsum in physical chemistry lab, and here is an example research paper). The boundary layer is on the order of hundreds of microns for aqueous room temperature systems. In the case of ice melting, having a temperature gradient within the boundary layer makes estimation of its magnitude more complex as the viscosity depends on temperature.
In any case, you could measure radius vs time for a melting ice sphere, and check whether a single magnitude of the boundary layer describes the kinetics well (or not). From my real-world experience, the 4.5 minute estimate in Porphyrin's answer seems much too long (here is a video to get the order of magnitude). Because I don't have a clue what the size of the boundary layer would be, I can't make an estimate with the model I am proposing.

It is simple. Drop an ice cube into a pot of boiling or just hot water and time it. The temperature gradient is from 0 to 100 Celsius because the ice was equilibrated at zero and the water boils at 100.
This could be a calorimetry experiment. Run the process in a calorimeter measuring the energy needed to maintain the liquid T at 100 [or other temperature] as a function of time. [Or measure the liquid temperature drop.] There are many variables to consider; that is the work of the experimenter, to identify and address the variables.

The question is how to model the process i.e. on a computer or if sufficiently simple solving the relevant equations analytically, not how to do an experiment


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First order kinetics
Finding time t in a first order reaction (JEE Adv 2019)
In reference to the above question,taking into account the stoichiometric coefficient, the standard equation of first order reaction time changes by a factor of 2.
I came across a question which i find similar to the one above.
However when i tried solving it the same way, i couldn't get the right answer. Can someone point out where did i go wrong?
The question is attached below:
The decomposition of $\ce{N2O5}$ in carbon tetrachloride was followed by measuring the volume of $\ce{O2}$ gas evolved:
$$\ce{2 N2O5(solv) ->[CCl4] 2 N2O4(solv) + O2(g)}$$
The maximum volume of $\ce{O2}$ gas obtained was $\pu{100 cm3}$. In $\pu{500 min}$, $\pu{90 cm3}$ of $\ce{O2}$ were evolved. The first order rate constant (in $\pu{min-1}$) for the disappearance of $\ce{N2O5}$, is:
Here is what i did
$$t=\frac{1}{2k} \cdot \ln{\left(\frac{P_0}{P}\right)}$$
\begin{aligned}
K &=\dfrac{2.303}{2×500} \log _{10}\left(\dfrac{200}{200-(2)90}\right) \\
=& \dfrac{2.303}{2×500} \log _{10}\left(\dfrac{200}{20}\right) \\
& \\
K &=\dfrac{2.303}{1000} \mathrm{~min}^{-1} \\
\end{aligned}

It is not a good idea to use 0 instead of O in chemical formulas.

You should derive the equation for k directly for the 1st order rate equation.   And general advise - avoid using literal values too early. Do not use them when you do not have to yet.

@Poutnik I have derived the equation and that is the reason why i am asking the error in the answer

minor point - upper case $K$ is typically reserved for equilibrium constants (thermodynamics), while lower case $k$ is used for rate constants (kinetics). And no matter which you use, you should be consistent across the problem and not switch between them.

As oxygen is reportedly being created by the first order kinetics, we can write:
$$- \frac{1}{2} \frac{\text{d}[\ce{N2O5}]}{dt}  = r = k_\text{r}[\ce{N2O5}]$$
Let substitute $y = [\ce{N2O5}]$:
$$\frac{\text{d}y}{y} = -2k_\text{r}\cdot \text{d}t$$
$$\int_{y_0}^{y_1} {\frac{\text{d}y}{y}} = -2k_\text{r}\cdot \int_0^t {\text{d}t}$$
$$\ln {\frac{y_1}{y_0}} = -2k_\text{r}t$$
$$y_1 = y_0 \cdot \exp{(-2k_\text{r}t)}$$
$$y_0 - y_1 = y_0 (1 - \exp{(-2k_\text{r}t)} ) = \left( \frac{y_0}{V_\mathrm{\ce{O2}, fin}} \right)  \cdot  V_\ce{O2}$$

$$V_\ce{O2} = V_\mathrm{\ce{O2},fin} \cdot \left(1 - \exp {(-kt)} \right)$$
$$ \exp {(-kt)}  = 1 - \frac{V_\ce{O2} }{ V_\mathrm{\ce{O2},fin}} $$
$$ k = \frac 1t \ln {\left(\frac{V_\ce{O2,fin} }{ V_\mathrm{\ce{O2},fin} - V_\mathrm{\ce{O2}}} \right)} $$
$$ k = \frac {1}{\pu{500 min}} \ln {\left(\frac{\pu{100 mL} }{ \pu{100 mL} - \pu{90 mL}} \right)} \approx \frac{2.303}{500 } \pu{min-1}$$
$$ k_\mathrm{r} = \frac{k}{2} = \frac{2.303}{1000 } \pu{min-1}$$

It is clear that $k = 2 k_\text{r}$, what meant the reaction kinetic rate constant is not the same as the first order rate constant, with the proportionality constant equal to stoichiometric coefficient. It is the direct consequence of how the reaction rate is defined.


text:
How to calculate the ground state molecular potential curve of Lithium in different spin configuration?
I am a physics student and interested in Atom, and molecule optical physics. I am studying Feshbach resonance and I need to know the potential curve of the molecule, then I ask question in physics stack exchange but I could not get any reply.

I saw the above picture on the German Wikipedia of Feshbach resonance. They mentioned the ground state of Lithium diatomic molecule($Li_2$) has this the molecular potential curve in different spin configuration($S=0,S=1$).
However because I am not familiar with the molecular physics and quantum chemistry, I don't understand how I can calculate/simulate them out. If someone can provide me some information, or list the formula(Hamiltonian), I will have some directions of this issue.
Thank you

What is 'the lithium molecule'?

@isolatedmatrix diatomic molecule of Lithium

The LI$_2$ molecule has been well studied, the potential energy curves  should be in this reference,   I. Schmidt-Mink, W. Muller, and W. Meyer, Chem. Phys. 92, 263 (1985).

@porphyrin Thanks for the information.


text:
Where does the relation Λm=κV come from? is it experimental?
I was learning about molar conductivity in electrochemistry and this relation
                                      Λm=κV

came up in my textbook. it somehow 'derives' this relation from a known relation except it does not do it very well. (or am i lacking something to understand that?.in that case please let me know which topic).

I just do not get how it equated area to volume. please educate me on this bit too. I will be very thankful.
Thank You!

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Numerical value 1 does not mean being unitless, Two quantities with the same numerical value but different units are not equal. B the numerical value and the unit are integral part of physical quantity value. If L = 1 m and m = 1 kg then L <> m.

These equations are definitions:
\begin{align}
\kappa &= G \cdot \frac{L}{A}\\
\Lambda &= \frac{\kappa }{ c} = G \cdot \frac{L}{Ac}\\
c &= n/V\\
\Lambda &= \frac{\kappa V }{ n} = G \cdot \frac{LV}{An} = G \cdot \frac{LV_\text{m}}{A}
\end{align}
$$\Lambda[\pu{S m2 mol-1}]=G[S] \frac {L[\pu{m}]}{\pu{1 m2} \cdot c [\pu{mol m-3}] }$$
If $L=c$ numerically then $\Lambda = G$ numerically.
If $A=1 \pu{m2}$  then $L = V $ numerically.
The formula  $\Lambda m = \kappa V$ would be correct, if there was $n$ (molar amount) instead of $m$ (mass).

Symbol
Quantity

G
Conductance

$\kappa$
Conductivity

$\Lambda$
Molar conductivity

L
Length

A
Area

c
Amount concentration

V
volume

n
Molar amount


text:
Organic Chemistry: Numeric prefix for 9 carbons doesn't follow pattern of greek numbers
In the nomenclature of organic compounds, a numeric prefix is used for denoting the number of carbon atoms in the main structure. From "pent" onwards, they seem to follow a pattern of greek numerals, however, at 9 carbons, there is an exception: "non" instead of "ene". Why is that? Is it to avoid confusion with double bond insaturations, like "eneene" would create? Is it just to make it sound better?
PS: Forgive my lack of knowledge on the subject, I only recently got an introduction to organic chemistry.
Thank you in advance!

In this answer we find that in fact it is the Latin prefixes which were original. The Greek prefixes are the intruders, except for nine. I still favor "enneagon" for the polygon, though.

Yeah, it does, Oscar had sent the same thing earlier. Thanks :D

Naming conventions are a hodgepodge, e.g., methanol, from Greek "μέθυ + ὕλη", i.e., "wine + wood", ethanol, directly from "μέθυ ", propionic acid, from "πρῶτος + πίων", "first fat". And methanoic acid is known as formic acid, because it made by ants, and ethanoic acid is acetic acid is from vinegar in Latin, acetum.  Hope you don't mind taking up your thyme (θυμώνω, which is sour??) with etymology.


text:
Naming of cyclic acid anhydrides
For some cyclic anhydrides, the -dioic anhydride suffix seems to be used.
E.g. butanedioic anhydride (succinic anhydride).
How does this apply to anhydrides which are not symmetrical about the oxygen atom bonded with two carbons?

Is it correct to call this compound "Hept-2-enedioic anhydride"?
According to ChemDraw online (JS Sample Page), its name is "(Z)-4,5-dihydro-2H-oxocine-2,8(3H)-dione". I don't understand this.

You could call it like that, but IUPAC has general nomenclature for heterocyclic compounds. Oxocine is parent unsaturated ether.

Cyclic anhydrides are generally named as heterocyclic pseudoketones; see my answers here and here.

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-65.7.7 Cyclic anhydrides
P-65.7.7.1 Cyclic anhydrides formed from two acid groups attached to the same parent hydride structure are named in two ways:
(1) as heterocyclic pseudoketones;
(2) by changing the class term ‘acid’ to ‘anhydride’ in the systematic or retained name of the dibasic acid.
Method (1) generates preferred IUPAC names

Using Method (1), the first step is naming the heterocyclic component. Heteromonocyclic rings with three through ten ring members having the maximum number of noncumulative double bonds (called mancude ring systems) are used as parent components. Since there is no retained name for the parent component of the compound that is given in the question, a systematic Hantzsch-Widman name is used, in this case oxocine.
The original parent structure oxocine ($\ce{C7H8O}$) has the maximum number of noncumulative double bonds for this kind of structure. Compared to an ideal unsaturated compound, however, oxocine has one extra hydrogen somewhere. The position of this extra hydrogen must be indicated. A few different isomers are possible; for example, the original parent structure could be 2H-oxocine.

The corresponding pseudoketone is named using the suffix ‘one’ in accordance with rules expressed for ketones, which gives the name (4Z,6Z)-2H-oxocine-2,8(3H)-dione.

Finally, removing the extra double bond yields the complete name (6Z)-4,5-dihydro-2H-oxocine-2,8(3H)-dione.


text:
Possible error in textbook question? (Heat transfer)
I am dealing with either (a) a severe misunderstanding of the question, or (b) a wrong textbook answer.
Here is question verbatim:

A jacketed vessel is used to heat a water stream using steam
available at $110°C$, as shown in Figure $10.10$. The steam temperature remains unchanged as its latent heat is removed. The jacketed
vessel has a circular cross section with internal diameter $0.4 m$ and
height $0.9 m$. It is fitted with an appropriate stirring device such
that its content is well mixed. The water-side resistance for heat
transfer is $0.4545 m^2 °C/kW$, while the steam-side resistance is
$0.14 m^2 °C/kW$. The vessel wall is $2 mm$ thick and is made up of
copper with a thermal conductivity of $0.4 kW/m °C$. What is the
maximum hourly flow rate of water that this vessel can handle to
provide hot water at the specified outlet temperature? The specific heat capacity of water is $4.2 kJ/kg °C$ and heat transfer
through the top and bottom surfaces of the vessel may be assumed to be negligible.

(a) $2880 kg/h$
(b) $720 kg/h$
(c) $1440 kg/h$ (this is the textbook answer)
(d) $2160 kg/h$

Here's my working:

$2mm = 0.002m$
$R_H=0.4545+0.14+ {ln({r_o\over r_i})\over 2L{\pi}K} $
$R_H = 0.5945 +{ln({0.4 \div 2 +0.002\over 0.4 \div 2})\over 2(0.9){\pi}(0.4)} $
$R_H = 0.598899$ $(6 s.f.)$
$(\frac{T_{out} - T_w}{T_{in} - T_w})=exp(\frac{-\pi{d}L}{m{C_p}{R_H}})$
$ln(\frac{70-110}{25-110})=\frac{-\pi(0.4)(0.9)}{m(4.2)(0.598899)}$
$-0.753772={-0.449624\over m}$
$m = 0.596499$ $kg\cdot s^{-1} = 2147.40$ $kg\cdot h$

Seeing as my answer is totally different from any of the presented options, I likely made a total logic problem, so what was my mistake?
Textbook:
Khan, S. A., Farooq, S., Swee, K. Y., Jangam, S. V., & Wee, E. C. L. (2021). Heat Transfer II. Chemical Engineering: Visualizing Foundational Concepts (Rev. ed., pp 134-135). Department of Chemical and Biomolecular Engineering, National University of Singapore.

Your answer doesn't seem "totally different" from the presented options, your answer is the same as (D) if you had rounded 0.596499 kg/s to two digits.

Yeah, but it's totally different from Option C, 1440 Kg/h. That's the textbook answer.

I don't have a textbook in front of me to go over the formulas you're using, but your $R_H$ equation seems to have unit inconsistencies.  The first two terms are given in $m.m.C/kW$ while the third term looks to be in $C/kW$ judging by the numbers from the problem you were plugging in.  So it's possible that this is throwing off your $R_H$ value.  Units look like they match properly for the equation with the exponential, though.  You could work backward and solve for $m\cdot R_H$ to see what the textbook expects $R_H$ to be, to try to figure out the above unit issue.

Your calculation of the wall resistance is incorrect.  It should be 0.002/0.4.

Please include a proper citation for the textbook.

Khan, S. A., Farooq, S., Swee, K. Y., Jangam, S. V., & Wee, E. C. L. (2021). Heat Transfer II. Chemical Engineering: Visualizing Foundational Concepts (Rev. ed., pp 134-135). Department of Chemical and Biomolecular Engineering, National University of Singapore.

I figured out the problem with my working. I used the following equation earlier:
$(\frac{T_{out} - T_w}{T_{in} - T_w})=exp(\frac{-\pi{d}L}{m{C_p}{R_H}})$
This was the wrong equation to use because it assumes that there is a temperature gradient within the jacketed vessel, which isn't the case because it is well-mixed and thus the internal temperature can be assumed to be constant (in this case, at $70°C$).
Correct working:

$R_H = 0.4545+0.14+\frac{{\Delta}X_W}{K} = 0.5945+\frac{0.002}{0.4} = 0.5995$ (Thanks @Chet Miller)
Rate of heat transfer from hot water to cold water in the vessel = rate of heat transfer across the jacketed vessel wall from hot steam to water
Let $m =$ volumetric flow rate of cold water into the vessel
Let $A =$ area of vessel wall
$m*(T_{out}-T_{in})*C_P=\frac{T_w-T_{out}}{R_H}*A$
$m = \frac{(T_w-T_{out})*A}{R_H*(T_{out}-T_{in})*C_P}=\frac{(110-70)*{\pi}(0.4)(0.9)}{0.5995*(70-25)*4.2} = 0.399265kg{\cdot}s^{-1}=1440kg{\cdot}h^{-1}$ (3 s.f.)

This is not done correctly.  The right hand side should be using the log-mean temperature difference (59.7 C), not the wall temperature minus the outlet temperature.  The latter is not the average driving force of heat transfer.

Sorry, I don't get it. I've never encountered 'log-mean' before and it's not in my textbook. Admittedly that the reason may be that I'm only in my first semester as a year 1 student. Care to elaborate?

See my added Answer

The log-mean temperature difference arises from your original equation:  $$(\frac{T_{out} - T_w}{T_{in} - T_w})=exp(\frac{-\pi{d}L}{m{C_p}{R_H}})$$or$$\ln{\left(\frac{T_{out} - T_w}{T_{in} - T_w}\right)}=\frac{-\pi{d}L}{m{C_p}{R_H}}$$This can be arranged to $$Q=mC_p(T_{out}-T_{in})=\frac{\pi dL}{R_H}\frac{[(T_w-T_{in})-(T_w-T_{out})]}{\ln{\left(\frac{T_{w} - T_{in}}{T_{w} - T_{out}}\right)}}$$where Q is the heat load and where the log-mean temperature difference is given by:$$(\Delta T)_{LM}=\frac{[(T_w-T_{in})-(T_w-T_{out})]}{\ln{\left(\frac{T_{w} - T_{in}}{T_{w} - T_{out}}\right)}}$$Then we get:  $$Q=mC_p(T_{out}-T_{in})=\frac{\pi dL}{R_H}(\Delta T)_{LM}$$

The reason I was given as to not use that formula for this question is that this formula assumes that the liquid isn't subject to bulk motion from stirring, which would give a temperature gradient in the liquid (cooler nearer to the intake, hotter nearer to the outlet). However the liquid in the solution is well-mixed (see diagram) which gives roughly even temperature distribution in the liquid.

Good point.  That explains it.

Overall resistance $R= 0.4545+0.14+\frac{0.002}{.4}=0.5995\ (m^2C)/kW$
Heat Transfer coefficient $U=\frac{1}{R}=1.6681\ kW/(m^2C)$
log mean temperature difference $=\frac{85-40}{\ln{(85/40)}}=59.7\ C$
Area $=(0.4)\pi (0.9)=1.131\ m^2)$
Heat Load = (1.6681)(1.131)(56.35)=112.6 kW
So $$\dot{m}C (70-25)=106.3$$
and $$\dot{m}=.5959\ kg/sec=2145\ kg/hr$$


text:
Product of oxymercuration-demercuration reaction

What will be the major product of this reaction?
I know that this is the OMDM reaction where Alkene is converted to alcohol following the Markovnikov rule. This question is also atypical because all the previous questions used NaBH4 as the demercuration (reducing) agent instead of NaBD4. According to my reasoning, the answer should be:

But my answer is incorrect. Please help me out.
PS- Can someone suggest me some website/app where I can draw chemical structures?

I agree. The the given answer is incorrect.

Well, the answer is correct except instead of -OH it must be -OMe because the reagent used is Hg(OAc)2/ MeOH and not Hg(OAc)2/H2O.
Here's an illustration of the rection using methanol

Ahh! Yes thank you. That is the right answer.

Hey so if we were using D2O instead of MeOH (along with NaBD4) would we get OD instead of OMe?

Unless there is another proton source in the mix, probably. If there any somewhat acidic protons available, the $\ce{OD}$ would rapidly exchange with them.


text:
How many molecules in DNA?
This seems to be a simple question but I have failed to find an answer by searching.
Let's consider human DNA so 23 chromosomes.
The first question is on the double helix, is that considered as one or two molecules?  This is the less interesting question to me as I know the structure and it is mainly a matter of the definition of the term molecule.
The bit that is more interesting to me and I cannot find a clear answer on is the chromosomes.  Is each chromosome a separate molecule or pair of molecules (hence 23 or 46) or is the whole of the DNA a single molecule or pair (hence 1 or 2).  If each chromosome is not a separate molecule then how are they connected?  If the whole DNA is a single molecule or a pair then what happens at the chromosome boundary?
I am trying the chemistry stack exchange as I am just interested in the chemical structure and not its biological function.
A clarification: I am not suggesting that a chromosome consists solely of DNA.  I mean the DNA within the chromosome.

When you take a single somatic cell that is not replicating, lyse it, remove the membrane lipids, remove all the protein, you would be left with just nucleic acids.
If each DNA strand were intact, and none of them are covalently cross-linked, you would expect 23 x 4 molecules.
However, DNA undergoes single-strand breaks. They can be repaired fairly easily when the second strand is intact. However, there might be thousands of single-strand breaks at steady state, so thousands of molecules.
Inter-strand crosslinks are less common but possible. For each of those between two as-of-yet unconnected strands, you reduce the number of molecules by one.
So if you take a chemical view, the number of molecules of DNA is largely undefined. If you take a biological view, you would just simply say that humans have 23 chromosomes, and learn about what they do in various stages of the cell cycle.

Is each chromosome a separate molecule or pair of molecules (hence 23 or 46) or is the whole of the DNA a single molecule or pair (hence 1 or 2).

I don't think anyone would have a good argument for describing the DNA in a human cell as a single molecule, or two.

Why 23 x 4 and not  23 x 2?

Mom's and dad's copy (pairs of homologous chromosomes, except for X and Y in males), and each is a double helix.

Is each chromosome a separate molecule or pair of molecules (hence 23 or 46) or is the whole of the DNA a single molecule or pair (hence 1 or 2).

None of the options.
Chromosomes are organelles, consisting of a DNA molecule (depending on a view, a single molecule or the complementing two ones, bound by multiple hydrogen bonds), wrapped in various histone and chaperone proteins. By other words, they are much more complex than a virus, which is not a molecule either.
Human genome consists  from 22 chromosome (organelle)  pairs plus chromosomes X and Y, so 46 wrapped parts of DNA as dual helixes. Each pair represents two variants of the (almost) same information.
Additionally, there are multiple short mitochondrial DNA segments in typically 1000 mitochondrions. BTW, scientists among Star Wars fans must have noticed similarity between a mitochondrion and a midichlorian.
More in Wikipedia.

Well the two strands together make the DNA molecule. you have 46 chromosomes that half of them are taken from your biological father and the other half from your biological mother so you have 2 copies of each chromosome that are like two exact houses but with different people inside so the two copies have different genes. you can not see your chromosomes unless your cells are duplicating that's when chromosomes start to form. chromosomes are completely compressed in metaphase which you can see them in a karyotype. in a normal cell chromosomes are not visible because the whole sequence is tangled together.


text:
What is the product and mechanism of reducing a cyclohexanone derrivative with LAH?
I've been trying to figure out the product and mechanism of the reaction pictured below for a while now:

I'm fairly sure that the first step is the reduction of the ketone by LAH, which results in a negative charge on the oxygen, that then forms a salt with the positively charged lithium. If this was just cyclohexanone, the acidic workup would simply yield a cyclohexanol.
However, from the given chemical formula of the product, I can tell that additional reactions/rearrangements must be happening: The educt has the formula C20H20O2, so the product must have one O, two Cs and 4 Hs less than the educt. I'm pretty sure that this ammounts to the loss of the ethoxy group, which would in turn mean that the product gains one H from some other source (since it loses 5 Hs from the ethoxy group, but has 16 Hs).
This is about as far as I got, and haven't yet found a reaction mechanism that would yield a product with the correct formula.
Here's one of my attempts, with the idea being that it's sort of analogous to how LAH reduces Esters first to Aldehydes and then Alcohols.

The problem here is that there's two Hs too many. I'm also very unsure about the validity of the electron pushing arrows.
What is the correct mechanism in this case?

I think there's an error in there.

LiAlH4 reduces the ketone to alcohol. The acidic workup hydrolyses the enol ether to ketone AND dehydrates the alcohol to alkene forming the enone.


text:
Creation of a Solution of Liquid Latex + Sodium Caseinate with Polyvinyl Alcohol
I am going to be experimenting with creating different types of liquid latex formulations, primarily for the purposes of increasing shear strength and tear resistance, as well as just general exploration of various modifications of plastic properties that can be achieved.
Unfortunately, I am a hobbyist, and have zero training or background in chemistry. I have purchased liquid latex and various other ingredients, and am going to be creating various different mixtures of each very soon. However, I also want this project to serve as a jumping off point for my own study of chemistry in general. e.g. I don't simply wish to remain a hobbyist making random mixtures, but want to understand more of the chemical theory behind what I'm doing.
Issue is, I don't know what technical terms I need to know to begin looking up the proper subjects for this project online. To that end, I have some very specific questions I would like to ask:

Is Polyvinyl Alcohol likely to be suitable as an emulsifier between liquid latex and sodium caseinate? I've done research on emulsions, but most of what I can find involves oil-water, oil-wax, or wax-water type emulsions.
Is there some sort of basic theory of emulsions or resource of that nature that would help me determine emulsion suitability between any number of different classes of chemicals? What is it called? Without the proper terms, I'm basically flailing blind online. :(
In previous test procedures, I extracted micellar casein from heated milk using vinegar, and dissolved the casein into a solution of sodium hydroxide and calcium chloride. However, I just put general amounts and hoped for the best. I know that there is a way to calculate how much calcium chloride and sodium hydroxide would be needed to precisely dissolve the micellar casein with no excess chemicals in the solution.
Unfortunately, I do not know what this process is called or how to about researching it. But I have seen strange equations with characters on one side, an arrow in the middle, and characters on the other side. I would like to start learning what this is called.

I would very much like to thank you guys for your time. I may be making a lot of very wrong base assumptions, but one has to start somewhere. Any and all help is appreciated. Thank you.


text:
metallic mercury is shaken with a solution of mercury(II) nitrate

Hi there, I am reviewing equilibrium.
About this question, I wonder if the chemical equation is wrong.
Because the description says 'a solution of mercury(I) nitrate is formed'.
However, in the equation, the product is Hg2 2+?
I thought the product should be Hg1+.
Here is my answer, but my answer seems to be wrong.

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Your expression would be correct if the mercury(I) ions were individual, separate atoms like most metals. But they are actually paired up, forming $\ce{Hg_2^{2+}}$ with a covalent bond between the metal atoms. Thus, properly,
$\ce{Hg(l) + Hg^{2+} <=> Hg2^{2+}}$
with $K_c$ then equalling $\ce{[Hg2^{2+}]/[Hg^{2+}]}$ as given in the textbook.
This behavior of forming diatomic metal(I) ions is actually known with several elements in Group 2 and Group 12 (or if you are using an older text, Group 2A and Group 2B), but mercury is the one that most commonly has metal(I) ions and not always metal(II). So your textbook (presumably) identifies specifically mercury as forming $\ce{Hg2^{2+}}$.


text:
Is concept of Weak and Strong Field Ligands applicable in Valence Bond Theory?
I am high school student studying Coordination- Chemistry. I have read about Valence bond theory and Crystal Field Theory as well.
My teacher taught me that Valence bond theory is based on experimental value of magnetic moment whereas CFT focuses on splitting of orbitals in presence of Weak and Strong Field Ligands. My confusion is that can we explain VBT configuration on basis of SFL and WFL?
Like in one of my theory exams, we were asked to write configuration of [Fe(NH3)6]3+ on basis of VBT. But no magnetic moment data was given in question. Then how should I have written the configuration?

Yeah buddy I think you were thinking correctly, since NH3 is strong field ligand it will cause pairing, there are 5 electrons in 3d orbital of fe(3+) so there will be two pairs and one lone pair (you can find B.M using (n(n+1))^1/2 ) also since 2 d orbitals are vacant along with one s and three p hybridization will be sp3d2 ie octahedral. But one point remains VBT doesn't consider SFT or WFT as per my memory, and it's also counted as it's limitations. But generally this way is used to solve exam questions.


text:
crystalline form of HCl
HCl is a covalent compound.
I do not understand why HCl cannot exist as an ionic compound with the combination of hydrogen ions and chloride ions. Is it a matter of electronegativity?
I have once wondered if it is possible to obtain HCl crystals from hydrochloric acid, since I was told that HCl(g) ionizes in water to produce hydrogen and chloride ions. However, I have overlooked the fact that the so-called hydrogen ions exist as hydronium ions instead of pure hydrogen ions.
Does anyone have some thoughts on the reason why HCl should be covalent, and whether it is possible to prepare HCl salt?

H+ as a naked ion in a bulk compound is quite definitely impossible. Hydronium is another thing; solid salts of H3O+ are known, but it seems this is not what you want.

H+ alone is not like other ions = atoms with nonzero net charge. It is just a proton, an elementary particle  with size roughly 10^5 smaller than atom, with intensity of electrostatic field roughly 10^10 stronger than e.g. for Li+. Any free proton encountering Cl- ion would form HCl.

https://chemistry.stackexchange.com/questions/73254

The crystalline state «does not mind» if the content of a unit cell are neutral atoms (as in metals), molecules (water as ice, iodine), ions (NaCl), or a mixture of them ($\ce{CuSO4 * 5 H2O}$). Simplified: the crystalline state only is one form (or in case of polymorphism, a couple of forms) where objects are arranged in a somewhat regular pattern that once you know the unit cell, you equally know the properties «here and thousands of unit cells away» (because of periodicity, in either 1D, 2D / tiling, 2.5D, or 3D / space). It still can become a bit more complicated down the road, though.

Pure HCl is a gas at room temperature and its melting point is ~-114° . The pure compound is a covalent molecule as a gas and as a solid, apparently. It doesn't exist as an ionic compound. The reasons why it doesn't form an ionic at room temperature is because it is a gas; in the solid state explanations are likely more complex but possibly because "naked" protons are very unfavourable isolated and always react to avid being isolated.

A purely ionic HX compound is not impossible, but not with H+ alone as the cation. We would need the hydrogen ion hydrogen-bobded to what would otherwise br neutral molecules. Fluoronium salts precipitated from fluoroantimonic acid, with empiricalbformulae HSbF6 and H3Sb2F13, would qualify.

HCl can form crystals if there are ammonia around especially in laboratory conditions. See these Reddit posts: https://www.reddit.com/r/chemistry/comments/cratr6/strange_crystals_on_hcl_bottle_what_is_this/?rdt=54675&onetap_auto=true and https://www.reddit.com/r/chemistry/comments/1m6xkl/found_a_bottle_of_hcl_in_an_old_lab_with_neat/

As has been mentioned, a compound with a bare hydrogen cation cannot exist. Salts derived from strong acids generally contain hydrogen ion covalently bonded to a neutral molecule, so the charged species is multiatomic.
Ironically, the best approximation to an ionically bonded hydrogen cation in a stable compound is not derived from an acid at all. Inverse sodium hydride[1], described as a section of this Wikipedia article, contains natride ion ($\ce{Na^-}$) combined with a protonated $3^6$-adamazane ($\ce{C18H36N4 * H+}$). The hydrogen ion represented by $\ce{H+}$ is not covalently bonded to any one nitrogen atom, but tetrahedrally coordinated to all four of them like the central metal ion in a typical chelate complex. The hydrogen-nitrogen interactions must be a combination of a delocalized covalent bond to the $1s$ orbital of the hydrogen with electrostatic attraction.
Reference

Redko, M. Y.; Vlassa, M.; Jackson, J. E.; Misiolek, A. W.; Huang, R. H.; Dye, J. L. "Inverse Sodium Hydride": A Crystalline Salt that Contains $\ce{H+}$ and $\ce{Na−}$. J. Am. Chem. Soc. 2002, 124, 5928–5929. doi: 10.1021/ja025655+. PMID 12022811.

Note, one of the nice details of mhchem is the following: instead of a lengthy $\ce{C_{6}H_{12}O_{6}}$ subscripts in glucose's Hill formula simply type as $\ce{C6H12O6}$ to yield $\ce{C6H12O6}$.

You may be right, but in computer programming class I learned to put extra parentheses to make things doubly clear (e.g. (3×6)+5 versus 3×6+5). Same principle here.

This requires three additional characters per subscript – in a simple Hill formula, in absence of any charges? My preference is to use curly braces in expressions (test page) of higher complexity. On the other hand, yes, especially in a hurry late arvo, «explicit is better than implicit» (PEP20 in Zen of Python) has its point (an echo of implicit none in Fortran?). Even calculators can be a cause errors here if not used well (see photo in Wikipedia's order of operations, for instance).


text:
Ionic bond formation
When two separate pieces of sodium chloride crystals are physically put together, no wonder that they will not fuse with each other.
But I don’t understand why. Ionic bonds are described to be non-directional and bonds should be formed between the sodium ions and chloride ions in the two pieces of sodium chloride crystals.
Why must I melt the two pieces of crystals to fuse them together/ evaporate a solution in which the two pieces of crystals have dissolved?

If the surfaces are scrupulously clean, atomically smooth and placed in a vacuum, then two separate crystals will probably fuse after being contacted. It'd be a form of cold welding.

@ Nicolau Saker Neto Why is the fusion impossible at room temperature and pressure?

Consider surface adsorption of molecules of gases and bad alignment of surface ions of the touching crystals.

Because you need to place the ions less than a nanometer apart, and in ordinary conditions there'd be jagged edges, grime and air in the way.


text:
Force contribution of image atom using periodic boundary conditions
This is maybe a stupid question but lead to a discussion between me an my colleagues. If we use periodic boundary conditions and an atom has itself as neighbor do we need special care of this contribution to the Force?
In all cases the Energy depends on the distance $r_{ij}$ between two atoms $i$ and $j$. In case of the periodic image the distance will be the position of the image translated by the Lattice vector and thus just $L$. Since the movement of the atom and its image is correlated the distance between them will never change and thus the derivative $\partial r_{ij}/\partial x_i$ vanishes and this also leads to zero Force. However, I think there is a mistake in our assumption. I feel uncomfortable to just setting this Force to zero. The Force also enters the virial and thus the pressure and stress. If the contribution would be zero there would be no stress in a single atom system with PBCs.
I just hope that some one can clear my thoughts on this matter.

The force of an image atom on the original is not zero - it is a different atom. Think about a physical crystal, an atom in one cell is different from one in another cell even if they are related by a translation which is a lattice vector. Whether to have such a small cell that such forces are significant is another matter all together and depends upon what you are modelling. I suspect if the correlated motions are worrying you then you need a bigger simulation cell, one of sufficient size that the term you are worrying about are negligible.

To put it another way if you want one atom to move in one way and another translationally symmetry related one to move in a different one, then your cell is too small; symmetry constraints due to PBC make such a calculation impossible.

Might want to consider moving this post to matter modeling SE.

Thank you @IanBush I also think that the image atom should be treated as a new independent atom, but I'm still a bit struggling with it mathematically. My colleagues made the valid point that the movement will always be correlated. If I account for this in the distance $\vec{r}_{ij}=\vec{r}_i-\vec{r}_i-n\vec{L}$ then the gradient of this distance w.r.t. coordinate change vanishes. Probably one should just not account for the correlated movement, but I wish I had a more clear mathematical proof Buck Thorn How can I move the post?


text:
Self heating of Hydrogen
I have read that hydrogen can undergo self-heating, through a catalytic process, without the presence of oxygen or air. Is that true?
I understand that this is not "burning" or "ignition", since this occurs without oxygen, and I don't think this is due to decomposition or rearrangement, as happens to $\ce{NI3}$ or 1,2,3-trinitroxypropane (nitroglycerine), since there's just molecular hydrogen, $\ce{H2}$.
Could someone explain if that occurs, and if so, where does the heat (energy) come from?

In case of hydrogen, it is not possible without oxidant, that may not be oxygen. Unless you mean nuclear reactions, like catalytic CNO cycle in massive stars.

Jack, I've taken some liberties in editing your question to make it more meaningful, i.e., changing "ignition" to "heating". Please let me know if my understanding is correct.

Perhaps you're thinking of para-ortho conversion of hydrogen, which is certainly not combustion, but does release a bit of heat.
Nuclei (e.e., protons) in a hydrogen molecule can have parallel or anti-parallel spins, called spin isomers. Weitzel at al state, "Freshly liquefied hydrogen... consists of a 3-to-1 ortho-para mixture." When hydrogen is liquified, the para spin isomer is energetically preferred. Slowly, the protons reorient, releasing about 1.5 kJ/mol as heat -- which causes hydrogen to boil off, even though the container is insulated from outside heat sources.
Clearly, it's not desirable for rocket fuel, for example, to be lost due to that energy release. During production of liquid hydrogen, catalysts, such as $\ce{Cr2O3}$, $\ce{Fe2O3}$ and $\ce{Fe3O4}$, can speed up the transition, so that the work of the extra "cooling" is done during production, rather than in storage.


text:
Can Seaweed Be a Thermoplastic?
I was on the Internet, and I saw this post. This post shows a photo of thermoplastic seaweed, which I thought was impossible. `

BTW, if you're wondering what the difference is between a thermoset and a thermoplastic is, read this.
I am well aware of seaweed-based bioplastics from sources like agar. To my knowledge agar creates a thermoset plastic. I think this because of this study I found online. One of the figures in the study shows the crosslinking of the agarose molecules with hydrogen bonds, effectively creating a thermoset plastic, at least to my knowledge.

I know that there is such a thing called Thermoplastic Starch, or TPS. This is made with amylose and amylopectin, with concentrations of 15-25% and 75-85%, respectively. Agar powder, on the other hand, is comprised of agarose and agaropectin, with concentrations of 65% and 35%, respectively. I believe you can make a thermoplastic out of starch because the large amounts of amylopectin, which is a highly branched molecule, limits the intermolecular bonding between the starch molecules. My source for this claim is from Quora (yes, I know it's not the most trustworthy site).
So, I guess I'm just wondering whether I'm melting the bioplastic wrong, or whether thermoplastic agar is fundamentally possible to create, or if the pictured thermoplastic seaweed is using some dark magic.
Any help would be appreciated! I am also open to other ideas about creating bioplastic from seaweed, too.

Sea weed=plant is not agar=polysacharide gel.


text:
Convert daltons (Da) to molar mass (g/mol)
This is probably a stupid question...
In converting Da to molar mass (g/mol),
Since 1 Da = 1 g/mol, is it logical that 64,776 Da = 64,776 g/mol or am I missing an intermediate calculation?
Thank you!

For the most part, you are right. There is a complication though, when converting masses obtained from mass spectrometry (which is the usual context of the unit Dalton) to molar mass, and that is isotopic abundance. For a few elements, there is only one naturally ocurring isotope, but most are a mixture of isotopes with different weights. This has to be accounted for when calculating molar mass, meaning you need the molecular formula.

Thank you. This is for HbA1c (glycated hemoglobin). I am not sure the molecular formula, but I have been told this: "Looking back to the definition of the measurand reported by the IFCC Working group on the standardization of HbA1c (i.e. HbA1c - N-(1-deoxyfructosyl)-haemoglobin-ß-chain) and considering that HbA is composed by two alpha chains, two beta chains and four hemes (MW 64,452 da), the MW of HbA1c can be estimated to be 64,776 daltons (i.e. Mw of HbA - mw of 2 water molecules + mw of 2 molecules of glucose)." If you could help me interpret how this translates to the molecular formula?


text:
Methanisation in the Haber Porcess
I've been reading up on the Haber process lately.
In one such industrial case of ammonia synthesis, hydrogen is created through steam reforming, which creates some carbon monoxide. As the reaction continues, there exists both carbon monoxide and carbon dioxide.
In the final reaction chamber before the actual synthesis chamber, the Carbon monoxide and carbon dioxide are removed through methanisation - so it the majority (due to equilibrium) turns into water and methane.
My question is: how is the water and methane specifically removed without disturbing the remaining hydrogen and nitrogen?


text:
What is the temperature at which sodium hydroxide decomposes at? And how would you collect the condensation?
I don't think this is an extremely tough question, but I can't really find that much information online on this reaction. I found a video about the decomposition of calcium hydroxide, though I'm not sure how much the decomposition temperature differs between these two metal hydroxides. I also saw that the person was using a blow torch (I'm assuming) to decompose the calcium hydroxide, but I was told that test tubes weren't to be used under strong flames.

Forgot to mention that I'm also looking for a way to collect the water vapour that comes from the reaction, or prove that there was H2O formed from the reaction, e.g. using cobalt chloride indicators.

We see from the currently posted answer this won't work with sodium hydroxide, but why were you looking at that hydroxide? Telling that may neutralize the downvote and close vote I am seeing, and perhaps allow us to identify alternatives that will work for you.

NaOH has reportedly the boiling point 1388 deg C, so the condensate would be NaOH again.

Probably never, unless perhaps you throw the stuff onto an incandescent light bulb filament (are there still those?) or the surface of the Sun.
In thermodynamic terms, hydroxides decompose on heating because the steam that's generated is a favorable entropy effect when the hydroxide is still condensed and thus more "ordered". But alkali metal hydroxides present a problem. Since the ions are all singly charged and a strong enthalpy effect from the exothermic reaction between the (strongly basic) oxide and water must be overcome, the alkali metal hydroxides may actually boil before they decompose. Then you lose the entropy advantage.
Let's look at the numbers, as rendered by Wikipedia as of November 2023. Lithium hydroxide melts at 462°C and then decomposes at 924°C while still liquid, so with that alkali all is well for decomposition thermodynamics (albeit at a higher temperature than most other hydroxides). Sodium hydroxide, which is a stronger base itself and forms from a more strongly basic oxide, melts at 323°C and boils at 1388°C — with no reported decomposition before boiling, at least from Wikipedia. Not surprisingly, the same result carries over to the heavier alkali hydroxides.
And so, it appears that you can't realistically expect sodium hydroxide (or heavier alkali hydroxides) to decompose thermally at all, at least not by the mechanism that works for lithium hydroxide and for hydroxides outside the alkali group.


text:
NaCl Unit Cell - Chloride Ions on the edges or corners?
When I search online for an NaCl unit cell I often find a significant number of images, such as the one I have attached to this post, showing the chloride ions in the "edge" and "body centered" position. I also find many where the chloride ions are on the "corner" and "face-centered" position.
It made me realize that depending on where you cut the cube you could technically illustrate either configuration with accuracy, however, I'm unaware if there is a specific standard for illustrating the NaCl unit cell.
I'm hoping someone on here can chime in and confirm the accepted position of the chloride ion (or lack thereof) and why this designation is the accepted position if such an accepted position exists.
Thank you to everyone who responds with their thoughts on the matter.

For a real particle/crystal it could be either ion on the outside corners. Also I wouldn't expect a real particle/crystal to have absolutely flat faces.

The above picture is not the conventional unit cell for NaCl - it is 8 unit cells.See e.g. http://lampx.tugraz.at/~hadley/ss1/crystalstructure/structures/nacl/nacl.php and read the text associated with the picture

The commonly accepted unit cell is a face center cubic (FCC) chloride structure with all octahedral sites filled with sodium cations. Typically, we construct the framework of the unit cell with the larger ion or, in the case of unit cells with formulas having different stoichiometric quantities, the ion with the charge of greater magnitude.
When using the radius ratio rule, remember that the ratio of radii has equal a quantity less than 1, with the ion whose radius is represented by the numerator is the ion filling the sites in the unit cell scaffold constructed by the ion whose radius is represented by the denominator.
In NaCl, the radii for Na+ and Cl- are 102pm and 175pm, respectively. This gives a cation:anion ratio of 102/175, which equals 0.58, corresponding to octahedral holes.
As a result, we have (1/8)(8)+(1/2)(6) = 4 Cl- ions (each corner is 1/8th in the unit cell, and the centers of the faces have 1/2 in the unit cell) and (1/4)(12)+1 = 4 Na+ ions per unit cell (each edge has 1/4th in the unit cell, and there is one full Na+ ion in the center of the unit cell), for a formula of Na4Cl4 per unit cell.

Very insightful. I am too new to the site so the system will not allow me to up vote but I hit the checkmark for whatever that may be worth. Thank you.

NaCl crystallizes in the Fm-3m space group (where the underlying lattice is F, namely face-centered).
Na atoms are usually located (in the structural description) at the 4a Wyckoff site (coordinates: 0,0,0; 0,1/2,1/2; 1/2,0,1/2; 1/2,1/2,0); namely, their positions correspond to the lattice points of the cubic face-centered lattice.
Consequently, Cl atoms are usually located (in the structural description) at the 4b Wyckoff site (coordinates: 1/2,1/2,1/2; 1/2,0,0; 0,1/2,0; 0,0,1/2); namely, their positions correspond to the octahedral voids of the cubic face-centered lattice.
Nonetheless, the 4a and 4b sites have exactly the same site symmetry, so you can interchange the atomic positions. For this reason, it is possible to describe the crystal structure of NaCl in two ways:
(1)
Na @ 4a
Cl @ 4b
=> Na atoms located at the cubic face-centered lattice with Cl atoms in the octahedral voids
(2)
Na @ 4b
Cl @ 4a
=> Cl atoms located at the cubic face-centered lattice with Na atoms in the octahedral voids
These descriptions are perfectly equivalent, it depends on you and your scope what to choose.

Very insightful. I am too new to the site so the system will not allow me to up vote but I hit the checkmark for whatever that may be worth. Thank you.


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What chemical makes the ‘smoky’ flavour in smoked paprika?
I really detest ‘smoky’ flavours eg smoked paprika. It tastes like fumes to me. What chemical/ substance do these contain that causes this please

How should it taste like? BTW, there will be dozens to hundreds different chemicals, contributing to the smoky smell/taste.

Maybe you don’t like phenols.

Could be other substances like ketones or carboxylic acids, or a combination.

Thank you. I wasn’t assuming anything. I know nothing about it, that’s why I asked the question. I don’t have a scientific background hence I came here to find out. Very interesting answers. Thank you all

The National Institute of Health has provided a good list of chemicals found in "smoke" flavoring. Here are just a few:
Phenolic compounds
  Phenol
  2-methylphenol
  4-methylphenol
  2-methoxyphenol

Aldehydes and ketones
  Propanal
  1-hydroxy-2-butanone
  2-methyl-2-cyclopenten-1-one
  1,2-cyclopentanedione
  2,5-hexanedione

Furans and pyrans
  Furan-3-carbaldehyde
  Furan-2-carbaldehyde
  Tetrahydro-2,5-dimethoxyfuran
  1-(2-furanyl)-ethanone

and the list continues ad nauseam...
As to which of these you find objectionable, you'd have to get samples and test-taste (well, test-smell) each. Or you could arrange to be in the lab when these are fractionated, sniffing the output stream of a gas chromatograph, and have the most bothersome analyzed.
De gustibus non est disputandum.

Very interesting. Thank you.

@Jane If you like this answer, please consider accepting it by clicking on the checkmark.


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VBT and MOT at the same time!
In the SNAr reaction, it is said that the electron withdrawing groups should be at the ortho or para position for the effective electron delocalization to take place, and we draw Kekule's structure to justify it.
But can we justify it on the basis of Molecular Orbital Theory?

Yes. VB and MOT are different developments of the same theory, in the limit they are identical. So yes, if it can be justified in VBT it can also be justified in MOT.  I leave the details to the reader.


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Collapse of Butler-Volmer equation into the Nernst equation at equilibrium potential
How does the Butler-Volmer equation collapse into the Nernst equation when applying the equilibrium potential current? I don't know where to start, and this source I used (https://www.sciencedirect.com/topics/engineering/butler-volmer-equation#:~:text=It%20agrees%20with%20the%20Nernst,true%20under%20high%2Dresistance%20conditions.) uses Bard and Faulkner as given, but I would like to know how to derive it properly. Thanks in advance
Edit: I have to do the first task:

Maybe you are confused with the concept of equilibrium potential current, which seems strange. BV equation collapses to Nernst equation when the current density is zero, i.e., is thermodynamically consistent since it is an extension of transition theory to electrochemical reactions. An analogy would be recovering the law of mass action ($K = \prod_j a_j^{\nu_j}$) when setting the rate $r = 0$ for a chemically reversible reaction in a homogeneous phase.
When this is the case, the potential of the electrode with respect to the bulk solution is named the equilibrium potential. The math is immediate from the BV expression and setting $j = 0$
\begin{align}
  j &= j_0\left\{\exp\left(-\frac{\alpha F \eta}{RT}\right) -
          \exp\left[\frac{(1 - \alpha) F \eta}{RT}\right] \right\} \\
  0 &= j_0\left\{\exp\left(-\frac{\alpha F \eta}{RT}\right) -
          \exp\left[\frac{(1 - \alpha) F \eta}{RT}\right] \right\} \\
  0 &= \exp\left(-\frac{\alpha F \eta}{RT}\right) -
          \exp\left[\frac{(1 - \alpha) F \eta}{RT}\right] \\
  \exp\left[\frac{(1 - \alpha) F \eta}{RT}\right]
   &= \exp\left(-\frac{\alpha F \eta}{RT}\right) \\
  \exp\left(\frac{F \eta}{RT}\right) &= 1 \\
  \frac{F \eta}{RT} &= 0 \\
  \eta &= 0 \\
  E - E^\text{eq} &= 0 \rightarrow
  \boxed{E = E^\text{eq} = E^\circ - \frac{RT}{nF} \ln
  \left(\prod_j a_j^{\nu_{j}}\right)} \tag{1}  \\
\end{align}
Although we cannot directly measure the potential of a conducting solid phase, nor the potential of a liquid phase, the meaning of Eq. (1) is as follows. In an electrochemical cell in which no current is circulating between the electrodes, the electrode potential $E = \phi_\text{electrode} - \phi_\text{solution}$ in any of the electrodes, is only a function of the activities of the species participating in that electrochemical reaction in the bulk electrolyte solution.
I suggest reading chapter 2 from the book you cited, but if you need more precisions, feel free to comment and I will add more information.

Edit
Let us study the academic electrochemical reaction
$$ \ce{\nu_0O + ne^- <=> \nu_RR} \tag{2} $$
We employ the current-potential equation, equivalent to the BV equation, but before making manipulations in order to put in in terms of the overpotential and the exchange current density
\begin{equation}
  j = Fk^\circ\left\{
  a_O^{\nu_O}(0,t)\exp\left[-\frac{\alpha nF(E - E^\circ)}{RT}\right] -
  a_R^{\nu_R}(0,t)\exp\left[\frac{(1 - \alpha) nF(E - E^\circ)}{RT}\right]
  \right\} \tag{3}
\end{equation}
When the electrode potential is the equilibrium potential, i.e. $E = E^\text{eq}$ and $j = 0$, from Eq. (3) we obtain
\begin{gather}
  0 = Fk^\circ\left\{
  a_O^{\nu_O}(0,t)\exp\left[-\frac{\alpha nF(E^\text{eq} - E^\circ)}
  {RT}\right] -
  a_R^{\nu_R}(0,t)\exp\left[\frac{(1 - \alpha)nF(E^\text{eq} - E^\circ)}
  {RT}\right] \right\} \\
  0 = \color{blue}{Fk^\circ
  a_O^{\nu_O}(0,t)\exp\left[-\frac{\alpha nF(E^\text{eq} - E^\circ)}
  {RT}\right]} - \color{red}{Fk^\circ
  a_R^{\nu_R}(0,t)\exp\left[\frac{(1 - \alpha)nF(E^\text{eq} - E^\circ)}
  {RT}\right]} \\
  0 =
  a_O^{\nu_O}(0,t)\exp\left[-\frac{\alpha nF(E^\text{eq} - E^\circ)}
  {RT}\right] -
  a_R^{\nu_R}(0,t)\exp\left[\frac{(1 - \alpha)nF(E^\text{eq} - E^\circ)}
  {RT}\right] \\
  a_O^{\nu_O}(0,t)\exp\left[-\frac{\alpha nF(E^\text{eq} - E^\circ)}
  {RT}\right] =
  a_R^{\nu_R}(0,t)\exp\left[\frac{(1 - \alpha)nF(E^\text{eq} - E^\circ)}
  {RT}\right] \\
  \dfrac{a_O^{\nu_O}(0,t)}{a_R^{\nu_R}(0,t)} =
  \dfrac{\exp\left[\dfrac{(1 - \alpha)nF(E^\text{eq} - E^\circ)}
  {RT}\right]}
  {\exp\left[-\dfrac{\alpha nF(E^\text{eq} - E^\circ)}{RT}\right]} \\
  \dfrac{a_O^{\nu_O}(0,t)}{a_R^{\nu_R}(0,t)} =
  \exp\left[\dfrac{nF(E^\text{eq} - E^\circ)}{RT}\right] \\
  \ln\left[\dfrac{a_O^{\nu_O}(0,t)}{a_R^{\nu_R}(0,t)}\right] =
  \frac{nF(E^\text{eq} - E^\circ)}{RT} \\
  \frac{nF(E^\text{eq} - E^\circ)}{RT} =
  \ln\left[\dfrac{a_O^{\nu_O}(0,t)}{a_R^{\nu_R}(0,t)}\right] \\
  E^\text{eq} = E^\circ\ + \frac{RT}{nF}
  \ln\left[\dfrac{a_O^{\nu_O}(0,t)}{a_R^{\nu_R}(0,t)}\right] \rightarrow
  \boxed{E^\text{eq} = E^\circ\ - \frac{RT}{nF}
  \ln\left[\dfrac{a_R^{\nu_R}(0,t)}{a_O^{\nu_O}(0,t)}\right]} \tag{4}
\end{gather}

Eq. (3) comes by studying a free energy diagram vs the reaction coordinate. You can relate the changes in free energy with changes in the potential $E$. You may want to see chapter 3 from Bard and Faulkner.
In said book you will see that instead the standard reduction potential $E^\circ$, Bard writes the formal potential $E^{0'}$. This is because, almost in all cases, we don't know the activity coefficients. So the $E^{0'}$ incorporates its dependence (somewhat loosely), so we can write Eq. (3) with concentrations instead of activities. You can write it with activity coefficients as I did, to have a more general form.

Nice explanation. Is there a way of deriving the Nernst equation from Butler-Volmer directly instead of knowing that E=Nernst equation?  For example, setting E=E(eq) and going from here? That's what my sheet asks me to do

@Mäßige Hi!I think this is the way of deriving it. I imagine you know that the definition of the overpotential is $\eta \equiv E - E^\text{eq}$. So if the overpotential is zero, then the electrode potential reduces to the electrode potential dictated by the Nernst equation. However, maybe I am mistaking your question. If you can upload a photo of the exercise in the comments, I can see what might be going on.

Wait I think you might be right again. Wait let me upload it as an edit

Woww that's a very nice derivation thanks a lot :D

One more question; why is ΔE=E(eq)-E(0) necessarily? Why the standard reduction potential, I thought the Butler-Volmer equation applied to any electrode, i.e. the redox pair not necessarily at standard conditions, just how the potential differs from equilibrium potential of the redox pair (which does not need to be at standard conditions)?

@Mäßige Hello! You mean why "$E - E^\circ$" appears in both exponentials in Eq. (3)?

I mean, why is the overpotential defined as $E-E^{0}$? From what I remember, the Butler-Volmer-equation describes, how the real potential at the electrode (when a current flows), is different to when no current flows (open-circuit-potential). An example; a silver electrode in 0.1 M silver nitrate has an OCP of 0.741 V (which is $E_{eq}$). Now assume we want to draw a current of 50 mA, the potential on this 0.1 M $Ag/AgNO_{3}$ will then be somewhat higher (which we call E), and then ΔE=E-$E_{eq}$. Or am I wrong?

Or equivalently, when we apply a bigger potential on this electrode than the OCP, a current flows. But I don't get why we incorporate the standard electrode potential $E^{0}$ in this.

@Mäßige Hello! Eq. (3) is equivalent to BV equation, as I said in the post. It does not imply a definition of the overpotential as $E - E^\circ$, rather it is an equation some steps prior to arriving to BV equation. To convince yourself, I invite you to do the following:

Both parts of the new colored equation I added is $j_0$.
Divide Eq. (3) by $j_0$. On the LHS just put the symbol $j_0$. On the RHS divide with the blue part or the red part, accordingly to both terms.
See that you arrive to BV equation.

If I wasn't clear, comment again.

Hmm I don't quite get how one would arrive to the BV equation, maybe I just don't see it, where in Bard and Faulker is it?


text:
growing flawless NaCl crystals
Simple experiments on the internet show how to grow salt (NaCl) crystals, but the crystals are small and flawed. Several companies sell flawless optics, like windows, made of NaCl, which are presumably cut and polished from large crystals. Rectangular windows of size 50mm x 25mm x 2mm are available at fairly low cost. Where would I find the procedure for growing such large, flawless NaCl crystals?

I would guess that salt is first dried, then pressed into a disc under a vacuum. I highly doubt that the commercial windows are made from a single crystal.

I remember having read once somewhere  that adding urea to the salt solution helps growing nice big crystals of NaCl.

First, you need a good source of comparatively pure $\ce{NaCl}$; table salt won't do. Not only are there accidental impurities, such as salts of other alkali metals, but anticaking agents and iodides often are added intentionally, and the anticaking agents are intended to reduce crystallization.
Second, if you're growing crystals from water solution,

It must be done slowly, over weeks or months,
Water must be allowed to get out without dust getting in, and
The container must be undisturbed, free of vibration and temperature change or gradients.

However, for (near) perfect crystals, don't grow them from solution, but rather from a melt under controlled conditions, such as is done in the Czochralski process and in the similar Bridgman–Stockbarger technique.
While those techniques might be beyond the average home chemist's resources for crystallizing $\ce{NaCl}$, it's not hard to get good $\ce{Bi}$ crystals by the similar process of withdrawing the seed from the melt by hand, as in this video. Of course, they have hopper crystal defects, as could $\ce{NaCl}$.
See Crystals and Crystal Growing by Morrison and Holden for far more information.

To complement the answer (+1), about NaCl windows for Lasers, there is the conference proceeding «Polytran NaCl Windows For LASL Antares $\ce{CO2}$ Laser System» by V. E. Straughan at Los Alamos Conference on Optics '79 (doi 10.1117/12.957732) which outlines the state of the art already available then. This basically is: melting (ca $\pu{800 C}$) and slow cooling of NaCl (though not the household commercial one) to obtain ingots of 18" ($\approx \pu{45 cm}$) diameter by Stockbarger technique, cutting, polishing, selling. And by now, we're 4 decades later.