text:
How do I express concentration of a substance at the interface between two membranes of different diffusivities and lengths at steady state?

I am lost on this question from my textbook:
"A solute diffuses across two membranes, A and B. The membranes' thickness are $\Delta x_A$ and $\Delta x_B$ respectively. The membranes' diffusivities are $D_A$ and $D_B$ respectively. The concentration of the solute decreases linearly from $C_A$ at the left surface of A to $C_i$ at interface, then to $C_B$ at the right surface of B. Given a steady state diffusion process, derive an expression for $C_i$ in the terms of the other parameters of this system."
The answer sheet as provided in the textbook indicates that the correct solution is:
$$C_i = {{D_A C_A\over \Delta x_A}+{D_B C_B\over \Delta x_B}\over {D_A + D_B\over \Delta x_A + \Delta x_B}}$$
I tried to solve this by making the following equation:
$${C_A - C_i\over {\Delta x_A \over D_A}} = {C_i - C_B\over {\Delta x_B \over D_B}}$$
This was wrong because it can't be re-arranged into the correct equation as given by the textbook. My prof hinted that since the system is at steady state, the rate of diffusion through A is equal that of B. How do I solve this?

The answer sheet as provided in the textbook indicates that the correct solution is:
$$C_i = {{D_\text{A} C_\text{A}\over \Delta x_\text{A}}+{D_\text{B} C_\text{B}\over \Delta x_\text{B}}\over {D_\text{A} + D_\text{B}\over \Delta x_\text{A} + \Delta x_\text{B}}}$$

The textbook solution is wrong. Let test it by a trivial example, using numerical values $D_\text{A}=D_\text{B}=2$, $Δx_\text{A}=Δx_\text{B}=1$ for 2 identical membranes.
Then, $C_i= \frac{2C_\text{A} + 2 C_\text{B}}{2}=C_\text{A} + C_\text{B}$.
This is obviously wrong and the textbook provides a wrong equation.

Your initial equation for equal diffusion rates for the steady state is correct.

$${C_\text{A} - C_i\over {\Delta x_\text{A} \over D_\text{A}}} = {C_i - C_\text{B}\over {\Delta x_\text{B} \over D_\text{B}}}$$

If we express $C_i = f(C_\text{A},C_\text{B})$:
$$C_\text{A}\frac{\Delta x_\text{B}}{D_\text{B}} - C_i\frac{\Delta x_\text{B}}{D_\text{B}} = C_i\frac{\Delta x_\text{A}}{ D_\text{A}} -  C_\text{B}\frac{\Delta x_\text{A}}{ D_\text{A}} $$
we get:
$$ C_i  =  \frac{C_\text{A}\frac{\Delta x_\text{B}}{D_\text{B}} +  C_\text{B}\frac{\Delta x_\text{A}}{ D_\text{A}}}{\frac{\Delta x_\text{A}}{ D_\text{A}} + \frac{\Delta x_\text{B}}{D_\text{B}}}   $$
This fits the general expectation the result should be in the form of a weighted average:
$$z = \frac{ax+by}{a+b}$$
Using our small test for identical membranes:
$$ C_i  =  \frac{C_\text{A}\frac{1}{2} +  C_\text{B}\frac{1}{2}}{\frac{1}{2} + \frac{1}{2}}=\frac{C_\text{A} + C_\text{B}}{2}$$
what is obviously correct.

If we multiply the numerator and the denominator of the equation by   $\frac{D_\text{A}D_\text{B}}{\Delta x_\text{A}\Delta x_\text{B}}$, we get the alternative version of the equation. It is almost identical to the textbook solution, but it has the denominator the sum of the fractions and not the fraction of the sums.
$$ C_i  =  \frac{C_\text{A}\frac{D_\text{A}}{\Delta x_\text{A}} +  C_\text{B}\frac{ D_\text{B}}{\Delta x_\text{B}}}  {\frac{ D_\text{A}}{\Delta x_\text{A}} + \frac{D_\text{B}}{\Delta x_\text{B}}}   $$
Using our small test for identical membranes:
$$ C_i  =  \frac{2C_\text{A} +  2C_\text{B}}{2 + 2}=\frac{C_\text{A} + C_\text{B}}{2}$$
what is obviously correct.

Thanks! Just checked back with my groupmate in my module, my textbook is an older edition which has been rectified in newer ones.

The textbook equation would be correct, if they used in the denominator the sum of the fractions and not the fraction of the sums.


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Temperature change in isobaric processes with and without change in number of molecules
Hello I was thinking about two thermodynamics problems and I wanted to get some insights into how to solve them.
The first problem was: Suppose we have one mole of ideal gas under constant external pressure (1 atm) conditions, and let it conduct a reaction (for example a photoinduced isomerization) where there is no net change in number of molecules. This reaction for example has an enthalpy change which is not known. We can measure the total heat Q evolved due to the reaction (like in a constant pressure calorimeter).

So because there is no change of molecules this can be treated as a formal isobaric change (can it?). As enthalpy change is equal to heat Q in isobaric processes, we can say that the measured heat when reaction goes to completion is equal to the reaction enthalpy change, and therefore we can calculate the temperature change.

My doubt is: Is the work the system performs on its surroundings included in ΔH?
How can we determine the temperature change then if we have an reaction of the type:

Also at constant external pressure? Here, the particle number changes.

Is the work the system performs on its surroundings included in ΔH?    Remind yourself that the difference between enthalpy change  and internal energy change at constant pressure is the expansion work that the system does. ΔH=ΔU+pΔV.

When you measure the heat evolved in an isobaric  calorimeter, it is assumed that the process takes place nearly isothermally (for the reaction mixture) and the heat involved is equal to the standard change in enthalpy $\Delta H^0$.  Whether the total number of moles changes or not, $\Delta H^0$ includes the work done on the surroundings.

Yes I was aware that heat evolved will be equal to the change in standard enthalpy, at least in a mathematical sense, because for isobaric conditions ΔQ=ΔH, and ΔH=ΔU+pΔV. So in theory, we start recording ΔQ before and after reaction, and the difference in Q is then ΔH. But in the case of gaseous reactants, under isobaric conditions there is volume work. So during the reaction, the Q the reactants give off to our measurement apparatus gets smaller because of the volume work which removes energy from the reactants, so we need to measure Q and the volume work for determining ΔH or am I wrong?

Yes, you’re wrong.  The work is included in the $\Delta H$ and the Q.

Okay, I think I get it now. The ΔH=Q is bigger than ΔU by a factor of +pΔV. Suppose the Q is negative (for exothermic reaction). The consequence would be that the absolute value of ΔH is smaller than the absolute value of ΔU (as ΔU is negative, but +pΔV is positive). For example, we would get a value of -30 kJ/mol for ΔH, but ΔU would be -32 kJ/mol in reality right?

It depends on whether the number of moles between reactants and products changes.

Let’s use my first example where it doesn’t.

If it doesn't, then no work s done (for transformation between standard states) and $$Q=\Delta H^0=\Delta U^0$$

Why is no work done :/ I don't quite get it. Suppose we have one mole in the container and the gas does a reaction where mole number doesn't change (isobaric conditions). If the reaction is exothermic, the gas heats up and expands, doing work against the piston, in addition to the heat released to the calorimeter. Or can you tell me where I'm wrong? How does this reaction proceed

In a calorimeter measurement of heat of reaction, the amount of heat required to bring about the small change in temperature of reaction products is considered negligible compared to the amount of heat needed to raise the temperature of the surrounding water and calorimeter.

Is there a mathematical formulation for why dH=dU=Q?

For no change in the number of moles and constant temperature, $\Delta PV=0$

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Why is octane more volatile than water while having a higher boiling point?
Octane has a boiling point of 120 °C. Water has a boiling point of 100 °C. The definition of boiling point is, "the temperature which the liquid substance's saturated vapor pressure equals the atmospheric pressure". Volatile substances have higher saturated vapor pressure at a given temperature, than the lesser volatile substances. When temperature increased (heated), the saturated vapor pressure increases rapidly and non-linearly. So, the volatile substance's saturated vapor pressure reaches atmospheric pressure quicker than the lesser volatile substance, hence, reaching boiling pint at a lower temperature.
However, water which is a less volatile substance than octane (if you put two separate liquid spots from water and octane, octane evaporates faster) has a lower boiling point than octane at atmospheric pressure. Is there something I am missing? If my reasoning is wrong please correct me.

Maybe worth defining "volatile" more formally, eg https://www.epa.gov/indoor-air-quality-iaq/technical-overview-volatile-organic-compounds

One thing you are missing is that air contains water, but usually does not contain octane. So for water, the process is:
$$\ce{H2O(l) <=> H2O(g)}$$
and for octane, it is
$$\ce{C8H18(l) <=> C8H18(g)}$$
For octane, the partial pressure of octane is virtually zero except near the liquid surface. For water, it depends on the humidity in the room.
If you look up the vapor pressure at room temperature, you will find that water has a higher vapor pressure. At equilibrium, the partial pressure of water will be higher than that of octane. However, the kinetics determine how fast a drop will evaporate (and the water drop will not evaporate at all if the humidity is higher than 100%).

Does that mean we applied Le Chatelier's principle here in terms of concentration of water and concentration of octane?

No, the system remains out of equilibrium. The forward and reverse rates depend on the rate constant (which might be very different for the two systems) and the partial pressure at the interface (which depend on bulk partial pressure and transport kinetics, which in turn depend on the shape of the drop and the object the drop is on). Overall, it is plenty complicated.

@Karsten Is there something to do with steam distillation? Octane that we talk here is petrol ( which is a mixture).  As far as I know, octane is also mixed with alcohols to increase petrol's efficiency as a fuel. Alcohol being a polar substance, might be less soluble in octane, which is non polar ( I tried to find solubility of alcohol in octane, but couldn't find. so this is just an assumption.)  So as in steam distillation, they boil at low temperature. (therefore, evaporate more at room temperature.)

Good point with mentioning existing humidity, I have forgotten to mention it.

Higher than100%! Equilibrium is a dynamic process.

@donthababakka "Octane that we talk here is petrol" – No. Typical gasoline contains less than 1 % octane.

The are three major factors, affecting observed relative volatility, ordered by assumed importance:

The molar mass. The liquid with higher molar mass evaporates faster than a liquid with the same vapor pressure but lower molar mass. This has two reasons:

That is because for the given vapor pressure, the mass concentration    of vapor is proportional (as ideal gas approximation) to its molar  mass:  $\rho = \frac{pM}{RT}$
The higher molar mass leads to lower specific enthalpy of evaporation, so it need lower heat transfer for the same evaporation rate.: $ΔH_\text{evap,spec}=\frac{ΔH_\text{evap,molar}}{m}$

Another factor Karsten mentioned (and I as a former enlisted military meteorologist should remember to write) is water vapor presence, which slows down evaporation. Air already contains typically $\pu{20 to 100\%}$ of possible water vapor amount, what slows down water evaporation. That is not the case for other liquids.

The third factor is the molar evaporation enthalpy.
The higher is the molar evaporation enthalpy of the liquid, the faster is the vapor pressure decrease with temperature.

The typical other case is toluene with the boiling point $\pu{111 ^\circ C}$, but significantly more volatile than water.

Note the van't Hoff equation about the dependence of liquid vapor pressure on absolute temperature in the integral form:
$$\ln{\frac{p_2}{p_1}=\frac{ΔH_\mathrm{evap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})}$$

Substance
Molar evaporation enthalpy ($\pu{kJ/mol}$)
Molar mass ($\ce{g/mol}$)

Water
40.65
18.02

Toluene
36.08
92.14

n-Octane
39.1
114.23

I suspect the molar mass difference really is the biggest factor at work here. Assuming we're talking about the vapour phases as ideal gasses (which we are), the natural quantity to talk about is number of molecules. Two puddles of water and octane with the same volume (what "feels intuitively right" to compare) actually refers to amounts of molecules which differ by a factor of $\mathrm{\frac{114.23}{18.02}\frac{0.997}{0.703}=8.99}$ times - there is simply just much less octane to evaporate.

@NicolauSakerNeto I do agree, assuming the effect is on the order molar mass > vapor saturation > evaporation enthalpy. The mentioning order was not meant as order of importance. I sil reorder that.

Since molar enthalpies are similar it is the Number o molecules that is pertinent

@jimchmst Sure, I have let it there for completeness. I have originally thought there is bigger difference until I have compared the values. I know there isTrouton's_rule of about constant molar entropy of evaporation 95-88 J/k/mol, but I have thought there are deviations for highly polar liquids with hydrogen bonds like water.

The OP does not refer to volatility in terms of a kinetic effect (maybe a shortcoming with the q in not being specific): "Volatile substances have higher saturated vapor pressure at a given temperature". Therefore the existing concentration of water in the vapor should be irrelevant to whether it is more or less volatile.

Also, other posts refer to area or number of molecules. This is also a kinetic effect. I think your thermodynamic description is the right one.

I took another look at the answer and have trouble understanding this: "The liquid with higher molar mass evaporates faster than a liquid with the same vapor pressure but lower molar mass." The mass in the vapor is greater for a larger Mw, but the amount, as described by the molar density (or concentration), is the same at a given pressure.

@BuckThorn But our perception is based on volume and mass, not amount.

I don't see where perception comes in. The two drops being compared containing the same volume, and roughly same mass and density? Is that the idea? And, evaporation of a drop with the same surface area, ie volume, but bigger (heavier) molecules is expected to be faster?

@BuckThorn I mean, you see 2 comparable drops of water and n-octane and you do not say  "Look, the n-octane drop contains 8 times less molecules.".  You say "Look, the n-octane drop has similar volume and mass as the water one.". // I also assume that in pseudostatic scenario is not the determining factor the rate of bidirectional mass transfer between phases, but mass concentration of saturated vapor.

Ok, then a specific enthalpy of evaporation seems the right thing to compare, since intermolecular interactions (vdW specifically) strengthen with molecular size in absence of H bonds.

That too, it is quite high for water. Like 2 effects of 1 cause ( pM/RT and  ΔH_evap,spec) of the molecular mass.

It is a case of careful observation and awareness of the construction of matter. Water and octane are molecular so a given volume with similar densities [1 vs 0.8] have very different # of molecules. The molar heats of vaporization are similar ~10Kcal/mole, [I leave it to you to look up the exact data]. Approximately similar masses with the same available energy and similar vapor pressures the fewer molecules will evaporate faster. As the MW difference increases the differences will change. Compare kerosene or cetane.
Also drop shape and surface area are important when comparing rates. Heat transfer is important.

Firstly, there are 18 different isomers of octane, each having a unique boiling point.
The gasoline standard is that "100 octane" corresponds to pure iso-octane (2,2,4-trimethylpentane).
Iso-octane has a boiling point of 99°C.
Secondly, the mass-rate of evaporation, according to the Hertz-Knudsen equation, given everything else is equal, is proportional to the square root of molecular weight. Therefore, if the vapor pressure of water and iso-octane are approximately the same, the mass-rate of evaporation of octane should be greater.
See also problem 1b here.
Thirdly, since the density of octane is about 0.7 relative to water, if equal-volume drops are being evaporated, the rate of evaporation will be even greater than the mass-rate.

I don't see why this is downvoted. There are indeed many isomers of octane. 2,2,4-trimethylpentane does have a boiling point of 99C, lower than water (while some other isomers, particularly n-octane, have a lot higher boiling point.) We should be sure we are comparing the same substance. Also, the per-mass latent heat of octane is indeed much lower than water. (They have similar per-mol latent heat, but the molecular weight of octane is much higher. )

It looks like 4 ups despite You complicated issues. the same argument applies regardless of the isomer. also the last sentence is puzzling. Molecules determine what is happening.

@jimchmst Each isomer is unique; lower boiling point than water is different than higher.   The premise of the question is "if you put two separate liquid spots from water and octane, octane evaporates faster", which sounds like equal volume (as opposed to equal mass or equal moles),

The boiling point is when it starts making bubbles inside.

"Liquids may change to a vapor at temperatures below their boiling points through the process of evaporation. Evaporation is a surface phenomenon in which molecules located near the liquid's edge, not contained by enough liquid pressure on that side, escape into the surroundings as vapor. On the other hand, boiling is a process in which molecules anywhere in the liquid escape, resulting in the formation of vapor bubbles within the liquid."

"a liquid will evaporate until the surrounding air is saturated"

Evaporation is a type of vaporization that occurs on the surface of a liquid as it changes into the gas phase. High concentration of the evaporating substance in the surrounding gas significantly slows down evaporation, such as when humidity affects rate of evaporation of water.

All things are not equal.

The boiling point is an important property because [all things being equal, e.g., in a vacuum] it determines the speed of evaporation. Small amounts of low-boiling-point solvents like diethyl ether, dichloromethane, or acetone will evaporate in seconds at room temperature, while high-boiling-point solvents like water or dimethyl sulfoxide need higher temperatures, an air flow, or the application of vacuum for fast evaporation.

"at atmospheric pressure" - at STP, which is 0% humidity and does not naturally occur on planet Earth? What you're "missing" is having conducted the experiment in a vacuum chamber.

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How does a dry shampoo work?
From my chemistry knowledge, I know that soaps are made from fatty acids and some sort of base and this reaction also gives us $\ce{H_2O}$ so the grease in hair dissolves and gets washed away in water. What I don't understand is how dry shampoos can have a cleaning effect on dirt or grease and how they work.

They work on the assumption that grease and oils stick more tightly to the cleaner flakes than to hair. It's the same as using adhesive tape to remove label stickum left on a container. Oh, and the cleaner likely tries to cover any bad odor with an overpowering "perfume".


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Why does hypophosphorous acid oxidation occur in acidic media?

When one mole of $\ce{H3PO2}$ is treated with excess $\ce{I2}$ in acidic solution, one mole of $\ce{I2}$ is reduced; on making the solution alkaline, a second mole of $\ce{I2}$ is consumed.

I am supposed to find the reaction corresponding to the description.
Here is the answer given to me by my teacher:
$$\ce{H2O + H3PO2 + I2 -> H3PO3 + 2H+ + 2I-}$$
I do not understand why this occurs in acidic media, as $\ce{H+}$ is formed in the products.

By this logic, reactions producing water should not happen in water solutions either. But they do happen.

Similarly, many reactions ongoing in alkalic media produce $\ce{OH-}$.

You answer your own question! Hypo phosphorous acid H3PO2 is readily oxidized to phosphorous acid, H3PO3 [really H2HPO3] even in acidic solution [there probably is a limit]. Phosphorous acid is less readily oxidized to phosphoric acid so the acid present or formed inhibits the reaction. Removing the acid generates the anion, the anion is more readily oxidized. You must write the Second equation[s a neutralization and oxidation] to see this.
This is a general trend to oxyanions such as MnO4-, CrO4= and Cr2O7=, NO3- even sulfate and phosphate. In acidic solutions they are [strong] oxidizing agents; In basic solutions they are much weaker and can be formed from their reduced counterparts by a sufficiently strong oxidizer such as O2 or peroxide or in the case of phosphite iodine, I2.

Using = as 2 negative charges is the invention I have not seen yet.

someone has to be innovative.

You can also replace the usual $\ce{PO4^3-}$   by Unicode U+2261 PO4≡ .  :-P  Who would use that boring mhchem anyway?


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Why does metal atom get an effective potential difference in voltaic cells
Why is it that metal atoms , experience this potential difference when immersed only in their salt solutions. When they are in contact in real life we cannot see current conducting through it.

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

Note that there is required fulfilling the basic prerequisite of search the site and general internet for the topic, to save effort of responders and avoid redundancy of internet content. // See also chemistry.stackexchange.com - resources-for-learning-chemistry

Seeing is believing? use all your senses. You are aware of lightning, static electricity and certainly avoid electrical sockets. Where do all these electrons come from! matter! Different metals even have a different taste

It is not that metal atoms get potential difference. Rather electrons experience different potentials.
Reactions $\ce{M(s) <=> n e- + M^{n+}(aq)}$ effectively serve as charge pumps, aiming for different metal potentials $E = E^{\circ}_{\ce{M/M^{n+}}} + \frac{RT}{nF}\ln{a(\ce{M^{n+}(aq)})}$
The potentials are in context of electrochemistry referring to the standard hydrogen electrode.
Kinetic reasons may cause troubles to reach the equilibrium potentials, sometimes even for open circuits.

At boundary of a metal and a solution of its salts, there are ongoing two chemical reaction in the opposite direction:

Metal dissolution $\ce{M(s) -> n e- + M^{n+}(aq)}$:

It is decreasing the metal potential by decreasing the metal net
charge.
The reaction rate is exponentially decreasing with the potential decreasing, establishing the negative loopback and slowing down.

Metal deposion $\ce{M^{n+}(aq) + n e- ->  M(s)}$:

It is increasing the metal potential by increasing the metal net
charge.
The reaction rate is exponentially decreasing with the potential increasing, establishing the negative loopback and slowing down.

As the result, the metal reaches such a potential both reactions have the same reaction rate and the net reaction rate is zero.
If the circuit is closed, the metal potentials are disturbed and there is ongoing the net reaction to reestablish the equilibrium potential.

So it is the electrons of the metal , that experience different potentials to possibly reduce the metal atom or oxidize the metal atom , and this value is intrinsic to the metal and when measured with reference to hydrogen electrode we tell its the Standard Electrode potential ?

It seems to me you may be very confused and may need to set some very basic foundations of electrochemistry, as each answer may easily lead to 2-3 questions.  Q/A sites are not best suited for that.  Consider reviewing the rich study material at libretexts.org, like Google search results for site:libretexts.org electrochemistry metal electrode potential // I will update my amswer, but consider online review, using this Q/A site rather just for filling the gaps.

@NaveenV - Remember this is for aqueous solutions. // Bare electrons aren't stable in aqueous solution so a single half cell reaction can't be measured. Hence the Standard hydrogen electrode as the reference point. // Some metals have more than one oxidation state. // The equation for E that Poutnik gives is the Nernst equation and is for an infinitesimal current flow. For a real battery the voltage drops as the current increases.

Thank you very much , for the answer as well as linking me to further material that is very kind of you , in my country the books just give the information and I find it hard to intuitively learn , but I will refer to the resources @Poutnik

@MaxW , is that due to the EMF of cell being a constant so , As $I$ increases the internal resistance contributes to the decrease in external potential difference

EMF of cells is not constant, it depends on multiple factors. It changes with local changes of cell chemical composition. Its decrease under load is complex function of many factors, like internal resistance, rate of diffusions, geometry, rate of chemical reactions at electrode/electrolyte interface.

oh okay thank you @Poutnik , one small clarification why is it that only when the circuit is closed metal potentials are distributed , is it because when these two metals are in contact , they feel a potential difference such that a reaction in anode (and cathode) dominates over the other ?

What do you mean by "metal potentials are distributed"? // Out of the cells, on the level of cell contacts and circuitry between them, the cell is nothing else but a power source, admitting with complicated electronic characteristics.


text:
Does -OTMS group transfer to a different carbon easily?
In this organic transformation question, (https://imgur.com/a/nr0LHUm) I understand that the intermediate below gets formed, but why does it lead to the final product?

Provide a better reference than a link to imgur; this link to an external reference may stop being functional at any time. Instead, add title and authors of a text book, or primary publication (in the later case, a link/doi is a useful addition). If this means to redraw some of the reaction sequence, but you don't have access to a sketcher like ChemDraw, there is a test page (with reduced functionality) to e.g. copy a screen photo. The drawing needn't be perfect, but share the relevant information in one spot here.

It is well established 1 that trialkylsilyl groups can undergo alcoholysis i.e. swapping between alcohol groups both inter and intramolecularly and Me3Si-O is particularly labile because of its small steric bulk. Once this happens the cyanohydrin anion that is formed is very labile with respect to loss of cyanide to form the ketone.


text:
Unusual Inconsistencies in Gibbs Free Energy Values at Specific Temperatures in Thermodynamic Simulation
I was using FactSageEdu software to simulate a metal with a composition of 67% titanium (Ti) and 33% aluminum (Al). During the simulation, I encountered an inconsistency in the calculated Gibbs free energy values (Gibbs free energy or G) at specific temperatures within the range of 933.45°C to 1166°C. These inconsistencies appear in the data as follows:
At 933.45°C, the Gibbs free energy was calculated as -3956630.1 J.
At 1166°C, the Gibbs free energy was calculated as -5565831.4 J.
These unexpected fluctuations in Gibbs free energy values within this temperature range are greater than typical variations and can have significant implications for our understanding of the thermodynamic properties and behavior of the material. Addressing and resolving this inconsistency is crucial for ensuring the accuracy and reliability of the simulation results and advancing our understanding of the system being studied.
I am seeking assistance and insights to understand the possible sources of this inconsistency and to identify potential solutions or adjustments in the simulation process to obtain more accurate Gibbs free energy values.

Well, the first thing to check is what formula(s) you used and compare to standard databases such as Dinsdale’s paper.

7 valid digits seem quite funny.

No, the issue does not appear to be solely related to the number of significant digits. I conducted a simulation using a composition of Al 53, Ti 46, and C 1, which adhered to the fundamental principles, and there were no apparent issues. Therefore, it is likely that the inconsistencies observed in the previous simulation are related to data entry or specific conditions associated with the Ti 67 and Al 33 composition.

I  have not said that significant digits are the reason of the issue. I  have wanted to say that presenting (internal using is OK) 7 significant digits is hardly justified, considering the accuracy of input data and rules of error propagation. // A kind of like presenting illustrative value 1.234567 +- 0.002


text:
How to treat enthalpy change and pressure in reactions involving a liquid and solid state
In physical chemistry, it is often stated that "at constant pressure, the change in enthalpy gives the amount of heat transferred into the system" and the justification for this claim is
$$dH = d(U+PV) = dU + PdV = \delta q + \delta w + PdV = \delta q$$
because $\delta w = -PdV$ (we are also assuming only reversible $p$-$V$ work). I understand this argument and why it must be true for a gas system in which the (well-defined) total pressure is being held constant (even if partial pressures of the individual components are changing). However, my textbook (Moore Physical Chemistry) makes use of this fact in discussing thermochemistry of reactions involving solids and liquids as well as gases. It is stated that in the lab, such reactions commonly occur under constant atmospheric pressure and therefore, we have $q_P = \Delta H$. I am unsure how to interpret this constant pressure condition for liquids and solids. For liquids, I agree that the pressure at the surface must equal the atmospheric pressure, but liquids generally have much greater pressure variation with depth than gases and we cannot, in general, assume that the pressure is relatively uniform. Furthermore, I don't see how we can even speak of the pressure of a solid as pressure is usually defined as force per unit area exerted by a fluid. I am hoping someone can explain how to carefully interpret pressure and enthalpy for reactions involving solids and liquids. Thank you in advance!

What matters is the external pressure, against which is done eventual expansion.

Because enthalpy must be a function of state, isn't the $P$ in the $H = U+PV$ referring to internal pressure rather than external pressure. I agree that the change in the quantity $U + P_\text{ext} V$ must equal the heat transferred but I don't see how we can translate this to a statement about enthalpy without properly defining the pressure within the system

There is usually considered simplified gravity less scenario with uniform pressure. For general scenario, W is integral p.dA.dl.


text:
How do I approach the problem if Cp is given as a function of volume?
Here is the problem I'm trying to solve:

A system undergoes Isothermal Expansion under the atmosphere. For the system, $ C_p = a + \frac{b}{V} \ $. The process is given by $ P_{ATM}, V_1, T_1 \rightarrow P_{ATM}, V_2, T_1 $. Calculate the change in internal energy

The first thing I notice is that this is clearly not an ideal gas. The first approach I could think of was to somehow use $ C_p $ to find $ \Delta H \ \text{(using} \ C_p = (\frac{\delta H}{\delta T})_P) \ $ and then use $ \Delta H = \Delta U + P \Delta V \ $ but I ran into a problem since $ C_p $ is given as a function of volume.
I realised later that this is equivalent to using the first law of thermodynamics: since pressure is constant, $ q = q_p = \int_{T_1}^{T_1} C_p \ dT$ which seems to just be zero, and so my answer $ \Delta U = q_p - P \Delta V = - P \Delta V $.
This is the best I could do, but I'm not convinced since I'm getting very confused and I think my concepts are extremely off. Any clarification/assistance is much appreciated.
Also, in general (since this was an exceptional case where my integral limits were the same) how do I evaluate the integral if $ C_p $ is given as a function of volume?
Thanks!

As it is isothermal AND isobaric expansion, the volume change is possible only due a chemical reaction or a phase change. And therefore there is unclear heat exchange, depending on the enthalpy change of reaction/phase change.

I think this made up task is badly defined.

Yes, it seems to violate the equation of state.

@Poutnik This was part of our homework assignment - our professor clarified that the system is not an ideal gas. From what I discussed with my peers, we couldn't really come up with anything substantial either. Again, the best we could think of doing is to just say that change in enthalpy is 0 (which is unconvincing (for the reasons mentioned above - I don't really know how to solve an integral from partial derivatives).
If you have any other advice, please let me know, thanks!


text:
is phenolate a strong or weak base? and why?
is phenolate a strong or weak base? I have read somewhere that it would be a weak base. However, since phenolate is the conjugate base of phenol, which is a weak acid, I would think that phenolate is a strong base. My general understanding is that strong conjugate base or acid is formed from a weak acid or base, respectively. Could someone help me?

"I have read somewhere that it would be a weak base." Please tell us where. Without that knowledge we can not know the context within which the comment was made, and this can not answer your question fully. Yes, phenolate would normal be consider a reasonably strong base due to the reason you give. But the original might have been a comparison with sodium hydride, or in a strange solvent, or something else - we need to know.

Evaluate it quantitatively. Take the phenol acidity constant and compute the phenolate basicity constant. Then make conclusion.

There are levels upon levels upon levels of different meanings for "strong" and "weak", and on top of that there are more levels. Are you big or small? A Stentor can grow to 2mm, which makes it a giant among ciliates. On the other hand, an adult Indian elephant is pretty small compared to Taj Mahal. Same thing here.

The old name for phenol is carbolic acid, does that sound like a weak acid?

Phenol is a stronger acid than water meaning its conjugate base is weaker than OH- but stronger than bases such as acetate, bicarbonate or ammonia? that have stronger conjugate acids. [ Do the math to be sure.] Your reasoning is fine but you forget that water is an acid too and acid strengths are relative to its strength.

Whether a base is considered strong or weak depends heavily on context. It is thus much better to consider the relative strength of acids and of bases rather than making absolute statements that one acid is strong or weak. This is best done by comparing the pKa values of acids (or conjugate acids in the case of bases). Phenol has a pKa of about 10.0 so it is a stronger acid than water (pKa 15.7), but weaker than acetic acid (pKa 4.8). You are correct that when comparing two acids, the stronger acid has the weaker conjugate base and vice versa. Therefore, we would expect phenolate (conjugate base of phenol) to be a stronger base than the acetate (conjugate base of acetic acid), but a weaker base than hydroxide (the conjugate base of water).


text:
Why does the second law of thermodynamics imply that processes with positive entropy change are spontaneous
My understanding of the second law is that for an isolated system, the entropy change accompanying any process is non-negative and is zero if and only if the process is reversible. I don't see why this implies that just because the entropy change of a process is positive, it must occur (even kinetically unfavorable processes occur after a long enough time). In other words, why is irreversible and entropically favorable synonymous with spontaneous?
Furthermore, it is often stated that the second law implies that entropy must be at a maximum for an isolated system in equilibrium. However I don't see how the statement that the entropy of a system is non-decreasing implies that entropy must increase to a maximum. Can't entropy remain constant at a non-maximal value as well?

Actually you can change entropy of a system even if you are providing heat in reversible way because entropy is a state function.

Spontaneous processes are irreversible, but irreversible processes need not to be spontaneous. E.g. if you suddenly press the piston of the gaseous system, it is irreversible, approximately adiabatic compression, but hardly spontaneous.

'why is irreversible and entropically favorable synonymous with spontaneous?'
If a system changes infinitely slowly it is capable of doing the maximum amount of work, and this process is, by definition, called 'reversible'.  The conditions for a reversible process are the same as those for being in a state of equilibrium. The change of internal energy $dU$ is the same however a change is made and this means that $dw_{rev} \lt dw_{irrev}$, because work done by the system is negative, consequently from the First Law, $dq_{rev} \gt  dq_{irrev}$. In a reversible change a maximum amount of work is done on the surroundings and a maximum amount of heat is absorbed from the surroundings.  A spontaneous/irreversible process is an observable process and absorbs less heat and does less work than a reversible one, i.e. a smaller change in heat along an irreversible path than a reversible one.
'...entropy must be at a maximum for an isolated system in equilibrium.'
The Second Law, like the First Law is based on experimental observation, for example that a perfect gas expands into a vacuum without doing any work, the observation of heat flow or that the work done by a chemical reaction is not accounted for by only summing enthalpies. The reason for these observations is that there is 'unavailable' energy present and this is quantified as $T\Delta S$ and the entropy, a state function,  is defined between two states $A,B$ as
$$\displaystyle \Delta S= S_B-S_A=\int_A^B\frac{dq_{rev}}{T}$$
As entropy is a state function it is independent of the path taken A to B and therefore entropy change is the same in going from A to B no matter how it happens. We always calculate the entropy using this eqn. which means that $q_{rev}$ has to be found somehow so that the integral $\int_A^Bdq_{rev}/T$ can be evaluated and this can only be done under reversible conditions. It is convenient to consider the derivative $dS=dq/T$ as a condition of equilibrium.
For a spontaneous i.e. observable process, $dq_{rev}\gt dq$  and so $\displaystyle dS \gt \frac{dq}{T}$
In an isolated system, one that cannot do work or absorb heat, $dq=0$ and the equilibrium conditions are $dS=0, S=constant$. For a spontaneous / observable change $dS\gt 0$. This means that for an isolated system spontaneous changes produce states of higher entropy until equilibrium is reached and thus entropy is a maximum at equilibrium. However, whenever a system can exchange heat or matter with its environment a decrease in entropy is entirely compatible with the Second Law.

Things happen for one of two reasons: Either there is a local decrease in energy or  increase in entropy. For this to be axiomatic processes must be broken into simple, hopefully elementary, steps. The Second Law simply states this; it does not imply anything. At equilibrium the energy and entropy effects are equal. For something to happen there are other necessities: there must be appropriate energy present, activation energy; there must be appropriate mixing; there must be a possible mechanism; and the system must be displaced from equilibrium. These are the "implications"; the Second Law simply states if Delta G is negative a reaction is possible. Notice this means displacement either direction from equilibrium.
Our environs are never in an isolated situation but in an energy flux of constantly increasing entropy, a steady state. When this changes there can be dire consequences; consider the dark side of the Moon or global warming for that matter. The Second Law explains it all.


text:
Under what conditions does ferric phosphate precipitate from solution?
I would like to know the conditions which cause ferric phosphate to precipitate from solution.
I am trying to culture iron bacteria, for which the culture media contain ferric citrate and phosphates, and my experience is that I persistently get a yellow precipitate which I believe is ferric phosphate.
Once the yellow precipitate has formed, the culture becomes inactive presumably because all the iron has been taken out of solution.
I assume that ferric phosphate precipitates above a certain pH, perhaps pH 7, but perhaps redox potential is also a factor, so I'm not sure, hence the question.
Update
My solutions typically contain the following things:
di-basic phosphate: 1.5 milli mole/litre
mono-basic phosphate: 0.2 milli mole/litre
ferric ammonium citrate: 25 micro mole/litre
(I always keep the pH around 7)
This didn't form a yellow precipitate when I first tried it, but does now.
Possibly because the iron precipitated out as ferric hydroxide so quickly that the ferric phosphate didn't have time to form.

What is the order of magnitude of the concentrations of Fe(III), of iron citrate and of phosphate ions ?

Under most conditions. Ferric phosphste is very insoluble. Citric complex with ferric ion may keep precipitation in check, kinetically and thermodynamically. The outcome very depends on ratio of all three components, together with pH. Provide details.

Is possible to use a citrate pH buffer instead of the phosphate one? or to provide more citric acid/citrate up to amount comparable to phosphate? As solubility of FePO4 is in order of 0.01 micromol/L, much less in presence of phosphate excess.

Thank you for this suggestion.  Since iron reducing bacteria can apparently use iron in the form of ferric oxide or hydroxide, I am now trying that, since it seems to avoid the phosphate problem.


text:
Bond length in square planar coordination complex of Platinum
(Q): Select the correct statements for $[Pt(NH_3)_2Cl_2]$ :
(a): Geometrical Isomerism
(b): Optical Isomerism
(c): All bond length are equal.
(d): Chelation effect is effect.
(e): Hydrogen bonding is present.
Pt is in +2 oxidation state in this compound. I know that that for 4d and 5d series transition metals, most ligands behave as Strong field ligands for them. Therefore the structure of this would a low spin, inner orbital complex or we can say a square planar structure is formed. Since it is of the type $[MA_2B_2]$. It will show cis trans geometrical isomerism. Hence option (a) turns out to be correct. Since it is square planar, it will show no optical isomerism. No chelation effect is present as there are no chelating ligands. Both amine and chlorido are monodentate. No hydrogen bonding is also present. The option I am struggling with is option (c). I know that bond lengths in square planar are equal only when the ligands are same. Here the ligands are not same and hence the  bond lengths are not equal. I want to know exactly which bonds are shorter and which longer and why. I have tried to search for it but could not find it. I need help here.

N and Cl: which is bigger?

The bond lengths in cisplatin are:
Pt-Cl: 2.298 Å to 2.325 Å
Pt-NH3: 2.04 Å
so clearly Pt-Cl bonds are longer
This maybe due to small size of Ammonia molecule which enables it to come closer to the central metal atom


text:
H-NMR Analysis of C15H24O

I was able to figure out that the structure of this compound was a benzene ring with 2 equivalent trimethyl groups to give the 18H singlet. Another non-equivalent methyl group to give the 3H singlet. The singlet (integration of 1) from a hydroxyl group and the singlet at approx 6.9 ppm from hydrogens attached to the benzene. However, when assembling the molecule, I found that my structure was incorrect as it was not symmetrical, why is it a requirement that this molecule has to be symmetrical?

Hydroxyl? I don't think so. As for symmetry, why, you need the trimethyl groups to be equivalent, right?

Looks very much like a Ar-OH hydroxyl to me.

On second thought, yes, you are right.

This is a good fit for 2,6-di-tert-butyl-4-methylphenol. See the 1H spectrum here


text:
What is more soluble, calcium sulfate or calcium sulfate dihydrate?
I think the dihydrate form is more soluble right? I expect because of the sulphate group attached to two molecules of water.
Also, do you know what's the solubility of each in g/L?
Thank you in advance for any explanation!

According to the Wikipedia Infobox, solubility products are given::

$4.93 × 10^{−5}\text{ mol}^2\text{L}^{−2}$ (anhydrous)
$3.14 × 10^{−5}$ (dihydrate) [1]

The reference is difficult to trace online, but the theory is sound. In the presence of water anhydrous calcium sulfate spontaneously converts to the dihydrate, so the dihydrate must have lower free energy and therefore less solubility in equilibrium with water. In effect, if you try to dissolve the anhydrous compound into water, the solution becomes oversaturated with respect to the dihydrate and thus the latter precipitates.
Reference

D.R. Linde (ed.) CRC Handbook of Chemistry and Physics, 83rd Edition, CRC Press, 200.

In mole par liter, $\ce{CaSO4}$ and $\ce{CaSO4·2H2O}$ have the same solubility : $0.018$ mol/L, according to J. Chem. Ed. $77$, Vol. $12$, p. $1558$, Dec. $2000$


text:
What is the effect of an electronegative atom being close to (but not directly bonded to) a carbocation?
Does oxygen in 1-methoxy-2-methylpropan-2-ylium affects the tertiary carbocation? Is it destabilized due to its electronegativity?

https://chemistry.stackexchange.com/a/65872/9961

since the oxygen in the compound is bounded to alpha carbon atom it will provide -I effect to the carbocation and destabilize it. As it is not directly bonded to the carbocation there will be no resonance effect. Had it been so, it would have stabilized it.
The electronegative atom can create dipole-dipole interactions with the nearby carbocation. These interactions can stabilize the carbocation by aligning the partial positive charge on the carbocation with the partial negative charge on the electronegative atom. This effect is more significant when the electronegative atom is close but not directly bonded to the carbocation.
If the electronegative atom or electron withdrawing functional group is of large size then it may cause steric hindrance (not sure about this one since its not directly bonded).


text:
How's the formula mass of a molecule calculated? Is it the mass in it's empricial formula or molecular formula?
Like glucose, butane and benzene. How would you calculate their formula mass? Would you calculate it as the mass of their empricial formula or their molecular formula (basically same as molecular mass)
I've seen some people say it's the mass of the molecular formula, while others say it's the mass of the empirical formula.

Benzene is $\ce{C6H6}$. Its formula mass is the molecular mass. It is $6·12 + 6·1 = 78$. Its unit is g/mol, or u = atomic mass unit. The same calculation is made for glucose and butane.

Thanks! So we I'm asked about the formula mass of any molecule, I write it's molecular formula mass instead of its empirical formula mass?

All these words are synonymous. Formula mass or molecular formula mass or empirical formula mass.

How are empirical formula mass and molecular formula mass synonyms?
Glucose, butane, and benzene each have a different molecular formula and empirical formula. So they have different values, right?
Like glucose has a molecular mass of 180 (C₆H₁₂O₆), but empirical formula mass of 30 (CH₂O)

OK. Sorry ! I forgot that empirical formula may be smaller than molecular formula, like $\ce{CH2O}$ for glucose.


text:
How to dissolve paraformaldehyde in solvent without using water?
In the chloromethylation reaction of polysulfone (PSU), I am required to use paraformaldehyde and trimethylchlorosilane as the chloromethylating agent while stannic chloride as the catalyst. I followed the same protocol of many papers such as this one.
During the experiment, PSU was first dissolved separately in solvent (e.g dichloromethane and chloroform) then paraformaldehyde (PFA) was added in form of solid. For both solvent and at two different dissolution temperatures (25 and 60 °C), the PFA was not fully dissolved.
I tried to dissolve PFA separately in solvent than add it to the dissolved PSU, this slightly improved the degree of chlore in the produced chloromethylated PSU (C-PSU), however gelation and non-homogeneity was obtained.
Moreover, the obtained C-PSU was not soluble in many solvents (e.g. NMP, DMAc, DMSO, Toluene, DMF), which can be caused by the low degree of functionalization and the precipitation of insoluble cross-linked polymer.
I am aware that PFA is

slightly soluble in alcohol
soluble via depolymerization by heating PFA in distilled water at
60°C followed by a titration with NaOH

Which I already tried the water and it was successful in terms of PFA dissolution. The problem is that I don’t want water in my reaction due to the future application of the system.
I would appreciate if you could help me to improve my system by providing some suggestions on how to achieve a good homogeneous PFA/solvent solution. Any ideas on how to increase the degree of chloromethylation of PSU are very welcomed.

You should be able to depolymerize it without water. Or maybe just extract H2CO from formalin into organic solvent?

Thank you for your comment.
If I understand correctly, I can dissolve PFA in solvent just by increasing the pH which involves the addition of NaOH? I would appreciate it if you correct me.
I would also appreciate if you could describe How to extract H2CO from formalin into solvent? Thanks in advance


text:
Solubility of fluoroalkanes
I know haloalkanes generally aren't soluble in water due to their inability to form hydrogen bonds in water, despite being moderately polar. I was wondering why this is the case with fluoroalkanes too as fluorine is quiet capable of forming H bonds.
Someone suggested to me that fluorine atoms bonded to carbon atoms cannot form hydrogen bonds but this logic seems flawed. Take the case of chloral hydrate. If chlorine can do it, why not fluorine? Please correct me if I'm wrong.


text:
Calculating Fractional Yield as a function of temperature for a parallel reaction in a plug flow model
(This question comes from a Chemical Engineering background, I hope it still falls in the scope of engineering.stackexchange.com If not, please move.)
I am trying to calculate and plot fractional yield as a function of temperature for a parallel reaction of first-order reactions forming 3 products R, S and P from reactant A.

I know the highest possible reaction temperature is 500k. S is the desirable product and I am trying to find the reaction temperature to maximize its production and calculate reactor outlet concentration of S for XA = 100%. Thus far, I have tried to construct the respective rate equations and thus corresponding Arrhenius equations (which I have data for), what I'm not sure is if I am plotting this equation against temperature, is this a valid solution to evaluate the fractional yield?
Here is my MATLAB code and corresponding graph:
CA0 = 100; % Initial concentration of A (mol/m^3)

V_reactor = 1; % Reactor volume (for example, 1 m^3)

T_range = linspace(0, 1000, 10); % Temperature range from 0 K to 500 K

YieldS = zeros(size(T_range));

for i = 1:length(T_range)
    T = T_range(i); % Temperature at this iteration
    R = 8.314; % Universal gas constant (J/(mol*K))

k1 = 5*10^8 * exp(-65000 / (R * T)); % Adjust activation energy (kJ/mol) as needed
k2 = 10^9 * exp(-50000 / (R * T)); % Adjust activation energy (kJ/mol) as needed
k3 = 10^7 * exp(-45000 / (R * T)); % Adjust activation energy (kJ/mol) as needed

dC = @(V, C) [-k1 * C(1) - k2 * C(1) - k3 * C(1); % dCA/dV
               k1 * C(1); % dCR/dV
               k2 * C(1); % dCS/dV
               k3 * C(1)]; % dCP/dV

% Initial conditions (at V = 0)
C0 = [CA0; 0; 0; 0]; % Initial concentrations of A, R, S, and P

[V, C] = ode45(dC, [0, V_reactor], C0);

CA_outlet = C(end, 1);
CS_outlet = C(end, 3);

YieldS(i) = (CS_outlet / CA0) * 100;
end

I don't fully understand your code but the result does not seem 'physical'. Is not the yield of S just $k_2/(k_1+k_2+k_3)$  where the $k$'s depend on temperature via the Arrhenius eqn.

I would like to point three things.

1. Mistake in mole balance As we frequently do in chemical engineering, we write the balances too fast and make some mistakes. In the following equations $F_j$ denotes the molar flow rate of species $j$ in $\pu{mol s^-1}$
The PFR mole balance, for a general species $j$ in the case for $k$ chemical reactions, yields
\begin{align}
  \frac{\mathrm{d}F_j}{\mathrm{d}V} &= \sum_k \nu_{j,k}r_{k} \\
  \frac{\mathrm{d}(v_0C_j)}{\mathrm{d}V} &= \sum_k \nu_{j,k}r_{k} \rightarrow
  \boxed{\frac{\mathrm{d}C_j}{\mathrm{d}V} = \frac{1}{v_0}\sum_k \nu_{j,k}r_{k}} \tag{1}
\end{align}
where $v_0$ is the volumetric flow rate in $\pu{m^3 s^-1}$. You forgot this guy in the mole balance. It is nice to remember that we can only take out $v_0$ from the differential when it is constant. For isothermal operations in liquid-phase reactions, this is an excellent approximation, because we assume that the fluid is incompressible. Make sure the species are liquids.
The four differential equations are in fact
\begin{align}
  \frac{\mathrm{d}C_A}{\mathrm{d}V} &=
  -\frac{1}{v_0}(k_1C_A + k_2 C_A + k_3 C_A)  \tag{2} \\
  \frac{\mathrm{d}C_R}{\mathrm{d}V} &=
  \frac{1}{v_0}k_1C_A \tag{3} \\
  \frac{\mathrm{d}C_S}{\mathrm{d}V} &=
  \frac{1}{v_0}k_2C_A \tag{4} \\
  \frac{\mathrm{d}C_P}{\mathrm{d}V} &=
  \frac{1}{v_0}k_3C_A \tag{5} \\
\end{align}

2. Definition of fractional yield Chemical engineering literature is not homogeneous in the definition of this quantity. For your case you are calculating the overall fractional yield, but you have to change the denominator. The overall fractional yield of species $j$ with respect to reactant $A$, denoted by $\phi_{j,A}$, is
\begin{align}
  \require{cancel}
  \phi_{j,A} &\equiv \frac{F_{j} - \cancel{F_{j0}}}{F_{A0} - F_{A}} \rightarrow
  \phi_{j,A} = \frac{\cancel{v}C_{j}}{\cancel{v_0}C_{A0} - \cancel{v}C_{A}}
  \rightarrow \boxed{\phi_{j,A} = \frac{C_{j}}{C_{A0} - C_{A}}} \tag{6}
\end{align}
for the case that: (1) no species $j$ enters the reactor, i.e. $F_{j0} = 0$; and (2) the fluid is incompressible, i.e. the volumetric flow rate is constant $v_0 = v$.
In your code you are only dividing by $C_{A0}$.

3. Solving If you are learning how to use MATLAB that is perfectly fine. However, Eqs. (2-5) are a system of ordinary linear differential equations, which are perfectly solved by pencil and paper. With some effort try to find that
\begin{align}
  C_A(V) &= \exp\left[-\left(\frac{k_1 + k_2 + k_3}{v_0}\right)V\right] C_{A0}
  \tag{7} \\
  C_j(V) &=\frac{k_j C_{A0}}{k_1 + k_2 + k_3}
  \left\{1 - \exp\left[-\left(\frac{k_1 + k_2 + k_3}{v_0}\right)V\right]\right\}
  \quad j = R,S,P \tag{8}
\end{align}
Try to manipulate Eqs. (7) and (8) to find what you want. Retaining analytical expressions may give you much more insight of the system, because you can obtain clear dependences of the variables. When integrating numerically, you lose track of this and, you resort to changing the parameters to see how the systems evolves differently along time.


text:
When does negative charge represent a pair of electrons?
Probably a stupid question but it has been bugging me for quite some time.
I was reading Clayden (organic chemistry textbook) and in it, the author gives an example of $\ce{OH-}$ and $\ce{BH4-}$ to tell that in $\ce{OH-}$ the - shows that it has lone pair of electrons whereas in $\ce{BH4-}$ , the - simply means that it has a negative charge on it. It has no lone pairs.
For simple molecules like this, I have no problem what the - is supposed to mean (by simply drawing Lewis structures). But when it comes to complex molecules, I have no idea what to do.
There's a similar problem with the positive charge as well. Sometimes, it means vacant orbital, and sometimes it just means a positive charge.

One negative charge always means one electron in excess with respect to electric neutrality, and not necessarily a pair of electrons. One positive charge always means one electron missing with respect to electric neutrality, and not necessarily a vacant orbital.

Well, cation or anion radical certainly exist, but are mostly found in MS, or as chemical curiosities. Pretty slim chance you're gonna get a bottle of a anion radical salt, but could happen.

[OP] When does negative charge represent a pair of electrons?

Never. The negative charge means that there is one electron in excess of protons. It refers to the entire species (sometimes even emphasized in brackets like for $\ce{BH4-}$ below), so it makes no statement about which atom that extra electron "belongs" to, or whether it is paired or unpaired, or part of a bond or not.

[OP ...] in $\ce{OH-}$ the - shows that it has lone pair of electrons whereas in $\ce{BH4-}$ , the - simply means that it has a negative charge on it

That does not make sense as written (by the OP or by Clayden, I can't tell). You are using the same symbol, so how would you know which of these two things it means. It is correct to say that hydroxide ions have lone pairs (3 in the conventional Lewis structure depiction), and the borohydride anion has none. You can't tell from the negative sign, though.

[OP] For simple molecules like this, I have no problem what the - is supposed to mean (by simply drawing Lewis structures). But when it comes to complex molecules, I have no idea what to do.

For complex molecules, it makes sense to learn about formal charges. If there is a formal charge of negative one on an oxygen atom, you expect three lone pairs and one bond (just like in the simple hydroxide ion). If there is a formal charge of positive one on an oxygen atom, you expect one lone pair and three bonds (just like in the simple hydronium ion). Finally, if there in no formal charge, you expert two lone pairs and two bonds (or a double bond).
Acids and their conjugative bases
If you start with a Brønsted–Lowry acid that is uncharged, and it loses one proton, you expect a conjugative base with a negative charge. You could say that the bond with hydrogen "turned into a lone pair", as the hydrogen ion does not carry any electrons. This is how you get from water (oxygen has two lone pairs and two bonds) to hydroxide (oxygen has three lone pairs, one bond and a formal charge of negative one).
$\ce{BH3}$ is a Lewis acid. It does not lose a proton, but instead can react with hydride ($\ce{H-}$) or a hydride donor. While $\ce{BH3}$ does not have an octet on boron in its Lewis structure, $\ce{BH4-}$ does (it has four single bonds).
Isoelectronic species
Sometimes, you can go from unfamiliar to familiar by looking at isoelectronic species. $\ce{BH4-}$ is isoelectronic to $\ce{CH4}$, which is not known for having lone pairs. $\ce{OH-}$ is isoelectronic to $\ce{HF}$ (although written in the different order), and you might know that the fluorine in the Lewis structure of $\ce{HF}$ has three lone pairs.
Indistinguishability
Often in organic chemistry, we pretend that we know which electron or electron pair does what. On the other hand, we can't label a single electron like we can for atoms (using isotopes), and electrons are indistinguishable in a molecule. So you can't "blame" an excess charge on a specific electron, and you can't really track an electron from reactant to product species, although we sometimes pretend we can.

The negative sign is a matter of accounting electrostatic charges even for complex molecules. Count the total number of protons in a given molecule or a given species, and count the total number of electrons. If the number is equal, that chemical entity is electrically neutral, if it has an excess proton, it is positively charged, and if there is an excess electron, there is a negative charge.
Don't confuse these signs with formal charges and signs associated with oxidation states.

consider any covalent compound. Say, hydrochloric acid (HCl). When we dissolve it in water, it breaks apart into H+ and Cl-. Why does a negative charge stand for an electron pair though? If you look at the bond in HCl, there is a shared pair of electrons constituting it (one from hydrogen and one from chlorine). When this bond breaks the chlorine atom takes with it the lone pair of electrons. I think this is the reason that the negative charge also represents a lone pair. As for borine, I think it acquires a pseudo-negative charge via coordinate covalent bonding with Lewis bases such as ammonia, which again, have donated a pair of electrons to it providing it with a negative charge.

BH4 does not have a "pseudo" negative charge. It has a negative charge.


text:
Why is silver(I) oxide sensitive to a long period exposure to the light?
I will do a reaction where I will use silver oxide as a catalyst (Lewis acid) that will help me to do a nucleophilic substitution. I read that it is photosensitive but I could not find the reason why it is.
I wrote this equation but I am not even sure this what happens when it is exposed to light:
$$\ce{2 \overset{+1}{Ag}_2O -> 4 \overset{0}{Ag} + O2}$$

Remember the analogical reaction in the classical film material with AgBr,

@Poutnik I don't know this analogical reaction...

Getting to know it does not take much more time than writing you do not know it. // $\ce{2 AgBr ->[light] 2 Ag + Br2 ->[gelatin][-Br2] 2 Ag}$

@Poutnik I did'nt find that anywhere that is why. This compound was actually used during a reaction with nitrobromobenzene and primary alcool. The goal of it was to accelerate the reaction. But they told us to keep it in dark so I was just thinking what could happen to my reaction if I let it in the light

Contrary to the fact that many silver compounds are photosensitive, silver oxide is not photosensitive, although it readily decomposes at temperatures above 280 °C.


text:
Addition of NaCl to the Anode Side of a Zinc-Copper Votaic Cell
I used a NaCl solution with a Zinc electrode in an otherwise standard voltaic cell with a Potassium Nitrate Salt Bridge and Copper Sulfate solution in the Cathode side. All the solutions were 1M.
I had assumed nothing wouldve happened but I got a result of 4mV. I used two different multimeters both with a uncertainty of 0.1mV so I am pretty sure that this was an actual voltage and this experiment did not work with water when I tried so I think that the NaCl solution is doing something. But I have no idea how it may have changed the half reactions.
I have read a similar post which talked about the creation of tetrachlorozincate but I am not sure if that is what is happening here so I am asking for some insight.

What was the dimension and the nature of the bridge between the two compartments ? Was it an agar gel ? Was it a blotting paper ? How long ? How thick ?  If the bridge had a strong internal resistance, you may have gotten only $4$ mV between the two electrodes.

The salt bridge was a U-Tube filled with a Potassium Nitrate solution with cotton balls at the end. To clarify, a wire was used to connect the electrodes and multimeter was connected to this wire. Would the internal resistance of the salt bridge affect the overall resistance?


text:
Can 'Beta-hexachlorocyclohexane' (C6H6Cl6) be Optically named?
I have not found RS configuration in Beta-hexachlorocyclohexane(C6H6Cl6) anywhere but tried to give it, although some books say that it can not be named in this configuration. So, is it named correctly, if not what is the mistake?

Welcome to ChemSE. This issue has been addressed previously on this site. If this hexachloride were hexahydroxyl it would be scyllo-inositol, which has the lower case "r"-configuration at each carbon. Go to this link for an additional explanation.

All the atoms have two identical substituents, so how could you get R/S?

If you rotate the molecule 120 degrees in plane, it is the same molecule which means any S/R assignments must alternate. If you flip the molecule through the horizontal plan, it is the same molecule however all the S configurations would now map to R and vice versa, therefore all centres are both S and R which means they cannot be assigned.

@Tragamor: Your R/S analysis is not in keeping with CIP rules. All carbons are of the r-configuration and are, therefore, assignable. Follow my links above.

This hexachlorocyclohexane (red box) is one of nine possible stereoisomers in analogy with the inositols. The isomer in question is of the scyllo-configuration and all carbons are of the same configuration. Imagine a molecular model lying on a table. It falls off. You pick it up and place it on the table exactly as before. There are six solutions. Configurations at $\ce{C1-C4}$, $\ce{C2-C5}$ and $\ce{C3-C6}$ are RR, SS and SR, respectively, and they are internally inconsistent because all carbons are identical. Given the planes of symmetry through  $\ce{C1-C4}$, $\ce{C2-C5}$ and $\ce{C3-C6}$, each carbon is stereogenic and achirotopic. The latter condition requires r/s assignments and not R/S.
To assign the r/s configuration, one constructs a digraph (vide infra) where $\ce{C1}$ is the non-duplicate atom and C(1) is the duplicate atom, which is attached to three atoms of atomic number zero.  Atoms $\ce{C2-C6}$ are given temporary (Ro/So) R/S configurations because, in the digraph, these carbons are both stereogenic and chirotopic. As an example, $\ce{C2}$ is assigned So because the priorities are Cl>long chain>short chain>H. Likewise, $\ce{C6}$ is Ro.   Having made temporary assignments for $\ce{C2-C6}$ and realizing that as R>S, as is Ro>So, the scyllo-isomer reduces to Cl>$\ce{C6}$>$\ce{C2}$>H (Cl>Ro>So>H). $\ce{C1}$ and $\ce{C2-C6}$ all have the r-configuration. Other stereoisomers require more complex rules for the final assignment.


text:
Molar heat capacity of water at equilibrium
Molar heat capacity is defined as the heat energy given to one mole of substance to raise its temperature by 1 degree celcius (or 1 Kelvin)
Mathematically, $$C_{m} = \frac {q}{n\Delta T}$$
Here's a question on Molar heat capacity:

Molar heat capacity of water in equilibrium with ice at constant pressure is :
a) 0
b) infinity
c) 40.45 kJ/K/mol
d) 78.5 J/K/mol

The answer is given as infinite. However, shouldn't it be impossible to determine, since if we provide heat to the system, it's temperature also changes? Can someone please explain to me why it is infinite?

Yeah, you're right.

Because you have a phase change going on:

IMHO, it is badly formulated question. Heat capacity of the system with water as substance is infinity due the phase change. But molar heat capacity of water as liquid phase remains the same as the one at 0 Deg C, roughly 4.2x18 J/K/mol. They explicitly ask about water as the liquid phase, in contact with ice.

So water is at equilibrium with ice at 273.15K or 0 degrees Celsius.
The difference between ice and water both at 0 degrees Celsius is evidently not their temperature (they’re both at 0 degrees). Yet there is clearly a difference in energy between the two species.
This difference arises as ice is lower in energy as it has formed bonds to make a solid packing structure without changing temperature. In other words when at the phase equilibrium temperature, the energy supplied/released is being used to form/break intermolecular bonds, not to change the temperature.
This can also be seen on a temperature vs phase diagram, when at the temperature of a phase equilibrium, this corresponds to the ‘flat line’ on this type of a graph.
Therefore the answer is zero, if you supplied energy to water at the phase equilibrium with ice then it would not change temperature, as all the energy would be used to break bonds, not change temperature.
Of course if you supplied an excess energy then the temperature would change, but in doing so the species would no longer be in the temperature relevant to the phase equilibrium any longer.

Thank you for the clarification, @Kirandk. The book has evidently marked the wrong option as correct, so I thought that the correct option is zero. I'll edit the question now.

The molar heat capacity of water in equilibrium with ice is the same as molar heat capacity of water without ice presence. The flaw of the question formulation is confusing water(I) ( the liquid phase of substance water(II)) with water(II). Water(II) cannot be in equilibrium with ice as the solid phase of water(II). Water(I) can be.


text:
Why are "vinylic" and "allylic" carbons named so?
I find it hard to remember which is "vinylic" and "allylic" carbon, so I  feel it would be easier to remember if I know the reason why they are named so, like their word root origin or something like that. What is the historic origin of the term "vinylic" and "allylic"?

Actually, word roots in organic chemistry aren't of much help because the origins have little connection with the actual molecular structure. There are entire books dedicated to these topics of you like chemical word origins.
Since allyl and vinyl are very common words, one cannot say that it was impossible to find the origins on the web. Search vinyl and etymology and you will get the desired results.
Allyl- implies coming from garlic. Still you will need to memorize the structure because those allylic compounds must have been found in garlic.
Vinyl- has the same fate. The name comes from German, derived from Vine, Vinegas (obsolete and uncharacteristic German: ethene) and -yl.
In short one has to memorize certain structures with common names.
Another puzzle in organic chemistry history is barbituric acid. It is allegedly stated that Barbara was the name of the German chemist's friend. Now how can one connect a lady's name with the structure of barbituric acid, if this were true?
Reference: Dictionary of Chemoetymology by A. Senning (Elsevier). The book is not online.

Knowing the origin of the terms won't help you to identify what carbon is "allyic" or "vinylic". A vinyl group is a functional group with the formula $\ce{−CH=CH2}$ and allyl group is vinyl group but an extra methylene bridge ($\ce{-CH2}$), so it is  $\ce{−CH2−HC=CH2}$. So, the carbons in vinyl group is "vinylic" carbon and the carbon in the extra methylene bridge is the "allylic" carbon.

(Picture source)
Regarding the origin of the word, the respective Wikipedia page has a good information. In short:

allyl comes from the scientific name for garlic, Allium sativum. In 1844, Theodor Wertheim isolated an allyl derivative from garlic oil and named it "Schwefelallyl".
vinyl comes from Latin vinum = "wine", and the Greek word "hylos" 'υλος (matter or material), because of its relationship with ethyl alcohol.


text:
Calculating the solubility product for a salt in a metal-metal insoluble salt electrode
When trying to find the solubility product of a salt in a metal - metal insoluble salt electrode, I find that I am getting a different answer if I consider the cell to be a concentration cell and if I do not.
For example, to find $K_\mathrm{sp}$ of $\ce{PbSO4}$ in the cell:
$\ce{Pb}$(s) | $\ce{PbSO4}$(s) | $\ce{Na2SO4}(aq)(0.01M)$ || $\ce{Pb(NO3)2} (0.1M)$| $\ce{Pb}$(s)
Given: $E_\mathrm{cell}$ at $298 K$ = $0.236 V$.
At the anode :

$\ce{Pb -> Pb^2+ + 2e-}$

$\ce{Pb^2+ + SO4^2- -> PbSO4}$
Overall: $\ce{Pb + SO4^2- -> PbSO4 + 2e-}$
So, $E_\mathrm{anode}$ = $E^0_\mathrm{anode}$ - $\frac{0.059}{2}$ $\log\frac{1}{[\ce{SO4^2-}]}$
At the cathode :

$\ce{Pb^2+ + 2e- -> Pb}$

$\ce{Pb(NO3)2 -> Pb^2+ + 2NO3^-}$
Overall: $\ce{Pb(NO3)2 + 2e- -> Pb + 2NO3^-}$
and, $E_\mathrm{cathode}$ = $E^0_\mathrm{cathode}$ - $\frac{0.059}{2}$ $\log{[\ce{NO3^-}]^2}$

$E_\mathrm{cell}$ = $E_\mathrm{cathode}$ - $E_\mathrm{anode}$
$E_\mathrm{cell}$ = $E^0_\mathrm{cell}$ - $\frac{0.059}{2}$ $\log{[\ce{NO3^-}]^2}[\ce{SO4^2-}]$
Thus, $E^0_\mathrm{cell}$ = 0.236 + $\frac{0.059}{2}$ $\log{(0.1)^2}(0.01)$ = 0.118 V
At equilibrium, $E_\mathrm{cell}$ = $0$, so $\log{K_\mathrm{sp}}$ = $\frac{(2)E^0_\mathrm{cell}}{0.059}$
So, $\log{K_\mathrm{sp}}$ = 4
$K_\mathrm{sp}$ = $10^{4}$
However, the answer comes out to be $10^{-11}$ when I consider it to be a concentration cell of $\ce{Pb^2+}$. Where am I going wrong or why does this method not work?

E_ cell=RT/(nF)ln([Pb2+][SO4^2-]/Ksp)

@Poutnik could you elaborate? also, my equation does not have Pb2+ in it, so is it wrong that I found Ksp from the E0cell value I got?

It is a concentration cell, where E_ cell=RT/(nF)ln(c1/c2) , where c1 is given by c of Pb^2+ from (dissociated) Pb(NO3)2 and c2 is given by Ksp and c(SO4^2-) from (dissociated) Na2SO4.

@Poutnik Even I thought it was odd, but that was what i got when i added the equations..

Equations are partly wrong.


text:
Esterification of phosphonic acid chloride is working with isoprenol but not with prenol
I am performing an esterfication reaction between an activated phosphonic acid (chloride) and a series of alcohols. Until now, I was not able to make the reaction work with prenol, while isoprenol is working fine. I am wondering, what could be the reason for such different chemical behaviour and reactivity? I tried different solvents like THF, DCM and acetonitrile and also bases like pyridine, NaH as well as DBU and DMAP, but until now, me only observed the isoprenol adduct. Isoprenol is working with absolute standard conditions. Is there anything you can recommend I haven't tried? Would you expect such different reactivity just by the positioning of the double bond?

I would not expect any difference in the reactivity of the two alcohols. Have you checked on the purity/identity of your prenol?

Yes, booth are fairly pure regarding NMR.

You could try adding 1eq of a silver salt e.g. AgBF4

If you use an aqueous workup, is solvolysis occuring?

@user55119 No, the isoprenol adduct withstands aqueous workup. During reaction control you the correct product mass increasing in counts, while the educt is consumed. I was asking myself if somehow the reaction with the prenol is somehow energetically disfavored.

@Waylander I am wondering how this would work mechanistically? After the acid chloride is formed, the pyridine substitutes the chloride and you have a good leaving group. At which point does the silver speed up the process? (I did not find any mechanistically review on that yesterday)

The silver removes chloride ion from solution increasing the rate of formation of the pyridinium species

The prenol phosphate is allylic and more readily prone to solvolysis than the isoprenol phosphate, which is just a normal primary phosphate.

@user55119 until now, I did not observe it in LCMS reaction control therefore I did not bother to do an workup on the prenol ester.

Tetrahedron Letters 1986,27,1225 describes the reaction of (R,E)-2-methylhex-4-en-3-ol with n-BuLi in ether followed by quenching the alkoxide with diethyl phosphorochloridate [(EtO)2POCl]. It appears that the allyic phosphate was used without isolation. The alcohol is secondary and allylic and perhaps prone to solvolysis, more so than yours. Have you considered phosphorus NMR to follow the course of the reaction prior to isolation?

NMR service at my institute is not very accessible. Therefore I try to avoid this as RC as good as possible and relay mainly on LCMS and TLC and do this only for final characterization....I was thinking of trying microwave radiation as well as the addition of a silver catalyst like @Waylander suggested.

@user5511 What else I find confusing is the fact, that that tetrabutylammonium hydrogen phosphat can react seemingly easy alike with prenol and isoprenol with trichloracetronitril.


text:
THF-H2O mixture at 80°C? isn't THF supposed to evaporate completely?
I was running a synthesis of TiO2 nanoparticles starting from a aqueous solution of TiCl4 2THF 0.125M at 80°C. Through a different experiment it was detemined that the maximum reaction yield was 96% at 30 minutes after the start (mixing of TiCl42THF with H2O at 80°C).
THF is used in this complex just to prevent a strong exothermic mixing with water, but it doesn't take part in the reaction.
Having said that, I thought that THF, after dissociating from the complex, was in the solution with a concentration of 0.25M. Since the temperature was 80°C and the boiling point of pure THF is 66°C, I considered the THF evaporating completely in 30 minutes.
My problem is trying to demonstrate by a theoric approach that THF leave the system.
I tried to use the raoult law : P(THF)= P°(THF)*X(THF) but I get a small value of partial pressure of THF.
I'll give you some more info: P°(THF) at 80°C = 1.54 atm , X(THF)=0.0047, P°(H20) at 80°C = 0.46 atm X(H2O) = 0.993
I am not sure about the right way to demonstrate the THF evaporation (30 min at 80°C), could you please help me?

Chem+Math Expression formatting reference:
MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.

It depends on how fast you remove the vapours, what is evaporation kinetics and what is the threshold THF content to be considered negligible (as it is exponential process, less THF = slower evaporation).

@Poutnik I don't how to estimate the evaporation kinetics. The system is a open one, so the equilibrium between the vapour phase and liquid phase it's only temporary. Do you know any mathematical equation that can explain the evaporation rate of the solution, including the THF?

As there are several unknown variables,  you have to rely on empirical parameters. You may measure the evaporation rate via mass or volume decreasing rate. From that you can estimate the rate of THF evaporation via molar fraction in vapour.

The (lab scale) synthesis of NPs often uses round bottom flasks because 1) it is typical equipment already in the shelves, 2) ease to heat/cool by baths, 3) ease to connect to Schlenk lines, 4) ease to mount additional equipment such as dropping funnels (with pressure equalization), or an efficient reflux condenser (Dimroth model).  What's the remaining problem?  Get a syringe pump if the addition by the dropping funnel still is too fast (but: this is more expensive than the funnel).


text:
TMSiBr and Oxalylchloride for ester cleavage
I am wondering what is the advantage of using TMSiBr do perform an ester cleavage in comparison to perform chlorination first for example with oxalylchloride or thionylchloride and than add water?

Oxalyl chloride to cleave an ester, are you sure about this reaction? Or Thionyl chloride?

TMSBr may be for specific esters. Do you have a reference?

Yes, I have multiple references: https://chemistry-europe.onlinelibrary.wiley.com/doi/abs/10.1002/slct.201900120 https://journals.sagepub.com/doi/10.1177/1747519819898142 https://www.sciencedirect.com/science/article/abs/pii/S0040403915000763 As far as I understand TMSBr is more "soft". And oxalylchloride is only working with specific esters. More interesting for me are phosphonic acid esters in general.

The ester/ether cleavage with oxalyl chloride is with p-methoxybenzyl esters and ethers, an SN1 process. A special case and likely not general.

I have used TMSiI extensively for cleaving phosphonic esters


text:
Is the comparison between absolute values of ∆S and ∆H for a given reaction meaningful?
How do you know if the reaction is dominated by enthalpic or entropic factor? Do you just compare the absolute values of them? If the absolute value of ∆S is bigger than the absolute value of ∆H, can you conclude that the entropy is a more potent factor in determining the reaction progression?

You cannot compare the values of ∆H and ∆S, because they don't have the same unit. The first is in Joule and the second in Joule/Kelvin

You need to calculate the change in Gibbs Free Energy: ∆G = ∆H - T∆S. Reaction is spontaneous if ∆G < 0.
You can see from this equation that knowledge of ∆H and ∆S without T is insufficient. For example, close to absolute zero ∆H dominates even if ∆S is large because T x ∆S will be very small.
Important Caveat: ∆G is only useful if temperature and pressure are constant. If the process involves a change in T or P, then ∆G cannot tell us if the reaction will be spontaneous.

Better to say that final T and p must be the same as initial ones. They can change during the reaction.

Delta G is useful with change in T and P The change in Keq is a known function of Delta H. Knowing Keq at a T enables determining spontaneity at a given set of reaction conditions. A more holistic view is that Delta G for a given set of conditions is a function of temperature depending on Delta H and Delta S and becoming zero when those conditions satisfy the equilibrium constant at the appropriate temperature.


text:
When does osmosis stop?
In the figure given below:-

Since only solvent particles can  move through semipermeable membrane, if we manage to keep pressure on both the containers equal say by keeping weights, will osmosis go on till pure solvent container reaches zero volume?

It depends on the geometry of the vessels. If tall enough, osmotic pressure will be balanced by hydrostatic pressure difference.

Osmosis phenomena produces a height difference which tends to a maximum ℎ
h. If both solvent and solution are at atmospheric pressure, this maximum is such that ℎ is equal to $1$ atm, for a $1$M solution. This height is also proportional the concentration of the solution.

@Poutnik I actually meant that if I apply and keep changing pressure on the other (solvent) container as well such that pressure remains equal on both the containers, will the osmosis take place?

If the membrane pressure difference (= external pressure difference  + hydrostatic pressure difference) balances the osmotic pressure, the net osmosis is zero.

@Poutnik Thanks I understood that. Just wanted like confirmation that osmosis will keep going on if there difference in membrane pressure difference  and osmotic pressure, right?

Let $H_l$ = Hydrostatic pressure in solution jar
Let $O_l$ = Osmotic pressure in solution jar
Let $P_l$ = Pressure from the weight applied to the solution jar
Let $H_s$ = Hydrostatic pressure in solvent jar
Let $O_s$ = Osmotic pressure in solvent jar
Let $P_s$ = Pressure from the weight applied to the solvent jar
Osmosis will take place until ...
$H_l + O_s + P_l = H_s + O_l + P_s$
We know $O_s = 0$ and $P_l = P_s$. So the equation for when the osmosis will stop simplifies to ...
$H_l = O_l + H_s$


text:
How does a transition metal(eg Zn,Pd,Pt etc) break the hydrogen bond in catalytic reduction
I want to know what exactly happens with the hydrogen molecules and transition metals in catalytic reduction leading to breaking of hydrogen bonds and forming two hydrogen atoms

I don't think it's well understood.

related https://chemistry.stackexchange.com/questions/15387/hydrogenation-stereochemistry-pd-pt-ni-catalysts


text:
Derivation of "activity coefficient form of the Gibbs-Duhem equation"
I'm trying to differentiate the excess Gibbs energy ($G^{ex}$) to get the activity coefficient ($\gamma_k$), which writes:
$$
\frac{\partial (G^{ex}/RT)}{\partial n_k} = \ln\gamma_k \quad (1)
$$
in the process of deriving the above equation, I saw some conditions are used, like:
$$
RT\sum_i n_i \left( \frac{\partial (\ln \gamma_i)}{\partial n_k} \right)_{T,P,n_{i\neq k}} = 0 \quad (2)
$$

$n_i$: mole numbers

or
$$
x_1 \text{d} \ln \gamma_1 + x_2 \text{d} \ln \gamma_2 = 0 \quad (3)
$$

$x_i$ mole fractions

I'm wondering how (2) and (3) are derived? I saw they were mentiond as "activity coefficient form of the Gibbs-Duhem equation"
but how Gibbs-Duhem equation
$$
\sum_i n_i \text{d} \mu_i = 0
$$
becomes (2) or (3) are unclear to me.
References:

(1) and (2) from Thermodynamics and Its Applications by Tester and Modell, P357-358
（3）from Wikipedia here

Remarks:

Equation (1) is by definition the natural logarithm of the activity coefficient, so you cannot derive it
$$ \left[\frac{\partial (nG^\text{E})}{\partial n_j}\right]_{p,T,n_{k \neq j}}
   \equiv RT\ln(\gamma_j) $$
Equation (2) is false, we will check it out later.
Equation (3) can be derived.

I will show you how to derive the Gibbs-Duhem equation and attack at the end the second point.

We postulate that the thermodynamic property $nM$ can be expressed as a total differential in terms of the pressure $p$, temperature $T$, and all amounts $n_1, n_2, ...$
\begin{equation}
  \mathrm{d}(nM) =
  \left[\frac{\partial (nM)}{\partial p}\right]_{T,n} \mathrm{d}p +
  \left[\frac{\partial (nM)}{\partial T}\right]_{p,n} \mathrm{d}T +
  \sum_i \left[\frac{\partial (nM)}{\partial n_i}\right]_{p,T,n_{j \neq i}}
  \mathrm{d}n_i \tag{2}
\end{equation}
here $n$ loosely means all the amounts, and the thermodynamic property $M$ is written per total amount, e.g., $\pu{J mol^-1}$. Even though we can choose any thermodynamic variables, these ones are only convenient for the Gibbs energy, but you can select any of them.
The last partial derivative, by definition, is the partial molar property $\overline{M}_i$ of species $i$ in a solution
\begin{equation}
  \mathrm{d}(nM) =
  n\left(\frac{\partial M}{\partial p}\right)_{T,n} \mathrm{d}p +
  n\left(\frac{\partial M}{\partial T}\right)_{p,n} \mathrm{d}T +
  \sum_i \overline{M}_i \mathrm{d}n_i \tag{3}
\end{equation}
where we also took out $n$ from the first two partial derivatives. Now we do two things: (1) open up the differential $\mathrm{d}(nM)$, and (2) use that $n_i = x_i n$ or $\mathrm{d}n_i = n\mathrm{d}x_i + x_i\mathrm{d}n$ (with these transformations, now our variables are in terms of molar fractions rather than amounts, so we change the indexes of the partial derivatives)
\begin{align}
  n\mathrm{d}M + M\mathrm{d}n &=
  n\left(\frac{\partial M}{\partial p}\right)_{T,x} \mathrm{d}p +
  n\left(\frac{\partial M}{\partial T}\right)_{p,x} \mathrm{d}T +
  \sum_i \overline{M}_i (n\mathrm{d}x_i + x_i\mathrm{d}n) \\
  \color{blue}{n\mathrm{d}M} +
  \color{red}{M\mathrm{d}n} &=
  \color{blue}{n\left(\frac{\partial M}{\partial p}\right)_{T,x} \mathrm{d}p} +
  \color{blue}{n\left(\frac{\partial M}{\partial T}\right)_{p,x} \mathrm{d}T} +
  \color{blue}{n\sum_i \overline{M}_i \mathrm{d}x_i} +
  \color{red}{\sum_i x_i\overline{M}_i \mathrm{d}n} \\
  \left(M - \sum_i x_i\overline{M}_i \right)\mathrm{d}n &+
  \left[\mathrm{d}M -
  \left(\frac{\partial M}{\partial p}\right)_{T,x} \mathrm{d}p -
  \left(\frac{\partial M}{\partial T}\right)_{p,x} \mathrm{d}T -
  \sum_i \overline{M}_i \mathrm{d}x_i\right] n =0 \tag{4} \\
\end{align}
Eq. (4) gives us two conditions for it to hold.
1. Summability relation The first one is the famous summability relation. It connect the property of the whole solution $M$ to the partial molar properties of all the species. We wil manipulate this one a little bit
\begin{align}
  M  &= \sum_i x_i\overline{M}_i  \tag{differentiate} \\
  dM &= \sum_i \overline{M}_i \mathrm{d}x_i +
        \sum_i x_i \mathrm{d}\overline{M}_i \tag{5}
\end{align}
2. G-D Equation The second one is the famous Gibbs-Duhem, once we combine it with Eq. (5)
\begin{align}
  \require{cancel}
  \mathrm{d}M -
  \left(\frac{\partial M}{\partial p}\right)_{T,x} \mathrm{d}p -
  \left(\frac{\partial M}{\partial T}\right)_{p,x} \mathrm{d}T -
  \sum_i \overline{M}_i\mathrm{d}x_i &= 0 \\
  \cancel{\sum_i \overline{M}_i \mathrm{d}x_i} +
  \sum_i x_i \mathrm{d}\overline{M}_i +
  \left(\frac{\partial M}{\partial p}\right)_{T,x} \mathrm{d}p -
  \left(\frac{\partial M}{\partial T}\right)_{p,x} \mathrm{d}T -
  \cancel{\sum_i \overline{M}_i\mathrm{d}x_i}  &= 0 \\
  \sum_i x_i \mathrm{d}\overline{M}_i =
  \left(\frac{\partial M}{\partial p}\right)_{T,x} \mathrm{d}p -
  \left(\frac{\partial M}{\partial T}\right)_{p,x} \mathrm{d}T& \tag{7}
\end{align}
Eq. (7) is general for any change in pressure, temperature, and amounts. For the case where the pressure $p$ and temperature $T$ are held constant, we have your Eq. (3) (if we divide by $n$), because the property $M$ is the Gibbs energy and its partial molar property is named the chemical potential $\overline{G}_i = \mu_i$
$$ \boxed{\sum_i x_i \mathrm{d}\mu_i = 0} \tag{8} $$

An easy way to prove that eq. (2) is wrong, is to test it in an activity model. For example, do the summation
$$ \sum_i n_i \left(\frac{\partial \ln \gamma_i}{\partial n_i}\right)_
   {p,T,n_{k \neq i}} \color{blue}{\neq} \tag{9} 0 $$
under the Margules $1$ parameter activity model, where
$$ \ln(\gamma_1) = \frac{A}{RT} x_2^2 \quad
   \ln(\gamma_2) = \frac{A}{RT} x_1^2 \tag{10,11} $$
and prove, in a beautiful manner, that you get instead $-(2A/RT)x_1x_2$.
I let you prove that if you manipulate your Eq. (2) a bit, you can obtain the summability relation. Try it over Eqs. (10) and (11), so you can recover the excess Gibbs energy for the Margules $1$ parameter model, which is $G^\text{E} = Ax_1x_2$.

I think I figured it out, I was trying to derive equation (2)(or3) from the Gibbs-Duhem equation. But if we change the direction, to replace activity coefficient with chemical potential, things will be easier.
Details
According to the Gibbs-Duhem equation, that the chemical potential of one component of a mixture cannot change independently of the chemical potentials of the other components. Thus:
$$
      \begin{align*}
        \sum_i n_i \text{d} {\mu_i} = 0
      \end{align*}
$$
Also, the defination of activity coefficient is:
$$
\begin{align*}
    \ln \gamma_i = \frac{\mu_i -\mu_i^{id}}{RT}
\end{align*}
$$
Therefore,
$$
\begin{align*}
    \sum_i n_i \left[ \frac{\partial \ln \gamma_i}{\partial n_k} \right]_{T,P,n_{i \neq k}} &=
    \sum_i n_i \left[ \frac{\partial (\mu_i - \mu_i^{id})}{\partial n_k} \right]_{T,P,n_{i \neq k}} \\
    &= \sum_i n_i \left[ \frac{\partial \mu_i}{\partial n_k} - \frac{\partial \mu_i^{id}}{\partial n_k}\right]_{T,P,n_{i \neq k}} \\
    &= \sum_i n_i  \left[ \frac{\partial \mu_i}{\partial n_k} \right]_{T,P,n_{i \neq k}}- \sum_i n_i \left[ \frac{\partial \mu_i^{id}}{\partial n_k}\right]_{T,P,n_{i \neq k}} \\
\end{align*}
$$
recall the Gibbs-Duhem equation $\sum_i n_i \text{d} \mu_i = 0$, therefore:
$$
\begin{align*}
    \sum_i n_i  \left[ \frac{\partial \mu_i}{\partial n_k} \right]_{T,P,n_{i \neq k}} &=0 \\
    \sum_i n_i \left[ \frac{\partial \mu_i^{id}}{\partial n_k}\right]_{T,P,n_{i \neq k}} &=0
\end{align*}
$$
so we have
$$
\begin{align*}
\sum_i n_i \left[ \frac{\partial \ln \gamma_i}{\partial n_k} \right]_{T,P,n_{i \neq k}} =0
\end{align*}
$$


text:
The separation of lanthanum oxide and chromium oxide
I have encountered a problem. How can I separate the alloy of lanthanum oxide and chromium oxide? I attempted to dissolve them in concentrated sulfuric acid with heating, but it didn't work. If I had dissolved them, I would then have added oxalic acid to the solution, and afterwards, lanthanum oxalate precipitate would have formed while chromium remained in the solution.

Try dilute Hydrochloric acid

@Waylander maybe. If the oxide has been annealed at high temperature, which is often the case with commercial Cr2O3, then any dilutecacid or alkali won't work. See https://www.sciencemadness.org/smwiki/index.php/Chromium(III)_oxide#Handling. We would then need something like what I found below.

Lanthanum oxide is soluble in dilute aqueous acids but chromic oxide may not be (depending on how it was processed). In a mixture the chromic oxide may inhibit dissolution of the otherwise more soluble lanthanum oxide. You need to:

Chromium(III) oxide (Cr2O3) can be reactively dissolved with sulfuric acid (~85 wt%; H2SO4/Cr2O3 ≥ 6 wt/wt) mixed with perchloric acid (HClO4/Cr2O3 ≈ ½ wt/wt) by ~175 ºC (≤ 4h; under reflux).

Under these conditions the lanthanum oxide will be dissolved too. So this would get everything dissolved.
Lanthanum oxalate may remain soluble in strongly acidic media (reacts to form oxalic acid and the lanthanum salt of the stronger acid).  With the chromium apparently oxidized to Cr(VI), you may want to dilute and neutralize the solution before applying the oxalic acid; this would give the lanthanum oxalate precipitate with chromate remaining in solution. Note that Cr(VI) is considered hazardous and carcinogenuc material.

Not only hazardous but carcinogenic ;)


text:
Why doesn't the ocean gradually turn into hydrogen and oxygen gas?
Maybe I'm wrong about this, but I thought I remembered from high school chemistry that all reactions are in equilibrium. Some equilibria are extremely far to the right or left, so they appear to react 100% or not at all, but even in those reactions there is a tiny amount on the unfavored side.
If that's not true, then the answer to this question will be short. But if it is true, then I have a question about the following reaction, and bodies of water like lakes and oceans.
$$
\ce{2H2O (liquid) <=> 2H2 (g) + O2 (g)}
$$
If I remember right, the energy change ($\Delta{H}$ or $\Delta{G}$) in that reaction strongly favors the left side, so water molecules are not breaking apart into gas in large numbers. However, it's a gas on the right side, so I imagine that if a water molecule at the surface of the ocean breaks into gas, those products will escape, and the $\ce{H2}$ and $\ce{O2}$ gas molecules are no longer in physical contact and have no chance to react again.
That makes me think that gradually the ocean would turn into hydrogen and oxygen gas, even if the left side of that equation were favored. That doesn't seem to be happening, so what am I missing? I don't normally think there is hydrogen gas floating around in the air, but perhaps there actually is enough that some of it is colliding with $\ce{O2}$ at the ocean surface and balancing the equilibrium.

Why would you think H2 wouldn't react with oxygen in atmosphere?

Because it's a gas that escapes upward. Is there enough H2 gas hanging around at sea level? I'm not a chemist. I see a vote to close the question, but I think it's a decent question from a layman's or beginner chemistry point of view, even if the answer is very basic.

There's plenty of oxygen upwards. All in all, there isn't really an equilibrium like that, or more precisely it's one of extremely far to one side that you mention.

H2 and O2, being gases, are in physical contact and have every chance to react again. Also, the equilibrium is shifted so far to the left that the expression "strongly favors" is not strong enough to do it justice. Had all this not been the case, your concern would be valid.

Something a bit similar happens on Venus, IIRC, making the planet lose water. It's not a matter of any equilibrium, though.

I think it's technically correct to say that it this is actually happening, but in such a miniscule amount that it's completely obfuscated by all the other planetary chemical processes going on. But if you're talking spherical cows in a vacuum, yes, if you wait long enough (many, many times the age of the Universe), the Earth would dry out by this process.

This is good thinking but poor followup. I wonder what did happen to all the water that supposedly was on Venus.

Should I delete the question? Or improve it somehow? It gets downvoted.

Maybe I'm wrong about this, but I thought I remembered from high school chemistry that all reactions are in equilibrium. Some equilibria are extremely far to the right or left, so they appear to react 100% or not at all, but even in those reactions there is a tiny amount on the unfavored side.

Yeah, that's true. But the word extremely is doing a lot of heavy lifting to make that statement correct. Let's explore just how extreme the situation is.
Before even talking about the equilibrium we should discuss reaction rates. Even if the reaction is not in equilibrium, it may take a very long time to get there. And I'm not talking about "leave it overnight." The Arrhenius equation relates the rate of a reaction to the energy barrier. At standard conditions (25 C), the speed of a reaction is halved for every additional 1.73 kJ/mol of activation energy. Familiar reversible reactions that happen on the hour scale have activation energies less than 100 kJ/mol or so. By contrast, the Gibbs free energy of formation of water is 237 kJ/mol. So the reaction will be at least $2^{(237-100)/1.73} = 7 \times 10^{23}$ times slower. Instead of proceeding in hours, it will take quintillions of years, billions of times longer than the age of the universe, to reach equilibrium.
Even allowing for the reaction to reach equilibrium, what is that equilibrium? LibreText's chemistry book gives an equilibrium constant of $2.4 \times 10^{47}$ at 500 K. It doesn't state the units, but when the number is that large, any sane unit leads to the same conclusion. Namely, the equilibrium concentration of hydrogen will be very small, parts per quintillion or so.
In summary, maybe there's an equilibrium. But it will take quintillions of years to reach it, and even when it is reached, the quantities of hydrogen will be measured in parts per quintillion. So don't expect it to matter over normal periods of time.