text:
Acid Dissociation
I have a small confusion
If suppose HA is a acid then ,  HA gives H+ and A- ions in water.
The presence of H+ implies that the solution is acidic.
OK fine... but I came across that this solution will have OH- ions too.
I am confused, from where is it coming from ? Sources tell me that it could be from self ionisation of water but I didnt quite understand the logic.
Can someone please explain this ? p.s still a newbie..Thanks in advance ;)

Everything has both H+ and OH- ions. It is their quantity that matters.

When I say $\ce{HCl}$ you'd imagine I'm talking about the acid, but I'm talking about the gas. When you dissolve a particular concentration of this gas in water it gives an acid of a particular molarity. Hence every acid will dissociate into $\ce{H+}$ & $\ce{A-}$ ions. The constituent ions of the water used to make this acid, similarly seperate into $\ce{H+}$ & $\ce{OH-}$ ions. Bonus point - the acid just has more number of $\ce{H+}$ relative to a neutral solution - like water has equal $\ce{H+}$ & $\ce{OH-}$

@Desai ok so basically water disassociates to H+ and OH- ions and the introduction of HCl gas gives more H+ ions and hence the acidic character is explained. Is water always present as H+ and OH- ions ?

Yes, you've got it. Uh what else do you think water could be present as? Tell me what's going on in your mind?

@puma You'll learn more about acids in the  'equilibrium' chapter NCERT Chemistry. How they dissociate and how the pH is affected etc.

@Desai I thought water will be present as H2O only and doesn't dissociate easily ... and yes that chapter is the one from where I am getting my doubts

No. The lone pair of electrons in $\ce{H2O}$ and the difference in electronegativity in H—Cl causes this.


text:
How to calculate the quantum yield of a photochemical process?

I have that $\pu{100 ml}$ of an aqueous solution containing $\pu{1.0 mM}$ $\ce{H2O2}$, $\pu{100 mM}$ $\ce{CH3OH}$ and $\pu{200 mM}$ DMSO are exposed to monochromatic light with a wavelength of $300$ nm. The incident light intensity is $\pu{2mW/cm2}$ and the exposed area is $\pu{4cm2}$. The absorbance of the solution at $300$ nm (which depends only on  $\ce{H2O2}$) was measured to be $0.25$. Only the homolysis of $\ce{H2O2}$ to two hydroxyl radicals needs to be taken into consideration. The hydroxyl radicals formed react with  $\ce{CH3OH}$ at a rate constant of $\pu{1.0E8M^{-1}s^{-1}}$ and with DMSO at a rate constant of $\pu{2.0E9M^{-1}s^{-1}}$. The product of the reaction with $\ce{CH3OH}$ is formaldehyde ($\ce{CH2O}$). The concentration of formaldehyde as a function of reaction time was measured and is given below. Based on the data below, determine the quantum yield for photoinduced homolysis of $\ce{H2O2}$.
$$
\begin{array}{rl}
\hline
\textrm{time}/\pu{h} & [\ce{CH2O}]/\pu{M} \\
\hline
0 & 0 \\
1 & \pu{7.7E-7} \\
2 & \pu{1.5E-6} \\
3 & \pu{2.3E-6} \\
4 & \pu{3.1E-6} \\
5 & \pu{3.8E-6} \\
\hline
\end{array}
$$

I hope this isn't closed off as "off-topic" since it is quite a long question but I genuinely don't know where to start on this. I feel that there is so much information and I can't quite determine what is useful and what has just been added to confuse us.
I wanted to start by writing all the reactions and then use the steady-state assumption to determine the rate equation for $\ce{CH2O}$ but even there I got stuck. From what I have understood two of the reactions are:
$$\ce{H2O2 -> 2OH-}$$
$$\ce{2OH- + CH3OH -> CH2O + 2 H2O}$$
But I don't know how the third reaction with DMSO will look like and can't seem to find anything online. I don't know if the way I am trying to solve this problem is the correct way since I don't really know if I am supposed to solve for  $\ce{CH2O}$ and then plot it to get something from the plot so I would truly appreciate if someone could push me in the right direction or link helpful examples or other useful information that might help me solve this. Thank you!
EDIT
With your help I have tried to being answering the problem but I come across some things that I don't know how to tackle. I began with calculating that the incident light intensity should be $2$ mW/cm$^2$ * $4$cm$^2$ = $8$ mW. By using Lambert-Beers law I calculated the intensity of the outgoing light as:
$$log\frac{I_0}{I} = A$$
$$I = \frac{I_0}{10^A} = \frac{8mW}{10^{0.25}} = 4.49 = 4.5 mW$$
Then the amount of absorbed light is: ΔI = I$_0$ - I = $8 - 4.5 = 3.5 mW$
So from this I can now calculate the denominator of the quantum yield fraction (I call it b for amount of absorbed photons) from the equation:
$$b = \frac{absorbed light}{energy per mol photons} = \frac{ΔIλ}{hcN_A} = \frac{3.5*10^{-3}Js^{-1}* 300 * 10^{-9}m}{6.625*10^{-34}Js^{-1}*3*10^8ms^{-1}*6.02*10^{23}mol}$$
$$= 8.78*10^{-9} smol^{-1}$$
I don't think this is the correct answer since the units don't add up. Should I maybe take the intensity times the number of hours ($5$h) to get the intensity only in Joules?
Then if I write out the rate equations:
$$\frac{d[CH_2O]}{dt}=k_2[OH^*]^2[CH_3OH]$$
$$\frac{d[OH^*]}{dt}=k_1[H_2O_2]-k_2[OH^*]^2[CH_3OH] → [OH^*]^2 = \frac{k_1[H_2O_2]}{k_2[CH_3OH]}$$
which then gives:
$$\frac{d[CH_2O]}{dt}=k_1[H_2O_2]$$
I don't really understand what use I have of this in order to calculate the number of molecules of the reactant which is used..

Well, not writing anion instead of radical would be a start... BTW why do you know how to make a table and not about mhchem or mathrm?

This is a messy problem. As to DMSO as long as it does not produce formaldehyde it reduces this yield by a lot $10^8/(10^8+20\cdot10^8)$. Next the product yield is tiny anyway so $[H_2O_2],[CH_3OH],[DMSO]$ are effectively constant. You know from absorbance and Beers law  what fraction of energy is absorbed, you should assume each photon breaks a bond. Write down the rate equations. Ignore product from DMSO. You will need to integrate to get total yield as I assume illumination is continuous.

If DMSO interfere I think it should also be counted. The quantum yield would be the practical one on the products of interest. How many of the product formaldehyde / how many photons.

@porphyrin Thank you for your help! I have added some of my new calculations and questions to my original question under edit. I would truly appreciate it if You could take a look.

You are at steady state so $d[OH]/dt =2 k_{abs}[H2O2] - k_2[OH][MeOH]=0$. You get $k_{abs}$ from the number of photons/sec converted into 'moles of photons'/sec (Einsteins?) and you can then use data in table. You can assume pseudo first order as the amount of OH produced is tiny compared to other species.

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text:
Synthesis of diethyloxalate from ethene

Outline the steps involved in the following transformation: $\ce{C2H4}$ to $\ce{(COOC2H5)2}$ (diethyl oxalate).

I am familiar with the transformation of $\ce{C2H4}$ to ethyl acetate $\ce{CH3COOC2H5}$ by first converting to ethanol using water and acid, then using an oxidising agent such as $\ce{KMnO4}$ to convert ethanol to ethanoic acid and subsequently reacting ethanol and ethanoic acid together.
I'm not sure if the process needed to get to the diethyl oxalate is similar to this.

For future reference:  Within the body of questions, answers, and comments, on chemistry.se using mhchem is recommended.  It is something special not all web browsers know to trade with, thus, do not use this syntax in the title of an question.

Oxidise ethene to ethylene oxide (peracids will do this), treat with aqueous acid to form ethylene glycol, oxidise to oxalic acid, esterify.

Or treat ethene with bromine $\ce{Br2}$, then hydrolyze dibromoethane to glycol with $\ce{NaOH}$ and oxidize glycol to oxalic acid, before esterification by ethanol.

Conversion  1

Convert ethene to ethanol.
Ethene is mixed with steam and passed over a catalyst consisting of solid silicon dioxide coated with phosphoric(V) acid. The temperature used is 300 °C and the pressure is about 60 to 70 atmospheres (chemguide —
The mechanism for the acid catalysed hydration of ethene).
$$\ce{CH2=CH2 + H2O <=>[H3PO4] CH3CH2OH}$$

React the ethanol with oxalic acid to get diethyl oxalate.
$$\ce{(COOH)2 + 2C2H5OH ->[CCl4,\Delta] (COOC2H5)2 + 2 H2O}$$
The details of this reaction is given on Organic Syntheses — Ethyl Oxalate and in the paper by Jewel and Butts [1].

Conversion 2
Convert ethene to oxalic acid by first converting to ethyne and then making oxalic acid from it. Use the oxalic acid to make diethyl oxalate by reacting it with absolute alcohol.
$$\ce{CH2=CH2 ->[Br2][CCl4] Br-CH2-CH2-Br ->[alc. KOH][\Delta] CH#CH ->[alc. KMnO4] (COOH)2}$$
Reference

Jewel, P. W.; Butts, J. S. A Method for the Preparation of Diethyl Oxalate. J. Am. Chem. Soc. 1931, 53 (9), 3560–3561. DOI: 10.1021/ja01360a052.

Don't you think the oxalic acid might also be produced from ethene ?

@Maurice Yes, we can. Added in my answer

I don't think alcoholic $\ce{KOH}$ will do the job in conversion 2, reaction 2. You will need $\ce{NaNH2}$ in liquid $\ce{NH3}$.

@NisargBhavsar Both works fine.

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text:
Thermal properties of 301 stainless steel
So I know that 301 stainless steel was chosen for the SpaceX Starship rocket as it can handle high temperature differences.
What about this alloy makes the steel harden when it gets colder?

" it can handle high temperature differences" – Note that this is not what SpaceX (or rather Elon Musk) has said. What Musk has said is that the specific material properties of stainless steel compared to carbon fiber composites, at the two temperature points at which Starship will operate (~90 K and ~3000 K), when you factor in all the other considerations (e.g. the amount of thermal protection needed), then stainless steel is actually lighter for the same strength (or stronger for the same mass) than carbon composites.

Also note that SpaceX switched from 301 to 304L a while ago, and SpaceX's metallurgy department is working on a custom 300 series alloy currently codenamed 30X.

301 is the lowest alloy content of the "18-8" family. The slightly lower nickel ( compared to 304) makes the austenite less stable so when cold rolled it work hardens more than 304. It is still basically austenitic so does not become brittle at low temperatures. So it was chosen for the higher strength ( when cold worked)  at all temperatures , relative to 304. Austenite is non-magnetic but as 301 is cold rolled it becomes somewhat magnetic depending on the amount of cold work. Presumable caused by formation of tiny grains of martensite ( the magnetic , strong, crystal of the iron family). The small martensite grains block the slip planes in the austenite matric making it stronger.


text:
Ideal CSTR mixing assumption
Sorry if this seems like a silly question but it has me confused a lot. So I have just started taking Chemical Reaction Engineering, and was studying about continuous-stirred tank reactor (CSTR).
So in my textbook it's stated that "It is normally operated at steady state and is assumed
to be perfectly mixed; consequently, there is no time dependence or position
dependence of the temperature, concentration, or reaction rate inside the
CSTR. That is, every variable is the same at every point inside the reactor.
Because the temperature and concentration are identical everywhere within the
reaction vessel, they are the same at the exit point as they are elsewhere in the
tank. Thus, the temperature and concentration in the exit stream are modeled
as being the same as those inside the reactor"
This has me confused a lot, if concentration isn't changing with time, then does the chemical reaction has to occur instantaneously?
So the reactant is instantaneously and uniformly mixed throughout the reactor upon entry, but what about it's conversion into a product?. Because if that's taking time then the statement "concentration isn't changing with time" doesn't make any sense to me.
Can someone please clarify?

The "steady state" is used a lot in chemistry because it yields a condition which has a unique, defined and calculable answer. Obviously the reaction doesn't happen "instantly" but rather in a time frame which is short compared to how long the material takes to go from the inlet to the outlet.

The mean residence time in the reactor is the tank holdup divided by the mass flow rate.  So the amount of reaction that occurs is the reaction rate times the mean residence time times the holdup volume.


text:
Is an element's "atomic electron configuration" EXACTLY equal to the term with the largest coefficient in the CI series of the neutral atom?
For atoms and molecules with more than one electron, the wavefunction cannot be written as a single Slater determinant, so the ground-state configuration of (the atom or molecule in question) is defined as the configuration whose Slater determinant has the largest (in absolute value) coefficient in the configuration interaction(CI) infinite series expansion.
My question is the following- let one set an element, say aluminium. Then, does the term with the largest coefficient in (the CI series converging to the EXACT wavefunction of a neutral atom of aluminium) EXACTLY consist of the 13 one-electron wavefunctions corresponding to 1s²2s²2p⁶3s²3p¹, without containing any 4s, 3d, 4p(etc) one-electron wavefunctions in the determinant?

I have a feeling: no.

After reading this again, I'm quite convinced that the first paragraph is incorrect to start with. For more than one electron you can write the wave function as a single Slater determinant, it simply won't be the exact wave function. CI is, even at is limit, technically just a way to express the ground state; the configuration with the largest coefficient doesn't really have any physical meaning. It only is the best single determinant approximation. From all that I have to revise my last comment. I have a feeling: yes.


text:
Apparent paradox in the formation of ice at room temperature
The formation of ice out of liquid water can be written down like this:
$$\ce{H2O (l) <=> H2O(s)}$$
We can calculate the change in standard Gibbs free energy (per mol substance) in the following way:
$$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$$
If we do this at room temperature $(\pu{298 K}),$ we get: $\Delta G^\circ = \pu{546 J mol^-1}.$
So we can see that this process is very non-spontaneous, which is what you'd expect, you never see ice forming at room temperature. However if we calculate the equilibrium constant $K_\mathrm{eq}$ for this reaction we get:
$$K_\mathrm{eq} = \exp\left(-\frac{\Delta G^\circ}{RT}\right) \approx \mathrm e^{-0.22} \approx 0.80$$
It's obvious that this is wrong, because at room temperature we never see liquid water in equilibrium with a decent amount of ice. I just don't know what I did wrong, I have checked my units and calculations but can't seem to find what I did wrong.

You did it right. That's the thing with pure liquids and solids: even if the constant is not very far from 1, the reaction goes all the way to the end.

How did you get the entropy change from liquid water to ice at 298 K and 1 bar?

First of all, you should drive away this misconception of yours that every reaction needs to be in equilibrium, secondly, even if we have an equilibrium that doesn't mean the concentration of product is any appreciable value.

It's best to avoid the phrase "very non-spontaneous". Reactions are spontaneous or not or at equilibrium. There is no degree of spontaneity.

The difference between phase equilibria and chemical reactions is that in phase equilibria, chemical potential (aka activities) is not really a function of concentration in the system of interest.  If "ice" as a separate species could dissolve or distribute freely in "liquid" water, or vice versa, then your calculations would make sense.  But, that isn't what is happening.

Now that you've got $K_{eq}$, you need to take a look at the expression for the equilibrium constant, which is $$K_{eq} = \frac{a_{\text{ice}}}{a_{\text{water}}},$$ where $a$ is the activity of each species.
For a pure liquid and a pure solid, the activity is defined as 1, so the expression on the right hand side of that equation (usually referred to as the reaction quotient $Q$) can only take on three values:
1 if ice and water are both present
0 if water is present but not ice
undefined if water is absent, whether or not there is ice.
So if a mixture of ice and water is put in a room temperature environment, the ice melts because the reaction quotient of 1 is greater than the equilibrium constant of 0.8. Only when all of the ice melts and it makes the step change to $Q=0$ does it get below the equilibrium constant. Then we have a situation where anytime any infinitesimally small amount of ice crystallizes, then $Q$ exceeds $K_{eq}$ and the crystal returns to liquid form.

But following the same train of thought, when the ice has melted and Q=0 the equilibrium constant is still 0.8. Doesn't that drive it to freeze again? After all, the system aims to be at $K_{eq}$?

@BuckThorn - Yes indeed. But as soon as the amount of ice is non-zero, the driving force immediately switches in favor of melting (b/c Q=1) so they never get beyond an amount that is functionally 0. If you observed every molecule in the liquid over time, you'd see frequent assemblies of a few molecules into a solid-like structure, but they'd immediately get hit by other molecules and broken apart. Only when the T gets down to 273 K is the average KE low enough that those assemblies can continue to grow without being broken up.

@BuckThorn or you consider a third state, nuclei. They are dissolved in water, have a concentration, and will be in equilibrium with liquid water.

@Andrew would equilibrium constant for same reaction at 273 K greater than or less than or equal to 1?

There is a fallacy in the discussion that in this context the notion of an equilibrium constant retains its usual meaning (as rightfully questioned in a comment). It doesn't. An equilibrium constant arises when you assume that various chemical components (in the same or different phases) can coexist. The condition for coexistence of two phases is that the chemical potential of each component in all phases is the same. For liquid water in equilibrium with ice, one would write
$$\mu_\ce{H2O(\textrm{l})} = \mu_\ce{H2O(\textrm{s})} $$
But such an equation holds only at the melting point. At other points the chemical potentials are not equal, so that while a hypothetical equilibrium constant can be written, it is in practice without meaning as a proper "equilibrium constant". Such a parameter might be used to interpret the direction in which the system will change, as suggested  in another answer, but it no longer represents a ratio of actual activities in the system at equilibrium. The interpretation is much simpler than that:
$$\begin{align} K<1 &\rightarrow \text{reaction proceeds completely to the right}\\ K>1 &\rightarrow \text{reaction proceeds completely to the left}  \end{align}$$

I don't understand why this is not a "proper equilibrium constant". If Keq =0.8 and the system is allowed to equilibrate and then held at that temp, the time average of the ratio of the activities will converge to 0.8, just as with any other system in equilibrium.  Your answer applies more to systems where Keq is so large that the reverse reaction is functionally impossible because it is so improbable.

@Andrew it's not an equilibrium constant in the strict sense because you can't have equilibrium between the phases (except at the boiling point, when trivially $K_{eq}=1$ ). I am being literal. The equilibrium constant follows from an initial assumption that the chemical potentials of the species in question are equal (in equilibrium), but in this case they can't be.

From a practical standpoint, the phases are not in equilibrium because there's never enough solid to matter, but from a literal thermodynamic standpoint, we certainly do have two phases in equilibrium at any temperature, with the caveat that we must consider the time average, which is a caveat that applies to almost any system with a finite number of molecules. Because we can only have integer numbers of reactant and product molecules, Q at a specific time instant for any reaction "at equilibrium" is rarely exactly equal to K, but the limit of the time average of Q is K.


text:
Will "soft nucleophiles" undergo 1,4-conjugate addition even if the β-carbon is sterically hindered?
In general soft nucleophiles do 1,4 conjugate addition to a carbonyl carbon under thermodynamic control. Will soft nucleophiles like organocuprates still preferentially undergo 1,4 addition if the β-carbon is more sterically hindered than the carbonyl carbon?
For example, if I consider the following case

should I just accept it as a rule of thumb that these "soft nucleophiles" undergo 1,4-nucleophilic addition to the carbonyl group, or does it vary depending on the structure and the situation?
In addition to the example above, what happens if there is a methyl group at C-2 of the ring?

I don't see much steric hindrance in the example case.

Regarding your edit: The terms 'ortho', 'meta', and 'para' are only applicable to benzene rings. You have a cyclohexane ring here, those terms are not correct.

@orthocresol Though I recall "ortho product" and "para product / selectivity"  when describing the outcome of asymetric dienes (e.g., methacrylate) with asymetric dienes (e.g., 2-methyl-buta-1,3-diene) in the course of Diels-Alder reactions, too.

@Buttonwood Hmm, I recall Clayden uses these terms, but I think it's meant to be just a helpful mnemonic, not a serious scientific terminology.

Your line of thought possibly is influenced by the Walden inversion which is characteristic for the nucleophilic substitution mechanism of $\mathrm{S_N2}$ where steric hindrance affects the rate when this umbrella-like structure flips upon entry of the nuclophile and departure of the nucleofuge around a formally $\mathrm{sp^3}$ hybridized center:

(credit)
The reaction shown by you however is an addition on to a carbon with molecular orbitals formally hybridized as $\mathrm{sp^2}$.  Similar to the Bürgi–Dunitz trajectory for the entry of a nucleophile on to a $\ce{C=O}$ bond a) it is plausible that the angle of attack for the 1,4-addition b) equally is tilted, rather than an attack from the back.

(composed with a picture from here)

My impression (I never did much real-life organic chemistry) is that depending on the size of R and R', nucleophilic attack on the alkene (even via the B–D trajectory) can be quite tough: it probably depends on how big the nucleophile is, too. However, I can't say that I'm an expert on this, so had better leave it to the proper organic chemists :-)


text:
Why doesn't acetic acid affect the formation of beta-nitroamine in nitro-Mannich reaction?
In a nitro-Mannich reaction, if we deprotonate a nitroalkane species with a strong base, then add acetic acid followed by an imine species, the reaction still proceeds giving the $\beta$-nitroamine.
Why is that the case and why wouldn’t we get protonation of the nitro anion species hindering a subsequent reaction with the imine?


text:
Why is enthalpy change = heat change only for constant pressure?
Please help me, a mathematician doing his elective course on Physical Chemistry, out with this very simple question: why does the relation between enthalpy and heat $$\Delta H = \Delta U + p\,\Delta V = \Delta Q$$ hold only under constant pressure? With the integral definition of work, couldn't one just as well write $$\Delta H = \Delta U + \int_{V_i}^{V_f} p(V)\,\mathrm dV = \Delta Q $$ in the case where the pressure is not constant during expansion?
I'm still having some trouble with chemical notation of derivatives/infinitesimals as well as with the physical realities behind the formulae, so there might any number of basic things I'm missing here.
Thanks in advance. Also as this is my first question here, apologies if I did anything wrong or if I missed a similar question.

As a mathematician, would you say that $\Delta (PV)=\int{PdV}$

See my answer here:   https://chemistry.stackexchange.com/questions/126218/need-help-understanding-enthalpy/126262#126262  Also, using $\Delta Q$ is incorrect; please see my answer here:  https://chemistry.stackexchange.com/questions/136237/difference-between-Δq-and-dq/136254#136254

@Chet Miller Thanks, I believe that is the critical error I was making: if $\Delta (PV)$ is defined as $P(V_2) V_2 - P(V_1) V_1$, then of course it can only be equal to the integral if the pressure is in fact constant. What an embarrassing mistake to make as a maths major, but here we are...

@Eriol Have you got an answer to accept ?

Firstly, $\int_{V_i}^{V_f} p\, dV\neq p \Delta V$ unless $p$ is constant. Hence the formula holds in that case only. Also, the definition of $H=U+pV$, hence if we wanted $\Delta H$, we should take:$$\Delta H=\Delta U +\int_{V_i}^{V_f}\,d(pV)$$
Here, of course, $d(pV)=pdV+Vdp$ using the product rule.

Thank you, as mentioned above my main error was indeed confusing $\Delta (pV) $ and $\int_{V_i}^{V_f} p dV$, but I appreciate your additional clarifications.

@Eriol If you believe, this answered your question than consider upvoting and accepting the answer.

@NisargBhavsar Oh sorry, first post, didn't know I could to do that...I accepted it now, I think.

The enthalpy may be calculated even when the pressure is not constant, for example if the volume is constant. But it is much easier to measure a heat effect at constant pressure, usually at atmospheric pressure. You don't need complicated apparatus like an autoclave. You just need a thermometer. This is why you find enthalpies of formation or enthalpies of reaction in tables, and not the corresponding values of internal energies.

While you can calculate the enthlapy change for any process, it will only be equal to the heat flow under three conditions: (1) closed system; (2) the only type of work is pV-work; (3) constant pressure.

OP the comment by @theorist is likely the most straightforward answer. The formulas aren't necessarily general definition but those of convenient practical use. That is why in chemistry more often you encounter Gibbs energy and less often Helmotz energy, too.


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

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

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

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

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

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

This is a good answer, but it might help even more to have on the diagram a second pathway with lower activation energy but higher energy final products to show more clearly how one product can be favored thermodynamically and another kinetically.


text:
Trial by fire: Safely heating sand without it agglomerating?
Wikipidia quotes a melting point of 1713C for sand.
I went on the beach to get some sand. I assume it's a mix of all kinds of stuff.
The finality is to prepare a big pile of clean sand to use as filtering aid above paper filter (among other uses).
I'd like to get rid of plastics & organics by heating it up.
What temperature can I bring it up to before the smallest particles start to agglomerate? What temperature would you recommend for pyrolysing as much of everything in there that is not SiO2 such that it either degrades or later more easily disssolves in water?

Assuming you only need a couple of kilos or so, any reason to not buy clean washed sand from one of the chemical supply houses? Also, that melting point is probably for pure silica (quartz) sand. Real beach sand is all sorts of stuff.

I don't know that sand would be perfectly clean anyways, so, since tons of sand is available for free where I'm at I'm thinking why not clean it myself.

Perfectly clean is not reasonable, but have you priced silica, such as chromatographers might functionalize and use as column packing? Anyway, enough said: best of success and avoid steam explosions from unexpectedly wet sand!

Sand is not available as a reagent grade chemical, but rather is mineral origin particles of a size between 0.074 and 4.75 millimeters. Gravel is bigger, silt is smaller.
Pure silica (SiO$_2$) may melt at 1713$^o$, but the melting point of real sand is given as 1500 to 1610, and you can probably find some numbers outside that range.
Perhaps more of an issue than the exact melting point is how much are you preparing to treat? And with what heat source? And what sort of vessel are you going to contain the sand in while you heat it up and cool it down? And for what purpose?
Thin layers of sand (small overall quantity) may burn off organics relatively easily, say in a stainless steel measuring cup, with a propane torch  - although, the handle will get hot, so arrange some sort of apparatus to hold the vessel.
Different materials will need different temperatures and times to burn off completely; perhaps you can heat till smoke stops coming off and the sand looks like it is a bright red or orange color. I doubt that you would have to get to a yellow.

Thank you for this great answer! So around 1000C should do fine for this purpose and it should be low enough that it doesn't stick together (assuming >95% Quartz)?

If you heat in a sturdy pan, heating to 1000 C with some shaking should keep sand particles separate. Beach sand may contain some salts which could melt and bind sand grains together, but water would dissolve that and free up the sand. The best way to find out is to try it - just be safe: wear protective gloves, keep away from other burnable material, etc, etc.

I love answering questions that have a significant downvote. The downvote points to very different viewpoints, which is the basic source of the question/issue/problem. The interesting question only exists because someone is thinking outside the box, or because some instructor has described either an incomplete or perhaps imaginary box.

It is the mark of a noble mind to, at times, intinctively go against the natural reflex of the masses/majority.

@Hans above 500C is good enough, hydrocarbons break down at that temperature, so typical open flame without anything extreame will do. If oxidiser(air) can access the stuff it more than enough, if not then 600-700C is more than enough.


text:
Concentration of weak acid neutralized by strong base
I've got a question that should be simple enough but I must be doing something very wrong.

A $\pu{10mL}$ sample of $\ce{H2SO3}$ is neutralized by $\pu{18.6mL}$ of a $\pu{0.10M}$ strong base. Find the concentration of the acid.

My solution:

Given the information, I calculate $\pu{1.86E-3 mol}$ of $\ce{OH-}$ used for this reaction, so an equivalent amount of $\ce{H+}$ ions is assumed. Therefore the $\ce{[H+]} = 1.86 \times 10^{-3} / \pu{0.01L}  =  \pu{0.186M}$.

$K_\mathrm{a}$ of sulphurous acid is $1.5\times 10^{-2}$. $K_\mathrm{a} = \frac{\ce{[H+][A-]}}{ \ce{[HA]}}$. Assuming that $\ce{[H+]} = \ce{[A-]}$, the $\ce{[H2SO3]}$ at end of reaction is $\pu{2.3064M}$.

However, this is obviously wrong compared to the given answer, $\pu{0.093M}$. Is the "concentration of acid" not $\ce{[HA]}$? Am I making any incorrect assumptions?

Your title is very wrong. Short of that, you have a common misconception that [H+] matters in the neutralization business. It doesn't.
Also, see https://chemistry.stackexchange.com/questions/60407/why-will-a-strong-acid-neutralize-as-much-base-as-a-weak-acid/60409#60409

A strong base would be something like sodium hydroxide which which would react to neutralize both hydrogens of the $\ce{H2SO3}$ molecule.

@MaxW so how should I change my process based on that information? Something like [H+]^2 [A-] / [HA] = Ka?

Forget Ka. Just divide 0.186 by two.

Assuming NaOH for the base, the overall chemical reaction would be: $$\ce{H2SO3 + 2NaOH -> Na2SO3 + 2H2O}$$

A one loose fault of the question is not giving the nature of strong acid. Thus, it is safe to assume that $\ce{NaOH}$ is the strong base, and the overall chemical reaction can be written as:
$$\ce{H2SO3 + 2 NaOH -> Na2SO3 + 2 H2O}$$
Thus, each $\pu{mol}$ of $\ce{H2SO3}$ need $\pu{2 mol}$ of $\ce{NaOH}$ to neutralize. You have correctly calculate  $\pu{1.86 \times 10^{−3} mol}$ of $\ce{OH−}$ used for this neutralization reaction. Therefore, amount of $\ce{H2SO3}$ presence in the $\pu{10.0 mL}$ of solution is $\frac12 \times \pu{1.86 \times 10^{−3} mol} = \pu{9.30 \times 10^{−4} mol}$. Thus, molarity of $\ce{H2SO3}$ solution is:
$$\frac{\pu{9.30 \times 10^{−4} mol}}{\pu{10.0 \times 10^{−3} L}} = \pu{9.30 \times 10^{−2} mol L-1} = \pu{0.093 M}$$


text:
Do organopolonium compounds exist?
Analogues of alcohols exist for all the heavier Group 16 elements, namely sulfur, selenium, and tellurium. Would polonium also be able to form a "polonol" like $\ce{CH3PoH}$?

It probably should.

it is monstrously hard to do chemistry on something as radioactive as polonium.  A 1g sample would produce about 140W of heat from its radioactivity and would glow red. Small samples slowly evaporate (probably driven by alpha-particle spallation) making them very hazardous to work with. And even the chemistry in dilute solutions is heavily affected by reactions driven by the radiation not the inherent chemistry of the metal.

@matt not just for polonium, as you know. As one indication, look at the list of known oxidation states and note that most elements with less than two known states are highly radioactive ones with all isotopes having short half-lives.

As one might expect for an element with no long-lived isotopes, little is known about organopolonium compounds. What is known may be broadly divided into two categories [1,2; cited by Wikipedia]:

Compounds where polonium takes the place of a lighter chalcogen, acting more nearly parallel to sulfur, selenium and tellurium than to oxygen. Examples include polonoethers $\ce{R2Po}$ and arylpolonium halides $\ce{Ar2PoX2,Ar3PoX}$.

Complexes similar to those formed with metals.  WP cites complexes with 2,3-butanediol and thiourea from Ref. [1].

All such compounds are known only in tracer amounts because of both chemical instability and rapid radioactive decay.
References
1.
Zingaro, Ralph A. (2011). "Polonium: Organometallic Chemistry". Encyclopedia of Inorganic and Bioinorganic Chemistry. John Wiley & Sons. p. 1–3. https://doi.org/10.1002/9781119951438.eibc0182.
2.
Murin, A. N.; Nefedov, V. D.; Zaitsev, V. M.; Grachev, S. A. (1960). "Production of organopolonium compounds by using chemical alterations taking place during the β-decay of RaE" (PDF). Dokl. Akad. Nauk SSSR (in Russian). 133 (1): 123–125. Retrieved 12 April 2020.


text:
Homemade diode working mechanism

Hello,
I will list the questions directly, if you want to know details please read below.
As in the figure, we have water+baking soda in a plastic cup, there are 2 versions in figure that affects LED's brightness and current that flows in circuit.

In 1st version of figure, LED is bright and more current flows
(measured 9.76 mA with multimeter)
In 2nd version of figure, LED is
dull and less current flows (measured 0.22 mA with multimeter)

So my question is, why changing aluminium's place from negative to positive affects current flow of circuit? is it related with aluminium's chemical properties?
Details that explain how i got to this question:
My main aim was to make a home made diode by following instructions in this wonderful site, i did one and it worked.To understand the working mechanism better, i decided to setup things as in the figure upper and faced with this question that looks like main working logic of this home-made diode.In site there was an explanation but i couldn't understand it.

Hopefully, what I wrote as a comment, since converted to an answer, will not turn out to be duplicative of an existing answer: I wrote it just before calling it the end of a long day finishing my basement echelle spectrograph construction (thanks, @M.Farooq !).

When aluminum is the anode (connected to the positive battery terminal), a thin layer of insulating aluminum oxide is produced via the “anodization”. It serves as a dielectric, so basically you have a leaky capacitor: current is low. But when the aluminum is connected to the negative terminal, it is the cathode and then you get reduction of hydrogen ions to hydrogen gas. So considerably more current can flow. This is essentially how tantalum and aluminum electrolytic capacitors have rectifying properties: their oxide films are the cause. Fun link, by the way.


text:
Comparing nucleophilic strengths of tetrahydropyran and (2R,6R)-2,6 dimethyltetrahydropyran

Compare the nucleophilic strengths of the following two compounds:

According to me it should be the second one, that is, (2R,6R)-2,6-dimethyltetrahydropyran because the methyl groups would exert +I effect which would increase the electron density on the oxygen atom and hence it would become more nucleophilic, but the answer in my book is given as the first one. I don't understand why is it so. Is it because the effects of the methyl groups would cancel as they are connected in trans manner? But then I don't understand why would the first one still have a higher nucleophilicity.

The steric hindrance is a much more significant factor here than the inductive effect

It's similar to SIP effect. Search for SIP, there are many reputable sources with information on it.


text:
What is the difference between bond enthalpy and bond formation enthalpy?
I came across the following while going through my textbook

The standard enthalpy of reaction is related to bond enthalpies of reactants and products in gas phase reactions as: $$\Delta H^{\circ}_r = \Sigma(\textrm{bond enthalpies})_{\textrm{reactants}}-\Sigma(\textrm{bond enthalpies})_{\textrm{products}}$$  If we use enthalpy of bond formation, then $$\Delta H^{\circ}_r = \Sigma \Delta_f H^{\circ}_{\textrm{bonds of products}}-\Sigma\Delta_f H^{\circ}_{\textrm{bonds of reactants}}$$

My Questions

How is bond formation enthalpy different from bond enthalpy?
Why does the equation gets reversed when we are using bond formation enthalpy?
Bond formation enthalpy and enthalpy of formation of a substance are different, right?

Indeed precisely as written the box means nothing. Stick to the answer you have got.

This "enthalpy of bond formation" you are referring to is simply "enthalpy of formation", that is, the enthalpy of reaction when one mole of a substance is formed in its reference state from its constituent elements.
On the other hand, "bond enthalpy" is the enthalpy required to break one mole of a particular type of bond, all of this under standard conditions.
These quantities represent opposite phenomena, hence the equation gets reversed.


text:
Reactivity of NBS

The given correct answer is (A).
According to me there would be 2 enantiomeric pairs.
My doubt is that it does not react with NBS but it does react with Cl/hv though both have the same initiation step.
My guess is that this may be because Br is present in very small amount i.e. as an impurity so this may have something to do with it.

Look here for the final product. It doesn't have active hydrogen (allylic or benzylic) to react with NBS.  C and D are also correct. But it would produce more than one enantiomeric pairs when reacted with $\ce{Cl2}/h\nu$ and therefore A is the correct answer.

Initially, I wasn't going to give an answer to this question but suggested the choice A is the correct answer in my comment. However, since Random Quark hasn't give a satisfactory answer (yet I appreciate the explanation), it prompt me to give my thoughts as an answer.
As I pointed out in my comment to the question, the reaction given by OP is an example of Mozingo Reduction:

Thus, the final product is methylcyclopentane, which does not contain any acidic hydrogen because there are no carbonyl or double bond on the structure. Also, although the starting compound, (2S)-2-methylcyclopentanone is an optically active compound, it would loose its chirality due to reduction (gaining plane of symmetry).

As shown in the above diagram (insert in the bottom box), methylcyclopentane will get chlorinated in the presence of $\ce{Cl2}/h\nu$ at 1, 2, and 3-positions, and on the methyl group as well (only 2- and 3-chloro products are shown because they are the compounds in concern). It is also possible to get di-chlorinated, tri-chlorinated, and etc. based on the conditions used. Thus, as shown in the scheme above there would be more than one enantiomer pair by this reaction. Therefore, the statement A is incorrect.
As pointed out above, the final product does not contain any acidic hydrogen because there are no carbonyl or double bond on the structure. Therefore it cannot react with the presence of NBS with or without initiator. Therefore, the statement B is correct (and methylcyclopentane is optically inactive as well).
Two optically active alkenes (they probably mean by (3R)-3-methylcyclopentene and (3S)-3-methylcyclopentene) would give the final product (methylcyclopentane) when reduced by $\ce{H2}$ $(\pu{1 mol})$.  Therefore, the statement C is also correct.
Finally, the reaction with $\ce{H2/Raney Ni}$ is a redox reaction. Thus, the statement D is correct as well.

Hence, the only incorrect statement is A and consequently, it is the correct choice as given.

Firstly, about (A), you are correct that two enantiomeric pairs are formed. You have misread the question, it says the INCORRECT option.
Then coming to why Cl2/hv reacts but NBS doesn't give its usual bromine substitution. As you already know, these reactions are free radical reactions that need a free radical initiator like light, peroxide, heat, etc. Cl2/hv has the presence of sunlight that is able to do the homolytic cleavage of the Cl-Cl bond.
NBS adopts the following mechanism.
The initiation step in allylic and benzylic bromination with NBS is the formation of a
bromine atom by homolytic cleavage of the N-Br bond in NBS itself.

Wondering where the Br2 came from? The image doesn't show that but HBr formed reacts with NBS to give N-Bromosuccinamide and Br2.
The unique role of NBS is to maintain the very low concentration of
bromine by reacting with HBr.The low solubility of NBS in CCl4 (<0.005 M) is important for bromination in NBS. Different solvents can show different reactions while it may be so that NBS all alone doesn't give any reaction until we have a free radical initiator. Hence, CCl4 must be used as the solvent in allylic or benzylic bromination (I mean not necessarily allylic and benzylic only) with NBS. During the reaction, the insoluble NBS, which is denser than CCl4, disappears from the bottom of the flask, and the less dense by-product, succinimide forms a layer on the surface of the CCl4. Possibly the further reactions occur at this surface (who knows!).
TL;DR NBS all alone won't give any reaction here. The presence of a suitable solvent is the key to its bromination.

This is Mozingo's Reduction, and product formed would be methylcyclopentane. The reason $\ce{Cl}$ free radical reacts with this compound, but NBS doesn't is because $\ce{Br}$ free radical is formed in low concentration in NBS, so it only reacts to form stable free radical, namely, allylic and benzylic free radicals only. Also, you are right, $2$ enantiomeric pairs would be formed, the question is asking for $\mathbf {incorrect}$ statements.


text:
What determines the product at the anode/cathode of a electrolytic cell with an aqueous solution?
If you perform electrolysis on Aqueous Sodium Chloride of various concentrations, how does the product at the anode change but the product at the cathode does not?
What is the mechanism behind this change?

At the cathode, only hydrogen is produced, and the solution becomes charged in $\ce{OH-}$ ions, because of the reaction : $$\ce{2 H2O + 2 e- -> H2 + 2 OH-}$$
At the anode, chlorine is produced by the reaction : $$\ce{2 Cl- -> Cl2 +  2 e-}$$ But this chlorine $\ce{Cl2}$ is relatively soluble into water. And the ion $\ce{OH-}$ produced at the cathode migrates towards the anode. So, at the anode,  the solution will soon contain new species like $\ce{OH-}$ and $\ce{Cl2}$ which do react with one another to produce $$\ce{Cl2 + 2 OH- -> ClO^- + Cl^- + H2O}$$ So a new ion appears in solution : $\ce{ClO-}$ . The trouble is that, if the solution is heated, the hypochlorite ion $\ce{ClO-}$ gets decomposed according to : $$\ce{3 ClO- -> ClO3^- + 2 Cl-}$$ So you see that the anodic compartment chemistry is more complicated than the cathodic compartment.


text:
Why is the 2-bond (Sn-H) coupling constant larger than the 3-bond Sn-H coupling constant for tetraethyltin?
I was looking at the NMR spectrum of tetraethyltin [$\ce{Sn(CH2CH3)4}$], and found that the $\ce{Sn-H}$ coupling constant between $\ce{Sn}$ and $\ce{CH2}$ is larger than that between $\ce{Sn}$ and $\ce{CH3}$.
$\hspace{4.1cm}$

$$\begin{array} {|r|r|}\hline {}^2J({}^{119}\ce{Sn-H}) & {}^3J({}^{119}\ce{Sn-H}) \\ \hline \pu{49.2 Hz} & \pu{68.6Hz} \\ \hline  \end{array}$$
This trend is confirmed by literature values[1].
However, I am confused as to why this is happening. The magnitude of the coupling constant is determined by the Fermi contact mechanism, so surely the coupling constant over two bonds should be larger than the coupling constant over 3 bonds? The ethyl groups also should have free rotation, so any type of angle effects should be cancelled out on average.
So why is ${}^2J$ smaller than ${}^3J$ here?

Reference:

G. Barbieri, F. Taddei, "117Sn, 119Sn–proton long-range coupling constants in the 1H nuclear magnetic resonance spectra of alkylhalogenostannanes", J. Chem. Soc., Perkin Trans. 2, 1972, 10, 1327

Presumably it’s not just Fermi contact interactions. Note that 2J and 3J of 13C-1H have the same pattern - there’s a paper on this which I will try to find...

@orthocresol Thank you. So, are there other mechanisms except Fermi contact for nuclear spin coupling? I was only aware of Fermi contact. And please mention the paper if you find it, that would be very helpful.

This is quite a broad area and one that I'm not qualified to discuss myself, but here are a few relevant papers: https://doi.org/10.1021/jp0353284 (talks about C–H), https://doi.org/10.1021/jo200513q (talks about H–H where Fermi contact is the most important factor, but also summarises some of the other factors), https://doi.org/10.1002/chem.200306065 (discusses factors), https://doi.org/10.1021/ed084p156 (maybe the most accessible)


text:
How to balance the reaction equation between copper (I) sulfide and nitric acid?

What is the sum of stoichiometric coefficients of redox reaction: $\ce{Cu2S + HNO3 -> S + NO2 + Cu(NO3)2 + H2O}$?

This exercise is from a multiple choice exam. I have balanced the equation:
$$\ce{1 Cu2S + 4 HNO3 -> 1 S + 2 NO2 + 1 Cu(NO3)2 + 2 H2O.}$$
Therefore I have calculated the sum of coefficients as $\sum a = 11$. This has not been offered as an answer and the correct answer is 20.

The exact equation is : $\ce{Cu2S + 8 HNO3 -> 2 Cu(NO3)2 + 4 NO2 + S + 4 H2O}$

It'd be beneficial if you show how you balanced the equation. That way we'd see where you make the error.

At first glance, your $\ce{Cu}$ is not balanced.

$$\ce{Cu2S + HNO3 -> S + NO2 + Cu(NO3)2 + H2O}$$
Let's balance this reaction using the half-reaction method.
The oxidation half reaction (O.H.R) would be:
$$\ce{\overset{+1}{Cu}_2\overset{-2}{S} -> 2Cu^2+ + S^0 + 4e-}\tag{O.H.R}\label{ohr}$$
The reduction half reaction (R.H.R) here is:
$$\ce{H\overset{+5}{N}O3 +e- -> \overset{+4}{N}O2} \tag{R.H.R}\label{rhr}$$
The final reaction would be $4\times\eqref{rhr} + 1\times\eqref{ohr}$.
Therefore the reaction would be partially balanced as:
$$\ce{Cu2S + 4HNO3 -> 2 Cu^2+ + S + 4NO2}$$
Now, we have $\ce{Cu(NO3)2}$ and not $\ce{Cu^2+}$, so we need to add the 4 more nitrate ions which implies 4 more molecules of $\ce{HNO3}$
So now, we have the equation as:
$$\ce{Cu2S + 8HNO3 -> 2 Cu(NO3)2 + S + 4NO2}$$
Now, we balance the hydrogens and oxygens, the left has 8 hydrogens and 24 oxygens, whereas the right has no hydrogens and 20 oxygens. Therefore, we need to add 4 water molecules to the right. Therefore, the final reaction becomes:
$$\ce{Cu2S + 8 HNO3 -> S + 4 NO2 + 2Cu(NO3)2 + 4 H2O}$$
Now, $\sum a = 1 + 8 + 1 + 4 + 2 + 4 = 20$ and not 23.

RHR = right hand reaction? OHR = other hand reaction? Use \label{something} and \eqref{something} to make clear what you mean.

@Martin-マーチン label{} and \eqref{} seems to  break in my answer when it is displayed, i've modified the answer to fix the issues you have pointed out.

Works like a charm; MathJax usually requires an update/refresh though.


text:
Need help to calculate results of an electrochemistry cell potential lab work
So, the lab work is this. I made an AgCl electrode of a silver wire by coating it with AgCl. This was done by electrolysis by immersing the wire in HCl and using platinum as a counter electrode and applying current to the system. After that, the AgCl electrode was immersed into HCl dilutions of different molalities and the cell potential was measured against a hydrogen gas electrode (normal hydrogen electrode NHE).
The measurements were done at temperature of 25 °C. The results of the measurements are as follows:
\begin{array}{llc} \hline
c\,(\pu{mol/kg}) &\sqrt{c}\, (\sqrt{\pu{mol/kg}}) & \mathrm{E\,(V)} \\ \hline
0.001 & 0.032 & 0.6322 \\
0.002 & 0.045 & 0.6815 \\
0.004 & 0.063 & 0.6497 \\
0.006 & 0.077 & 0.6280 \\
0.008 & 0.089 & 0.6153 \\
0.010 & 0.100 & 0.6049 \\
0.020 & 0.141 & 0.5700 \\
0.050 & 0.224 & 0.5030 \\ \hline
\end{array}
Because the cell potential can be calculated by (as the pressure gradient of the hydrogen electrode was 0):
$$ E(m)= E^0 + \frac{2ART}{F} \sqrt{\frac{m_{\text{HCl}}}{m^0}} $$
By plotting the table values we get (excluding the first data point as an outlier):

We can see from the equation that the $E^0$ would be about 0,708 V. However, literature references say the standard potential of the AgCl electrode is about 0,222 V. Any idea what I am doing wrong here?
Then I also should plot the HCl mean activity coefficient values calculated from the formula:
$$
E(m) = E^0 - \frac{2RT}{F} \ln \left[ \gamma_{\text{HCl}} \left( \frac{m_{\text{HCl}}}{1\ \text{mol/kg}} \right) \right] \\
\Leftrightarrow \ \gamma_{\text{HCl}} = e^{ \frac{(E^0-E)F}{2RT} } / \frac{m_{\text{HCl}}}{1\ \text{mol/kg}}
$$
Thus I get values:
\begin{array}{cc} \hline
m & \gamma_\ce{HCl} \\
\pu{mol/kg} \\ \hline
0.001 & 4371.98 \\
0.002 & 837.44 \\
0.004 & 777.51 \\
0.006 & 790.73 \\
0.008 & 759.33 \\
0.01 & 743.74 \\
0.02 & 733.46 \\
0.05 & 1080.77 \\ \hline
\end{array}
But literature references (CRC Handbook of Chemistry and Physics - 97th Edition, p. 5-101) gives greatly different values:
\begin{array}{cc} \hline
m & \gamma_\ce{HCl} \\
\pu{mol/kg} \\ \hline
0.001 & 0.965 \\
0.002 & 0.952 \\
0.004 & 0.929 \\
0.01 & 0.905 \\
0.02 & 0.876 \\
0.05 & 0.832 \\ \hline
\end{array}
Any help would be greatly appreciated.

You say hydrogen electrode. Is is a standard hydrogen electrode (SHE)? I have only ever met one person in my life, the late Prof. Roger Bates, who used an actual SHE, at the former NBS no less. They are notoriously finicky: freshly platinized platinum black on platinum, etc. Maybe try the Ag/AgCl electrode versus something easier and not finicky?

I used normal hydrogen electrode (NHE).

Apparently your measurements tend to the potential of silver and not of AgCl.

That is actually true, that the values tend to point in the direction of a silver electrode (standard potential of 0.800 V). But how could that be, the electrolyte used in the lab work was hydrochloric acid (HCl). It does not contain any silver cations to be reduced to silver metal.

May be your coating of $\ce{AgCl}$ is not perfect. It may slowly dissolve to give enough $\ce{Ag+}$ ions to the solution.

The silver-silver chloride is a half cell. When the silver wire coated with AgCl is dipped into HCl solutions with no Ag$^+$ cations, as confirmed by the OP, it is not a half cell with AgCl in equilibrium with silver and silver chloride, but a half cell that is a silver anode corroding in acid solution (with oxygen from air?).
After some time (hours?), with stirring of the solution, enough AgCl may dissolve, still leaving enough on the silver wire, dropping the 0.5-0.7 V to wind up with the desired 0.222 V. Perhaps the HCl solutions need to be saturated with AgCl and deoxygenated before the immersion of the wire electrode.

Basically OP worked with a quasi AgCl wire. It should not be expected to have right the Ag+/AgCl potential.


text:
Radial node in p or d-orbital
While studying atomic orbitals and nodes, the only illustration of a radial node that I could find is in an s-orbital (like here). How do the radial nodes look like in p or d-orbitals?
(I don't know how to plot the distribution myself.)

A Google image search will provide many different depictions.

I find the following picture from Chemistry Libretexts. I don't own the picture, so the credit goes to respective authors.

I believe the d orbitals illustrated on the diagram are only the $d_{z^2}$ orbitals. For the f orbital, I believe it is $f_{z^3}$. Please let me know of any problems/suggestions.


text:
Why is the reverse aldol cleavage of fructose 1,6-bisphosphate highly endergonic under standard conditions?
I have another question similar to this one.  I just don't have intuition for why reactions are thermodynamically favorable or unfavorable and I'd like to build it.  This time I'd like to ask about the reverse aldol reaction $$\ce{Fructose-1,6-Bisphosphate -> Dihydroxyacetone Phosphate + Glyceraldehyde-3-Phosphate}$$
The standard free energy of this reaction is +23.8 kJ/mol.[1]
I was hoping someone could help me understand why.  The larger molecule has lower entropy and charge-charge repulsion.  Why is it so thermodynamically unfavorable for it to break apart?
[1]
Cornish-Bowden, A. Thermodynamic Aspects of Glycolysis. Biochem. Educ. 1981, 9 (4), 133–137.

Take a look at the bonds being formed and broken. That difference in enthalpy overwhelms the entropic consideration.


text:
Conjugated diene + Na in liq.NH3
I learned that alkynes gets reduced on addition of $\ce{Na}$ in liquid $\ce{NH3}$ to form alkenes (which is not further reduced). Recently I came across a reaction where a conjugated diene was treated with $\ce{Na}$ in liquid $\ce{NH3}$ .The product was what you would get by 1,4 adding $\ce{H2}$ in the conjugated diene.
Is this reaction correct? If yes ,then what makes it more reactive than an ordinary alkene?

Thanks!!It does not entirely answer my question."then what makes it more reactive than an ordinary alkene?"-This part to be more specific.

I think it's the stability of intermediate radical.

@MathewMahindaratne yes!can be as it forms a cyclic intermediate with Na.


text:
Determining difference between co-ordination complexes of copper
What's the difference between the co-ordination compounds $\ce{Co.(H2O)6}$ and$\ce{[Co(H2O)6]^2+}$?

First one has $\ce{Co^0}$ and second one has $\ce{Co^2+}$??

$\ce{Co·6H2O}$ does not exist. But, if it would exist, il would be a special sort of metallic cobalt, surrounded by 6 water molecules fixed around each $\ce{Co}$ atom. This substance would be electrically neutral. It would exist in a sample containing no other substance.
$\ce{[Co(H2O)6[^{2+}}$ does exist. It is a charged species (cation) that exists, but does not exist alone. It must be accompanied by one or several negative ion, like $\ce{SO4^{2-}}$ or $\ce{2 Cl-}$. There must be the same number of positive charges and of negative charges in any sample containing this ion. So there must be as many $\ce{[Co(H2O)6]^{2+}}$ cations as say $\ce{SO4^{2-}}$ anions. In solution, these ions are independent and can react differently. But the number of positive charges must always be equal to the number of negative charges.

In general, how does $\ce{X.nH2O}$ differ from $\ce{X(H2O)n}$?

@Farhan S. If $\ce{H2O}$ is attached to the preceding metal $\ce{X}$, it means that there is a covalence between them. If you separate them by a point, it means that the attraction between them is weaker than a covalence, and only due to van der Wals forces.


text:
How do atomic orbitals interfere both constructively AND destructively to give two molecular orbitals?
I am an organic chemistry student learning how atomic orbitals interfere to give rise to molecular obitals. The image below suggests that each of the hydrogens' atomic orbitals interfere both constructively AND destructively. How is this possible? More fundamentally, how does the 1s atomic orbital of a hydrogen atom have both positive and negative phase in the first place? I realize that there is a similar question on this topic but I wasn't able to understand the answer to that post due to my limited background in quantum mechanics and was hence was looking for someone who could explain in simple terms.

As there are two atomic orbitals (a0)  to begin with there are two MO's, we make the hypothesis that they are made up as linear combinations $ao_1\pm ao_2 $ as you have drawn.  The total energy is the same as the two ao's, one MO up in energy the other lower, which makes a difference when only two electrons are added but not when four are. Most phys. chem. textbooks explain this in detail but that is the main idea.

While this does not explain why the orbitals combine that way, it might still be helpful if you take in mind that the molecule is hold by only one. In this sense the molecule is not formed by both constructive and destructive interference, that concerns the orbitals as functions, which include l a phase like in the sketch.

What does MO formation entail?
A very common misconception is that the formation of MOs involve addition, or subtraction, of two physical objects. And the confusion arises because, how can the same physical object have two different phases when added or subtracted? However, this is not an accurate picture. MO construction is not a physical process which occurs between two real objects which each possess a definite phase. There are at least two reasons for this:

Orbitals are not objects. The electrons that go into orbitals are objects, but the orbitals themselves are wavefunctions.

Wavefunctions do not have definite phases. When we talk about the phase of an orbital, what it refers to is a constant $k$ which the entire wavefunction is multiplied by. It turns out that a fundamental postulate of quantum mechanics is that that the wavefunction $\psi(x)$ and the wavefunction $k\cdot\psi(x)$ represent exactly the same state, so the phase $k$ is actually completely meaningless. The 1s orbital does not possess a particular phase; it can possess any phase and still be a 1s orbital.

So, what does it really entail? I think a better description is that orbitals are possibilities for where the electrons can go. Let's say you have a hydrogen atom with one electron. This electron can either be in a 1s orbital, or (if you excite the electron) it can be in a 2s orbital, or a 2p orbital, etc. etc. So, the electron has many different options as to where it can go. Of course, in ground-state hydrogen, the eventual choice is the 1s orbital.
Alternatively, let's say you have two hydrogen nuclei, but only one electron. (So there is a net positive charge.) This electron could be in the 1s orbital of hydrogen atom A, or the 1s orbital of hydrogen atom B, or the 2s orbital of hydrogen atom A, etc. etc. These different atomic orbitals represent the possibilities which the electron can take.
In this regard, the orbitals are not objects or boxes to be inhabited by the electron, but they represent a set of choices. If we adopt this mindset, then it already becomes clearer how both bonding and antibonding orbitals form: these are simply the two "possibilities" which can happen. It is not a question of how can they both be true at the same time, because that isn't the point: the point is that MOs themselves are different possibilities for an electron to adopt, and it isn't necessary that both of them be true at the same time. Indeed, they can't be, because an electron can't simultaneously inhabit two MOs.

Cars
I hesitated to include this analogy in the original answer, but will include it now, since it may prove helpful to some. There are some problems with it, however, so don't take it as a literal description, but rather a conceptual one.
Consider, then, that you're standing on a hill overlooking a two-lane motorway which runs north to south. However, you don't have a compass on hand, so you don't actually know which way is north or south. (This is analogous to not knowing the actual phase of an orbital $k$: you can't tell whether it's $+1$ or $-1$, for example.)
Suppose you see two cars driving along the motorway, one on each lane. If you focus on one car at a time, what information can you give about each of the two cars? You can't tell which way they're headed, because you don't know north from south. However, you can tell which lane they are driving on: for example, because one lane is closer to you than the other. So you can furnish two pieces of information:

There is a car on the nearer lane. It is going either north or south, but we don't know which.

There is a car on the further lane. It is also going either north or south, but we don't know which.

This situation is exactly analogous to having two hydrogen atoms, with one electron in each 1s orbital. In other words, this corresponds to the "pre-MO" description of the $\ce{H2}$ system:

There is a 1s orbital on hydrogen atom A. It has some phase, but we don't know which.

There is a 1s orbital in hydrogen atom B. It has some phase, but we don't know which.

Now, let's return to the cars. You might think that there actually is another way we can describe their movement, and that is based on not just zooming onto one of them at a time, but rather by comparing their relative motion. So, again we have two possibilities:

There are two cars on either lane, and they are moving in the same direction. (North/north or south/south, but we don't know.)

There are two cars on either lane, and they are moving in opposite directions. (North/south or south/north, but we don't know.)

This is precisely analogous to the MO description of $\ce{H2}$:

There is a bonding orbital formed from two 1s orbitals on either atom, and they have the same phase, although we don't know exactly what that phase is.

There is an antibonding orbital formed from two 1s orbitals on either atom, and they have opposite phases, although we don't know exactly what phases those are.

Note that we didn't somehow physically stick the two cars together, or force them to travel in any particular direction. What we have added together are not cars, but rather possibilities: we combined the two "base" descriptions, of one car moving along each lane, into two "possibilities" where they are either going in the same or opposite direction. At any given point in time, whenever you see two cars going along the motorway, either of these possibilities can be realised.
The same is true of MOs. The MOs don't represent physical addition of two orbitals, but rather different possibilities of how these orbitals can combine: they can either combine constructively, or destructively. For any given electron, either of these possibilities can be realised, too: and unlike the motorway, it's not mutually exclusive, because one (or two) electron(s) can be in the bonding orbital and another can be in the antibonding orbital.

Linear algebra
This is a more formal description of how MOs are constructed, and much closer to the truth. There are very few, and possibly no, quick and dirty ways to explain MO theory, in my opinion. However, if you have studied linear algebra before,* then the transition from AOs and MOs is the same as a change of basis. A basis refers to a minimal set of vectors, from which any arbitrary vector may be constructed through linear combination. If you have a vector $(a, b)$, you can express this as: $$\begin{pmatrix}a\\b\end{pmatrix} = a\begin{pmatrix}1\\0\end{pmatrix} + b\begin{pmatrix}0\\1\end{pmatrix}.$$
It doesn't matter what $a$ and $b$ are, you can always express $(a, b)$ as a linear combination of the two basis vectors. You can think of the basis vectors $(1, 0)$ and $(0,1)$ as being the 1s orbitals on both individual atoms.
The MOs correspond to using the basis $(0.5, 0.5)$ and $(0.5, -0.5)$.† Notice that this is still a valid basis, because we can still express any arbitrary vector $(a, b)$ as a linear combination of these:
$$\begin{pmatrix}a\\b\end{pmatrix} = (a+b)\begin{pmatrix}0.5\\0.5\end{pmatrix} + (a-b)\begin{pmatrix}0.5\\-0.5\end{pmatrix}.$$
If we had rejected the "antibonding" combination $(0.5, -0.5)$, and only taken the "bonding" combination $(0.5, 0.5)$, then we would not be able to form a valid basis. For example, we could not express the vector $(1, 2)$ as just a linear combination of $(0.5, 0.5)$.
From this point of view, AOs and MOs are simply different bases.‡ The existence of both the bonding and antibonding MO is not only logical, but also mandatory in order to form a complete basis.
The familiar AOs that you have studied (1s, 2s, 2p, ...) are simply a series of basis states for the hydrogen atom. In the $\ce{H2}$ molecule, there are two sets of basis states, one for each atom. The transition from "2 sets of AOs" to "1 set of MOs" is just a change of basis: we go from two sets of atomic bases to one single molecular basis.

Epilogue
The proper answer is, of course, to pick up a QM textbook and study it. By QM I do indeed mean quantum mechanics, not just MO theory. It is a long journey, and you may not necessarily be fully prepared for it right now, but it will reward you with a better understanding of how this works. MO theory fundamentally relies on quantum mechanics, and to learn MO theory without understanding quantum mechanics is what leads to confusion and questions like yours.

Footnotes
* The analogy I use above is slightly simplified, and might look slightly contrived, but the link between QM and linear algebra is actually very deep. QM is somewhat analogous to linear algebra but on an infinite-dimensional vector space (technically, a Hilbert space).
† Technically this should be $1/\sqrt{2}$, not $0.5$. In QM we are nearly always concerned with orthonormal bases, i.e. ones where each basis vector has a length of $1$ and forms a scalar product of $0$ with every other basis vector. (Mathematically, we have a basis $\{|i\rangle\}$ for which $\langle i | j \rangle = \delta_{ij}$.)
‡ "Bases" being the plural of "basis".

Good explanation. I had a rough time learning about SALC's and AO -> MO transitions, the identity tables etc - but I had never seen the maths before. After learning what linear combination actually IS, and what it basically DOES - it was much easier to understand what is going on. A learning that happened to me much later in life, but was very appreciated in the raw "A-HA!" value.

I appreciate the effort. But some sentences simplify to much and could be improved. "Indeed, they can't be, because an electron can't simultaneously inhabit two MOs." You literally show that electrons can be in superpositions of orbitals in your linear algebra example. In the H2 example It also sounds as if the electron could only be at A or B, neglecting the fact that the electron will be in a superposition of orbitals at A and B. Breaking the math down to concepts is hard, I don't think I could have done it better, but perhaps this point could be improved a bit.

@HansWurst Please feel free to edit if you like! I did struggle with writing this answer and I honestly don't think it's perfect. But I disagree with your first point: the electron may be in a superposition $2^{-1/2}(|\psi_1\rangle + \mathrm{e}^{\mathrm{i}\theta}|\psi_2\rangle)$, but that's not the same thing as simultaneously inhabitng the two states $|\psi_1\rangle$ and $|\psi_2\rangle$.

@orthocresol I guess it is a problem of semantics but arguing semantics is pointless. Colloquially I would say that a system in a superposition state can be viewed as being partly in each of two or more other states. But in the end all semantics are error-prone in comparison to the correct mathematical expression. I think "simultaneously inhabiting two states" could be used to describe superpositions, the other case where it could be used would be the density operator $\hat\rho=c_1|\psi_1\rangle\langle\psi_1|+c_2|\psi_2\rangle\langle \psi_2|$. Would you use it there ?

Orthocresol & @Hans Wurst the superposition should be a state, not really two taken simultaneously. But I admit that all this is rather obscure to me. H2 is a stuff, it is hold by the bond we known. I can't understand this continuous emerging of superposition concepts. Please help.

@HansWurst I'm not sure how I didn't see your comment until now. Sorry. Yes, it is mostly semantics, and you put it very well, I entirely agree that all of these descriptions are just better or worse approximations to the ground truth. As for the mixed state with the density operator, if you still care about my thoughts, I still think I'd shy away from "simultaneously inhabiting". I'm not sure why I'm so against this terminology(!), but to me it sounds like we have two particles, one in state $\psi_1$ and another in state $\psi_2$, which is not necessarily true even if $\rho$ has that form.

Unfortunately, in quantum mechanics there is rarely an explanation "in simple terms." Quantum mechanics is a mathematical construct that so far seems to predict the results of all experiments that have been done to test it, but explaining those maths in terms of anything we have learned from the larger-than-quantum scale "classical" systems has so far eluded us. In fact, attempting to understand QM by applying intuition from classical physics generally causes more confusion, as seems to be the case here.
It seems that you are considering the two H-atom 1s orbitals to be static standing wave systems that can only combine constructively or destructively, which would be the case if these were classical standing waves. But they aren't. They're quantum mechanical systems that live by different rules. When the two atoms come together, the whole system is changed, and the wave function describing the new system is not just an overlay of the previously separated "waves". Conceptually, it may be helpful to think of the new system as formed by constructive and destructive interference, but only in the vaguest conceptual sense. Pushing that metaphor too far gets you into the confusion you are in now.
As unsatisfying as it will be, it might be best at this point to accept that the conceptual description is qualitatively consistent with the mathematics and so is a guide to intuitively constructing molecular orbitals from atomic orbitals, but it isn't an explanation of why?. If you continue farther in your study of quantum chemistry, you'll learn how to construct these molecular orbitals more quantitatively, and you'll get better answers of how the math works to give certain answers, but you'll never get a good answer of why.

I will try to answer without using any mathematical formula, as your question is just made of words, without any mathematical symbols.
You should know that the wave function of an electron in the $\ce{H}$
atom is normalized. It means that the integral of the wave function over the whole space is equal to $1$. And it must be equal to $1$.
If you add two of these wave functions from two $\ce{H}$ atoms, the integral of this sum will be $2$. This is forbidden. It must be $1$. So you cannot simply add two normalized wave functions, like the $1s$ in the $\ce{H}$ atom. You have to take one-half of the wave function of the first atom and to add it to one-half of the wave function of the second atom. This would produce a new molecular wave function whose integral is equal to $1$. It is nice. This new molecular wave function describes a sigma bond, designed as $\ce{H-H}$ in the Lewis model.
Now you have to use and combine the other half of these wave functions, without adding them. The only thing to do is to subtract them. But such a subtraction produces a point between the two nuclei where the obtained wave function is equal to zero. As a consequence, this molecular orbital is a antibonding orbital. And the integral of this antibonding orbital is also equal to $1$.
Hopefully the pure theorists will pardon me for having simplified the description of the problem. I should have spoken of the factor √2/2
instead of $1/2$. I should have said that it is the square of the wavefunction that should have been integrated. I have tried to be understandable to a beginning student

Maurice: It is the square of the wavefunction that should integrate to 1. You are mixing a couple of things. I am not a QM expert, but please reconsider double checking the statements.

@M. Farooq. I know that it is the square of the wave function that should be integrated. I have been teaching it in my classes. But you know. The author of the post is a beginner. I haver tried to be as elementary as possible to give him an answer that he could understand.


text:
Calorimetry- Calculating change in temperature help
The reaction: Zn(s) + 2 AgNO3 (aq) → 2 Ag(s) + Zn(NO3)2 (aq)
takes place in a calorimeter.
30 cm3 of a 1.20 mol dm−3 solution of silver nitrate with an excess of zinc is used. If the value of ∆rH is −365.1 kJ mol−1 and all solutions are assumed to have a density of 1.00 g cm−3 and a specific heat capacity of 4.18 J K−1 g−1, what temperature change is produced?
I attempted to calculate the value of Q and got 13,143.6 J. Then substituted this into
Q = M x C x deltaT, to rearrange the equation giving delta T.
The answer I got is 105 but this is incorrect.

If you want to know where you made the error,  provide your detailed calculation procedure.

The $\pu{Delta H}$ value is $\pu{-365 kJ/mol}$. This is supposed to be "per mole of what happens in the equation with a coefficient $1$". So it is per mole of Zinc. It can also be rewritten so : $\ce{\Delta H = -365 kJ (mol Zn)^{-1} = -365 kJ (2 mol AgNO3)^{-1} = - 182.5 kJ (mol AgNO3)^{-1}}$ And here, you are using $\ce{0.036 mol AgNO3}$. So the heat effect is twice smaller. : $\pu{\Delta T = 52.3 K}$


text:
Mechanisms for Reimer-Tiemann and Kolbe-Schmidt reaction on deuterium labelled phenol
I was solving a bunch of reactions when I encountered the following:

So for the first question:

This is a Reimer-Tiemann reaction. I felt that the tautomerism would result in the deuterium getting attached to the phenolic $\ce{O^-}$. As it turns out, that was wrong. Oh well. Then I proceeded with the second question. Now this one is a Kolbe-Schmidt reaction, where the $\ce{CO2}$ liberated will react with the phenol. Going by the previous reaction, this time I decided not to add the $\ce{D}$ atom to the phenolic $\ce{O-}$. Consequently, this is how it panned out:

All of this has left me quite confused. So here's my question:

Why is it that the $\ce{D}$ atom during tautomerism adds on to the phenolic $\ce{O-}$ in the Kolbe-Schmidt reaction, but does not proceed to do so in the Reimer-Tiemann reaction?

Image credit: Created using Marvin JS

The carbanion in the first problem is relatively more basic than the phenoxide ion and thus abstracts the deuteron from it while in the second problem  the carboxylate is relatively less basic than phenoxide ion thus it doesn’t abstract the deuteron from it.

@Rishi Alright. But the carbanion in the first case has two -I effect groups stabilizing the charge on it. Is it still capable of abstracting the deuteron better than the phenoxide ion?

At best, it would be approximately as acidic as Chloroform($\ce{pKa=15.7}$)

I appreciate Waylander's answer. However, although the answer gives you the basic mechanisms for Reimer–Tiemann and Kolbe-Schmidt reactions, it does not address some question and OP's misconceptions such as:

The $\ce{CO2}$ liberated will react with the phenol.

Thus, I'd like to address these by using the same starting materials used in the question. Following is the step-by step mechanism of the first reaction:

The ethoxide ion is a strong base, strong enough to abstract acidic hydrogen from the starting phenol and bromochloroiodomethane $(\ce{CHBrICl})$. After the proton abstraction, the iodide ion then leave from the resulting carbanion to generate relatively stable bromochlorocarbene $(\ce{BrClC\!:})$ in situ. This carbene is the active nucleophile that reacts with the phenoxide anion (generated by another ethoxide ion) to give an anionic intermediate with the negative charge transformed to the halogenated carbon (second step of the mechanism). Technically, this intermediate is the keto-form of the corresponding phenol. Thus, next step, the tautomerism to the enol-form to gain aromaticity, is thermodynamically driven, but a critical one, because the deuterium ion $(\ce{D+})$ can now transfer to two centers, either dihalo-carbanion or phenolic oxygen. Since phenoxide ion is more stable than the carbanion (negative charge is more comfortable with higher electronegative atom), 1,2-$\ce{D+}$ shift happens to give thermodynamically more stable phenoxide intermediate with deuterium on the dihalocarbon (bromochlorodeuteromethyl). The bromine of this intermediate is then replaced by a water molecule followed by a subsequent rearrangement to eliminate remaining chlorine to give the intended deuterated 2-hydroxybenzaldehyde product. Thus, deuterium is not lost but transformed during this reaction.
Now, let's look at Kolbe-Schmidt reaction with given starting materials. Following is the step-by step mechanism of the reaction:

The hydroxide ion is also a strong base, strong enough to abstract acidic hydrogen from a phenol or a starting material such as chloroform $(\ce{CHCl3})$. However, our starting material, dibromodiiodomethane $(\ce{CBr2I2})$ has no acidic hydrogen, but it can act as an excellent electrophile. Thus, the phenoxide anion (generated by reacting with hydroxide ion) reacts with $\ce{CBr2I2}$ replacing one iodide ion to give a neutral trihalo-intermediate (second step of the mechanism). Again, technically, this intermediate is the keto-form of the corresponding phenol. The next step, tautomerism to the enol-form to gain aromaticity is again, thermodynamically driven. However, the deuterium ion $(\ce{D+})$ does not transfer to the phenolic oxygen, because the reaction is still in basic medium. Instead, the $\ce{D+}$ would be abstracted by an another hydroxide ion to give the corresponding phenoxide as shown in the mechanism.
The Iodine of this intermediate is then replaced by a water molecule followed by the subsequent rearrangement to eliminate one of bromines to give the corresponding acid bromide, which subsequently hydrolyzed to give the corresponding benzoic acid. Thus, deuterium is lost during the transformation of this reaction. Also, keep in mind that $\ce{CO2}$ is not liberated to react with the starting phenoxide ion, but $\ce{COO-}$ is generated as a consequence of hydrolysis after the electrophilic substitution of $\ce{CBr2I2}$ to the phenol.

Thank you for the answer! So the deuterium is taken away by a hydroxide ion istead of the phenoxide. It seems my book is wrong in that respect. Also, regarding the formation of $\ce{COO-}$; it (my book) wrote the following reaction: $$\ce{ CBr2I2->[4NaOH] 2NaI + 2NaBr + C(OH)4->[-4H2O] CO2}$$
Your thoughts on this?

@: C_Lycoris: I think it is erroneous as well. Look at the step: $$\ce{C(OH)4 -> CO2 + 4 H2O}$$ which is not balanced.

@: C_Lycoris: I cannot find any hydrolysis reaction online involving $\ce{CBr2I2}$. But I found close one: $$\ce{CHBr3 + Na3AsO3 + NaOH -> CH2Br2 + Na3AsO4 + NaBr}$$ This reaction is peer-reviewed published work: Org. Synth. 1929, 9, 56 (DOI: 10.15227/orgsyn.009.0056).

I'm afraid the balancing was an error on my part. The book has titled such reactions as Extension of Reimer-Tiemann. It further writes: "Phenols when refluxed with $\ce{CX4}$ or $\ce{CXX'X''X'''}$ with base yield o-&p-benzoic acid".
A couple of other reactions given are: $$\ce{CCl4->[+OH- ]C(OH)4->[-2H2O]CO2}$$ The section ends with the lines: "Thus reaction proceeds with the same mechanism as in Kolbe-Schmidt reaction"
While the reaction of an alkali with $\ce{CCl4}$ exists, I could not find a proposed mechanism similar to what the book claims. Do I create a separate post on this?

I think so. That would help others to intervene. I'm pretty sure $\ce{CCl4 -> CO2}$ need more than hydroxide treatment. Go for it.

This is the mechanism of the Reimer-Tiemann reaction (source Wikipedia here)

The polyhaloform reacts with strong base to create a carbene, a highly reactive species which reacts with the phenoxide anion through the alpha carbon to give an anionic intermediate with the -ve charge on the halogenated carbon. It is this species that picks up the deuterium.
In Kolbe-Schmidt reaction, the process does not proceed through a carbene. It is nucleophilic attack (through carbon) of the phenoxide on the polyhaloform (or carbon dioxide) and the deuterium shifts to the phenol oxygen as the product re-establishes aromaticity. Mechanism from Wikipedia here

I afraid the deuterium ion in Kolbe-Schmidt mechanism does not shift to phenol oxygen as you suggested. That's because the reaction is still in basic medium and deuterium can be abstracted by the base first to give the corresponding phenoxide ion.


text:
Is it possible to functionalize silica gel as Activated Carbon can be? How?
Activated carbon and Silica gel both have a high surface area.
If I understand correctly it is possible to trap stuff inside activated charcoal pores to give it specific retention properties (by exposing it to functionalizing conpounds while heated to open up the pores?).
Is it possible to likewise trap stuff in silica gel pores? What can be trapped in there and how is this done?

Yes, silica gel can be modified to make its adsorption more specific. E.G., epichlorohydrin dimethyl amine copolymer (Epi-DMA) can be used to make silica gel effective for adsorbing anionic dyes.

Can it be modified with the same stuff as AC (NaOH/KOH/CaCl2/ZnCl2/H3PO4/etc.)? Can it also be done by warming it up with the impregnated substance that is to functionalize it? If so at which temperatures?


text:
How does functionalized activated charcoal work?
If I understand correctly it is possible to trap stuff inside activated charcoal (AC) pores to give it specific retention properties (by exposing it to functionalizing compounds while heated to open up the pores?).
When a solution is passed through AC, what happens inside the pores with the functionalizing compound to trap stuff that otherwise wouldn't be trapped? Can it be said that the functionalizing compound precipitates stuff inside the pore by producing low solubility products?

No. It is not a precipitation reaction. It is physical reaction. The pores are made of the rest of the membranes of the cells. When wood is overheated in air, it burns. When this is done without too much contact with the air, the cellular water gets evaporated. The content of the cell (cytoplasm, proteins, etc.) is pyrolyzed, destroyed and mostly transformed into gases. The membrane is carbonized or charred, but remains here, forming charcoal. The rest of the cell being eliminated, the charcoal contains a huge amount of the previous cells which behave like pores. Big molecules can enter these shells, and have difficulty to get out later on.

Interesting! What about the functionalization? What happens in the pores that make functionalized AC retain more of some molecules?

@Hans Honestly I believe that nobody knows how many molecules the pores may retain.

I'm talking about the mechanisms: What mechanisms is at hand / what is going on inside the pores of AC that has been functionalized with a chemical? Surely part of the specific chemical used (NaOH/KOH/CaCl2/ZnCl2 /H3PO4/...) must be retained within the pores and exert some activity in there specific to that chemical that gives the AC specific properties compared to AC activated with another chemical?

@Hans. I don't have any idea about the mechanism


text:
Correct Rate Equation for a Reversible Reaction
I'm helping my son with Matlab as he is having trouble learning differential equations as they apply to chemical kinetics.   His task is to solve for the equilibrium concentrations for a reaction
$$\ce{A + B <=>[$k_{\mathrm{f}}$][$k_{\mathrm{b}}$] C}.$$
We are given the forward and backward rates and the initial concentrations of the reactants and products.   I helped him to set up the required equations and ran them through the ode integrator in R and Matlab but the results don't look right.   I'm pretty sure the integration process is working so I think maybe the actual formulae are wrong:
\begin{align}
\frac{\mathrm{d[A]}}{\mathrm{d}t} &= {k}_\mathrm{b}[C]-{k}_\mathrm{f}[A][B]\\
\frac{\mathrm{d[B]}}{\mathrm{d}t} &= {k}_\mathrm{b}[C]-{k}_\mathrm{f}[A][B]\\
\frac{\mathrm{d[C]}}{\mathrm{d}t} &= {k}_\mathrm{f}[A][B] -{k}_\mathrm{b}[C]
\end{align}
When I run this through the integrator with $k_\mathrm{f}=3$, $k_\mathrm{b}=1.5$,
$[A]=\pu{1M}$, $[B]=\pu{1M}$, and $[C]=0$,
the concentrations all reach equilibrium at the same level of $\pu{0.5M}$.  Is that correct?  I would have expected them to be different with the forward rate being bigger.
Really sorry for the basic question but as I said, I'm a programmer not a chemist.  I got the formulae from this How to set up an equation to solve a rate law computationally? but assumed that I didn't need to square any of the concentrations since the reaction is $\ce{1+1<=>1}$.

Your son can check whether it is correct. At equilibrium the changes are zero, and your system of differential equation did reach equilibrium. The higher forward rate is offset by being multiplied by 0.5 twice.

If all you want to find is the final concentration of all the species at equilibrium, then you don't need to solve these differential equations.
n equilibrium constant $K_\mathrm{eq}:$
$$K_\mathrm{eq} = \frac{k_\mathrm{f}}{k_\mathrm{b}}\tag{1}$$
and
$$K_\mathrm{eq} = \frac{\prod a_j^{\nu_j}}{\prod a_i^{\nu_i}},\tag{2}$$
where $a_i$ and $a_j$ are the activities of reactants and products, respectively; $\nu_i$ and $\nu_j$ are the stoichiometric coefficients of reactants and products, respectively.
So for your case we can set the equations as such:
$$
\begin{align}
\begin{array}{cccccc}
& \ce{A} & \ce{+} & \ce{B} & \ce{<=>} & \ce{C}\\
\text{Initial} & \pu{1 M} & & \pu{1 M} & & \pu{0 M}\\
\text{Final} & (1-x)\pu{M} & & (1-x)\pu{M} & & x\pu{M}
\end{array}
\end{align}
$$
$$
\begin{align}
K_\mathrm{eq}&=\frac{[\ce{C}]}{[\ce{A}][\ce{B}]}\\
\frac{k_\mathrm{f}}{k_\mathrm{b}}&=\frac{[\ce{C}]}{[\ce{A}][\ce{B}]}\\
2&=\frac{x}{(1-x)^2}\\
2x^2-5x+2&=0\\
(2x^2-4x)+(-x+2)&=0\\
(2x-1)(x-2)&=0\\
x&\in \{\pu{0.5M}, \pu{2M}\}\\
\end{align}
$$
But as $[\ce{B}]\geq \pu{0}$, $x$ has to be equal to $\pu{0.5 M}$.
Which gives, $[\ce{A}]=[\ce{B}]=[\ce{C}]=\pu{0.5 M}$.
This also shows that your solving of differential equations was also correct!

Thanks for your answer.  I agree a closed-form sulution is simpler but the exercise required that we use an ODE solver.

To complement Nisarg's answer, note that we have the following system of polynomial ODEs
$$\begin{bmatrix} \dot a \\ \dot b \\ \dot c\end{bmatrix} = \left( 3 a b - \frac32 c \right) \begin{bmatrix} -1\\ -1\\ +1\end{bmatrix}$$
Thus, given an initial state, the state will flow along a given line. Since concentrations must be non-negative, given an initial state, the set of all states is a line segment (one of whose endpoints is to be determined) in the non-negative octant. Equilibrium is found at the algebraic surface defined by
$$3 a b - \frac32 c = 0$$
Suppose that the initial state is $(1,1,0)$. Pictorially,

where the equilibrium point is $\left( \frac12, \frac12, \frac12 \right)$.

It is reasonable to be confused by this. The behaviour of these sorts of reactions is counter intuitive if you don't have a chemistry background, and it doesn't help that it is sort of obscured by the particular choice of numbers in the assignment.
Looking at it naively, you might expect that because the forward rate is twice the backward rate, the final equilibrium should be 2:1 products:starting materials. But this is only true when the overall number of molecules is conserved.
This reaction is bimolecular in the forward direction and unimolecular in the reverse. That means in order to react in the forward direction, a molecule of A and a molecule of B must find each other and collide, but in the backwards direction a molecule of C can spontaneously decide to split.
Ask yourself - what happens to this reaction when it is highly diluted, and molecules of A and B rarely interact? How easy is it to form C in the first place? Is the rate of C splitting apart affected?
Conversely, what happens when A and B are jam-packed in a concentrated syrup? What happens to the rate of formation of C when the solution is nothing but A and B all colliding with each other, along with the minimum number of solvent molecules needed to keep everything in solution?
This reaction is concentration dependent. The more concentrated it is, the faster A and B will find each other, while the rate of C splitting up remains the same. The equilibrium therefore shifts towards C with increasing concentration.
You can see it in the equation:
$$K_{eq}=\frac{[\ce{C}]}{[\ce{A}][\ce{B}]}\\\\$$
Assuming [A] = [B], the equilibrium is affected linearly with the concentration of the product, but with the square of the concentrations of the starting materials.
As it happens, a concentration of 1 M is quite concentrated, relative to what an organic chemist might actually use in an experiment. If all the starting concentrations were at 0.1 M instead, the equilibrium concentrations would be at [A],[B] = 0.085 and [C] = 0.015 respectively. A 5.66:1 ratio of starting material to product, even though we haven't altered K at all.


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Photochemical versus Photophysical
What is the difference between the terms photochemical and photophysical properties? What are some examples from each category?

Reminds me about Differences between chemical physics and physical chemistry?

So is it bonkers when I see a paper describing the photochemical and photophysical  properties of X?

Just a guess: photophysical can mean anything relating to the interaction of light and matter, simple absorption for instance. Photochemical implies a chemical reaction results from the interaction due to formation of excited species.

In photochemistry light is used to break a bond e.g. $\ce{I_2 +light -> 2I\cdot}$ or trans stilbene + light $\to$ cis stilbene, photophysics is the study of how this happens.

Photophysics digs a lot into details. They can be or not associated to a following reaction. While absorption can be a analytical chemistry tool, and might be relevant to conduct photochemical reactions, doing time resolved spectroscopy is certainly photophysics. I think that the comment by Buck Thorn helps in a sort of definition, but the point made by Martin is valid, too. In all cases, recording Absorption spectra, to give an example, certainly isn't photochemistry. Though is likely the most familiar physical technique among chemists.

@porphyrin the stilbene example is relevant. If we consider thermal isomerisation is that photophysical? But if we consider the cyclisation of Z-stilbene to the phenanthrene, would that be photochemical? Finally, if I characterise the pi-pi* transitions of a photochromic ligand + its thermal isomerisation kinetics could I accurately describe that as 'photophysical characteristics'?

Thermal by definition is definitely neither photochemical nor photophysical. The photo-induced events have to occur from an excited state produced by absorbing a photon.


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Stability of hydroxy aldehydes
I cannot understand if carbonyl adjacent to alkene stabilises or destabilises it. Generally, we say that more substituted alkene is stable. In those cases attached groups increase electron density of the alkene. In the case of carbonyl, it decreases electron density of the alkene so it should destabilize the alkene.

But I read about the stability order of the above compounds to be 1>2>3. The reason being that the first is conjungated and second's enol is conjungated.
So, how can both increasing and decreasing the electron density of alkene stabilise it?

Ketones and enols are completely different molecules. Sure, they can be converted into one another, but it's absolute rubbish to say that a ketone is more stable because its enol is more stable. You should get a better book.

On top of that, it doesn't make sense to compare stability of different molecules with different chemical formulae in absolute terms. Absolute thermodynamic stability can be measured using things like heats of combustion, but it's pointless comparing two (or three) non-isomeric compounds because they will have vastly different heats of combustion (for example, the bottom compound will produce a lot more CO2 than the top compound).

I concluded this from a question on rate of dehydration. OH group was attached on the left end of each of the above mentioned molecules. As their carbocations were almost of same stability so I thought that I should compare the rate based on the stability of products. Thus, I came up with these explanations. Is the order stillcorrect?

Well, that's an entirely different question. "Which of these undergo dehydration the fastest and why" can definitely be answered. "Which of these products is the most stable" cannot be answered.

How do I approach the rate of dehydration question?

You should start by considering which type of elimination mechanism (E1, E2, E1cb) is likely to prevail for each of the individual compounds. In order to do this, you also should specify reaction conditions so that any ambiguity is eliminated. However, the comment section isn't the right place to go into this. You can ask it as a separate question (better not edit this since you already have one attempt at an answer).

Should I delete this question as it is misleading?

You can't do that either, since there's already an answer. Bit unfortunate, but this one probably has to stay.

I think the key of your question is when does an oxygen atom donate electron density and when does it withdraw thereof.
According to Lumen Learning How Delocalized Electrons Affect pKa Values, "the oxygen atom does indeed exert an electron-withdrawing inductive effect, but the lone pairs on the oxygen cause the exact opposite effect."
My professor tells me that oxygen has "dual personality." When its electron density cannot be donated through resonance, its electronegativity withdraws electron density through the sigma framework.
The key to explain the correct stability ranking is perhaps the 1,3-butadiene-like p orbital alignment through two adjacent double bonds. When 4 p orbitals on two double bonds align with each other, butadiene gains a resonance energy of 17.28 kcal. See more here. When two double bonds are not adjacent, p orbitals cannot overlap to gain more stability. I used ChemDraw to draw the tautomerization and the p orbital alignments. Hopefully it is not too bad.
I just finished Orgo 1 this semester, so my Orgo knowledge is at the surface level. Please point out any problems.

Thanks alot for the efforts but you seem to be unfamiliar with the concept of positive and negative resonance. I am confused if alkene is stabilised by both positive and negative resonance.

The drawings are fine, but please see my comments on the question. On a side note, I recommend using the "ACS Document 1996" style in ChemDraw (File > New from Stationery...). The default is... pretty ugly, let's say.

I will look into positive and negative resonance, and update the answer as needed.