text:
Acetone Enthalpy & Observations
I was looking at the enthalpy of various liquids using WolframAlpha and noticed that at 30 °C, while water and ethanol have a positive enthalpy, acetone has a negative enthalpy.
If I'm understanding this correctly, this means that as acetone evaporates, the temperature of the acetone will increase (because the evaporation will release energy, thereby raising the temperature). I started searching up in temperatures to find the zero-crossing, and that appears to be around 56.074 °C.
This suggests that if a sufficient quantity of acetone is left around, it will eventually heat up to 56.074 °C (which interestingly happens to be when its vapor pressure is really close to the STP pressure).
However, if I get acetone on my hand, I get a distinctive "cool" feel — it, like other liquids, appears to be endothermic and cools down. I may be warm-blooded, but I'm not above 56 °C.
Is there something wrong with my understanding? If not, why does the theory seem to contradict my observations?

WolframAlpha does not know apples from oranges when it comes to chemistry. Acetone certainly has positive enthalpy of evaporation, like any other liquid.

@iAdjunct Apparently, you are confusing enthalpy $H$ and enthalpy of vaporization $\Delta H_\mathrm{vap}$.

@Loong - I believe you are right; see my comment below on your answer.

Congratulations: you put a question on hold after it was answered for being unclear what I'm asking, but kept this question open: https://chemistry.stackexchange.com/questions/73608/vapor-pressure-paradox

The absolute values of enthalpy H at a single state point are meaningless. It is only the difference between two different state points that matter. Thus, the value for a single state point can be set to any arbitrary value. Many handbooks set the arbitrary state point so that the values of these properties are positive for most liquid or gas states.
Apparently, WolframAlpha is setting enthalpy H to zero for the saturated liquid at the normal boiling point (NBP) for acetone but not for water and ethanol.
For ethanol, WolframAlpha is setting specific enthalpy to h = 200 kJ/kg for the saturated liquid at 0 °C (IIR reference state).
For water, the triple point is selected as the reference state, where the internal energy U of saturated liquid is assigned a zero value. Accordingly, the enthalpy H = U + pV of liquid water at the triple point is slightly larger than zero, at a specific enthalpy of about h = 0.6115 J/kg.

I don't understand how a difference is used. My understanding was that when something says, for example, 30 J/kg, that means if you want to transition 2 kg of that substance from liquid to gas, you must add 60 J of energy to it. With a negative number, it seems to imply the removal of energy. How do you use a number with units of J/kg as references for differences?

@iAdjunct The absolute value of enthalpy has no physical meaning. For example, the given value for the specific enthalpy of liquid water at equilibrium at 100 °C is 419.17 kJ/kg. This value alone doesn’t say anything. Since only the difference between two state points is relevant, you would need a second value; for example, the specific enthalpy of gaseous water at equilibrium at 100 °C, which is 2675.6 kJ/kg. Now you can calculate the specific enthalpy of evaporation at 100 °C as 2675.6 kJ/kg − 419.17 kJ/kg = 2256 kJ/kg.

Ohhhhhhhhhhh. I think that also answered why "Specific Heat of Vaporization" isn't just called "Enthalpy" - because the Specific Heat of Vaporization is ∆Hvap = Hgas - Hliquid at whatever temperature/pressure you specify? (therefor ∆Hvap = SpecificHeatOfVaporization?)

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
can you separate inorganic ions with chromatography?
For example,  if you run a solution of sodium and potassium chloride through a colum?
I am just wondering if i can get pure potassium chloride from a mixture of sodium and potassium chloride.

https://en.wikipedia.org/wiki/Ion_chromatography may be helpful to you. I've never done this before so can't comment on how it would work in practice.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Which ultra-high temperature ceramics make a good electrical insulator?
I am looking for materials with melting point above 3,000 °C and yet is a good electrical insulator. Generally metal carbides have high melting point but a lot of them are electrical conductors.

Thorium dioxide gets you 3390°C.  Have fun getting it, $$$ and also the thorium is radioactive.  You also risk some nonstoichiometry, due to thorium having variable oxidation states, and that could make the compound somewhat conductive.  Magnesium oxide weighs in at 2800°C; but is more stable, cheaper, nonradioactive and having no variation in stoichiometry.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What is the origin of the huge difference in the autoprotolysis constant of heavy water versus regular water?
Basically, the gist of this question is, why is the autoprotolysis constant of heavy water different from that of ordinary water?
The autoprotolysis constant of heavy water at $\pu{25 °C}$ is $\pu{1.35 * 10^{-15}}$, while at $\pu{30 °C}$ it is about $\pu{1.47 * 10^{-14}}$. From this, I calculated that the enthalpy change for the autoprotolysis of heavy water is about $\pu{360 kJ/mol}$, which is already a huge difference from regular water's autoprotolysis enthalpy of approximately $\pu{+56 kJ/mol}$. The difference in entropy is also huge as well: by using the formula $-RT \ln(K) = \Delta G$ and $\Delta G = \Delta H - T \Delta S$ I calculated that the autoprotolysis of heavy water has an entropy change of about $\pu{+920 J K-1 mol-1}$, which is a world of difference versus water's approximately $\pu{-81 J K-1 mol-1}$. Can someone tell me why this huge difference in values is occurring?
Also could someone point me to a database that would contain this kind of thermodynamic data / equilibrium data because my textbooks aren't going to cut some of the questions I have in terms of raw data crunching.

Because you made mistakes in calculations?

How did you calculate these enthalpies?

@Mithoron: Using Van t'Hoff I get $\Delta H^0=+358\ \mathrm{kJ/mol}$ for heavy water. And this resource: http://www1.lsbu.ac.uk/water/water_dissociation.html lists the value for light water as $+55.8\ \mathrm{kJ/mol}$. So it appears OP's values are correct, at least for the enthalpies.

However for the $\Delta S$ values I get different results: $-283.5\ \mathrm{J/mol}$ and  $-267.8\ \mathrm{J/mol}$ for heavy and light water, respectively.

For both light water and heavy water, my results from G = H - TS confirms the values that I got. My textbook also accounts for the change in entropy for light water ($\Delta S = S_{products} - S_{reactants}$) using the thermodynamic data that I got from it, but it doesn't have enough data for heavy water. Can you tell me how you got your entropy value for heavy water?

As you stated correctly: $-RT\ln K=\Delta H-T\Delta S$, so $\Delta S=\frac{\Delta H+RT\ln K}{T}$. I used the previously calculated $\Delta H$ at $25\ \mathrm{Celsius}$ (and the stated $K$ values). I'll check again. Update: just verified the numbers for $\Delta S$ and I get the same as I reported before.

As regards the reason behind the different $K$ values, this is quite briefly explained here: https://en.wikipedia.org/wiki/Heavy_water#Effect_on_biological_systems. Quantum effects make the $\ce{OD}$ bond stronger than the $\ce{OH}$ bond.

How does the deltaH value compare with other substances?

@OP, please [edit] in all relevant data to reproduce the values, so that no-one has to read a lengthy comment discussion


text:
Are children's sparklers based on a magnesium reaction?

We were letting our kids play with sparklers on New Year's Eve, and my friend's son asked his Dad:

What would happen if we threw the sparkler into the water? Would it keep burning under water.

My friend pondered that for a moment - and said it depended if the reaction was based on magnesium or not.
My question is: Are children's sparklers based on a magnesium reaction?

Probably not, but it would depend on the sparkler. The most common metal is iron powder.

Many sparklers use aluminium as far as I know.

Just going by my assumptions on how the world works: water is the standard household method to put out fires. If a product should not be extinguished with water, I would assume big, fat labels on the packaging ‘do not put burning product under water!).

https://en.wikipedia.org/wiki/Sparkler

Not only will water put out the sparklers, but its recommended to put them into water after use to both rule out any uncombusted material later catching fire and to cool them as a common source of injuries with them is children (and indeed adults) grabbing them by the burnt portion when they are extinguished but still hot. Dousing in water removes that risk.

What would happen if we threw the sparkler into the water? Would it keep burning under water.

Most likely, not. Water is an effective coolant, so a wet sparkler wouldn't be able to propagate a burn front. Pyrotechnic compositions often severely degrade in a wet air (some might self-ignite and some might loose the ability to burn).

Are children's sparklers based on a magnesium reaction?

The most common sparkler composition I'm aware about is

fine $\ce{Al}$ powder as a fuel,
$\ce{KNO3}$ as an oxidizer
dextrin or other combustible binder
Iron coarse powder for orange-ish sparks.

$\ce{Ti}$, $\ce{Al}$ or $\ce{Sb2S3}$ in coarse powder can be used for producing white sparks. Charcoal powder or potassium poly-sulfides can produce redder sparks. To my knowledge, there is no way to produce purple, blue or green sparks.
Exact composition is a delicate balance between gases produces (so sparks were thrown away at meaningful distance) stability and temperature of burn, amount and color of sparks produced, safety and cost.
Please, note. While a common sparkler cannot burn in water, a big slag with same composition and water-proof coating certainly can. But you don't want to be nearby when it happens. Like, at all. An example with a much tamer fuel here:
https://www.youtube.com/watch?v=czwBWB5u6Hg
Also, magnesium is usually avoided in pyrotechnic compositions in favor of aluminum. Magnesium usually produces a lot of thick, white smoke and is more sensitive when stored in real conditions. If aluminum doesn't burn good enough, $\ce{Mg/Al}$ alloy might be used.

It is easy to tell sparklers don't keep burning in water. Throw sparklers into water and they go out. The core reaction is an oxidiser + fuel and not a bulk metal burning in air. Oxidisers like potassium nitrate are not water-proof unlike some burnin metals.

The relevant part of the youtube video starts at 2:37. Ultra slow motion at 3:30.

To answer the question, let us understand sparkler chemistry. A sparkler consists of:

An oxidizer: potassium nitrate or potassium chlorate. They burn off a mixture and are part of oxidation-reduction reaction:
$$
\begin{align}
\ce{\underset{potassium nitrate}{2 KNO3 (s)} &-> \underset{potassium nitrite}{2 KNO2 (s)} + O2 (g)}\\
\ce{\underset{potassium chlorate}{2 KClO3 (s)} &-> \underset{potassium chloride}{2 KCl (s)} + 3 O2 (g)}
\end{align}
$$
Fuel: firework fuel is charcoal and sulfur. Both sulfur and carbon (charcoal) are reducing agents forming sulfur dioxide and carbon dioxide.
Metal powder: aluminum, iron, steel, zinc, magnesium, titanium or ferrotitanium dust or flakes. These metal flakes become incandescent and shine brightly in the presence of oxidizers. If temperatures are high enough, these metal flakes burn. The temperatures of these sparks range from 1800 °F to 3000 °F (1000 °C - 1600 °C)(Ref. I). Many elements heat up and emit at a characteristic wavelength (Ref. II).

A combustible binder: binders hold oxidizers, fuel and metal powders together. Common binders are dextrin (a sugar) dampened by water.
Colorants: aluminium, magnesium and titanium all give near brilliant white sparks. Iron produces orange sparks.

Putting all together: these chemicals are mixed together to form a slurry. The slurry is coated on to wire and allowed to dry. When ignited the metal flakes  aluminum, iron, steel, zinc or magnesium dust or flakes create the bright, shimmering sparks. As to your question, children sparklers are based on metal oxidizing reactions.
References

Ref. I https://www.wired.com/2014/07/the-awesome-physics-in-a-simple-sparkler/
Ref. II https://www.forbes.com/sites/startswithabang/2016/07/01/the-physics-of-fireworks/#3895e0711eee
http://www.compoundchem.com/2014/11/04/sparklers/
http://www.compoundchem.com/2013/12/30/the-chemistry-of-fireworks/
http://www.compoundchem.com/2014/07/02/the-chemistry-of-gunpowder/
http://www.skylighter.com/fireworks/how-to-make/sizzler-sparklers.asp

Does the oxidising agent mean that atmospheric oxygen is unnecessary? Including this auxiliary information would make this answer more useful.

@wizzwizz4 Yes, sparklers should be able to burn in absence of oxygen, but sparks might be a lot dimmer, since they are partially heated by reaction with oxygen from the air. Also, magnesium generally doesn't give sparks, it usually gives a bright, blinding flame. Magnesium (just like zinc) has rather low vaporization temperature and so is unsuitable for producing of small, bright sparks. Aluminum, iron and sometimes antimony sulfide or titanium are usually used for this. Magnesium usually also produces a lot of white smoke.

@wizzwizz4 ,depending on the composition of the sparkler ,raecting with atmospheric oxygen is possible.In which case ,it is a combustion reaction.Explosives ignited by gunpowder have  their own oxygen.If a sparkler mixture  has gunpowder with metal flakes , then oxygen is from Potassium nitrate.Thankyou.

Your conclusion ("children sparklers are based on metal oxidizing reactions.") doesn't match the content you wrote. The primary reaction is based on gunpowder (fuel+oxidant) and the metal burning is a side effect not the primary combustion taking place (at least that's what you described earlier in your answer). So it won't work under water.


text:
Why do we use the Schotten-Baumann synthesis method for the synthesis of amides?
Recently, I learnt about the Schotten-Baumann synthesis method. I know that this synthesis method was invented to overcome the problem of neutralising the acidic by-products of the amide synthesis reaction without introducing strong basic nucleophiles to compete with the amines. The alternative to this synthesis method would be to use two equivalences of the amine, allowing the 2nd equivalence of the amine to act as a base to neutralise the acidic by-product. I would like to enquire more about the differences between these two synthesis methods. In particular, I would like to know the advantages and disadvantages of each of these synthesis methods.

You do need a base (non-nucleophilic) to absorb the acidic proton, or the reaction doesn't proceed. It often uses a 2-phase solvent system, consisting of water and an organic solvent. The base within the water phase neutralises the acid, generated in the reaction, while the starting materials and product remain in the organic phase.

@getafix But if u add a 2nd equivalence of amines, then it could act as a base as well. Why do we need another base? Amines are basic enough.

Well, synthetic chemistry is a practical science. Your amine starting material might be expensive, or difficult to synthesise. You wouldn't want to waste it, or spend time recovering it from the reaction mixture.

This is quite correct, if you've spent 2 weeks synthesising your amine you will not use 2 eq of it. In the real world of the lab we rarely use S-B prefering to use a single phase system and an equivalent of a second organic base. Triethylamine or N-methyl morpholine are often used

@Waylander Wouldn't the second organic base be nucleophilic enough to attack the carbonyl carbon as well and result in the formation of unwanted side products?

No. Not if you use tertiary bases with significant steric hindrance around the basic nitrogen.


text:
Special case of Cannizaro reaction of formaldehyde with 100% NaOH/KOH follows 4th order kinetics
It is given that when we use $50{-}\pu{60\%}\ \ce{KOH}$ in a Cannizaro reaction of formaldehyde, the reaction follows third order kinetics, but under special cases of Cannizaro reactions it was given than when formaldehyde reacts with $\pu{100\%}\ \ce{NaOH/KOH}$ the reaction follows fourth order kinetics.
What is the reason for this change in the order of reaction with an increase in the concentration of base?

Doesn't https://en.wikipedia.org/wiki/Cannizzaro_reaction answer this?


text:
Can a substance donate a proton without the proton getting accepted by another substance?
I have read that the concept of Brønsted and Lowry is based on the fact that acid-base reactions involve the transfer of $\ce{H+}$ ions from one substance to another, and that an acid is a substance that donates a proton to another substance, but:
can a substance donate a proton without the proton getting accepted by another substance? So for example:
$$\ce{AH -> A- + H+}$$
without reacting with another substance?
If yes: Is this a Brønsted-Lowry acid?

The process itself should be possible depending on the nature of A. Kind of photolysis. But in absence of media it will go reverse.  Nothing in common conditions....

https://chemistry.stackexchange.com/questions/46882/do-bare-protons-exist-even-transiently-in-aqueous-solution https://chemistry.stackexchange.com/questions/41997/does-the-hydrogen-ion-actually-exist

The closest thing I can find to this involves a most unlikely proton donor.  Given an appropriate base, sodium hydride donates a proton for encapsulation in the base (rather than forming a localised covalent bond), leaving the valence electrons on sodium (!) in inverse sodium hydride.  The proton is not free (a completely free proton is possible only in plasma), but analogous to complexed metal ion or solvated electron.

"Encapsulation in the base"? Does it mean that the base receives/accepts the proton?

Yes, some bonding interaction (in this case between the protons and lone pairs on nitrogen atoms in the base) to drive the "reversed" oxidation states.  But you avoid the usual localized covalent bonding, the encapsulated species may be fairly called ironically bonded $\ce{H^+}$.

I don’t think this qualifies as the proton not getting accepted by anything since it does get accepted by the encapsulation host.

Hence, "the closest thing I can find".

It's not closest by any means and no idea why you put such example. BTW it would be good to use links instead of bare urls.

@Mithoron I am curious about your ideas.  Superacids still involve a localized covalent bond to the reactive hydrogen and I was looking for an interaction that does not reduce to such a bond.

I actually have a few, and this wasn't so dupey... There are things even more powerful then typical superacids.


text:
Nucleophilic Substitution rate comparison 1-bromobutane and 2-bromo-2-methylpropane with NaOH
I want to ask a question about the hydrolysis of haloalkanes to form alcohols.
I have been presented with a flowchart diagram of possible reactions of haloalkanes below:

Now, I have been presented with the question below:

I have identified the reagent I relevant to the question as NaOH (aq)
First, I need to compare 1-bromobutane and 2-bromo-2-methylpropane.

I can see that there are 3 alkyl groups in 2-bromo-2-methylpropane and only 1 alkyl group donating electrons to the Carbon atom attached to the Bromine.
So, during the reaction, the carbocation intermediate formed by reaction of NaOH with 2-bromo-2-methylpropane will form be more stable than the carbocation intermediate formed with 1-bromopropane (application of Markownikoff's rule) as alkyl groups are electron-donating groups and the three methyl groups will stabilise the positive charge on the carbon atom to a greater extent in 2-bromo-2-methylpropane.
However, I am slightly confused.
Why does this mean that the rate of reaction with 2-bromo-2-methylpropane is faster than the reaction with 1-bromopropane?

Your flowchart should include reagents. And phew, there’s a lot hidden in that question D=

@Jan they asked me in the previous question to identify the reagent but I have mentioned that I identified it just now in the edit above, and yes certainly there is a lot hidden, but it is a good thinking question, something I enjoy looking at!

With ‘question’ in my previous comment I meant the text you posted on this site. Meaning an answer will have to address quite a few things.

I can't exactly remember where I read this, but in case of substitution of halogen in a haloalkane by $\ce{OH-}$, the overall rate of reaction is tertary>secondary>primary. Even though tertiary reacts via $\ce{S_N1}$ and primary and secondary mostly via $\ce{S_N2}$.

Don't know if this is a general trend in case of substitutions on alkane, and no idea why this happens


text:
Why are MD (Molecular Dynamics) simulations performed at room temperature for the purposes of studying proteins?
All of the MD simulations which are cited in the literature seem to be performed at room temperature (~300 Kelvins), while enzymes are usually bioactive at body temperature (~310 Kelvins).
I am wondering why not just use 310 K for the MD simulation? I suspect it is because most water models are designed specifically for room temperature (ie. TIP3P waters).

Great question. I coauthored some MD studies between 2000 and 2006, where we looked at motor proteins (ncd, myosin, kinesin) and sure enough, we went with 300 K. In many cases, you're not dealing with a human system - myosin from $\it{Dictyostelium,discoidium}$, ncd from $\it{Drosophila}$ - so human body temperature is not appropriate. Also, at least in the studies I  was involved in, you're interested at normalizing computational conditions to experimental, which many times turn out to be at room temperature.

I'm no MD expert, but I don't think a reliable water model at 300K would be bad at 310?

Another reason why people don't bother to use 310K versus 300K is because any time you are using MD, you have an approximate force field describing the dynamics of the system which means you have an approximate phase diagram for the system. When comparing MD and experiment, it is a reasonable question if you really want to compare properties at the same temperature or compare properties at "equivalent" points on the phase diagram. As an example, some models of water which are very widely used have a phase diagram which is drastically different from the real phase diagram of water (i.e. they boil at 400K and other weird things) and you will see studies which describe the phase diagram of DFT water or TIP4P water so that you can make comparisons based on the temperature and on the phase diagram.
So, this is to say, if you submit a paper where you do MD at 300K when studying an enzyme in solution, it is unlikely you will get complaints that this is a biomolecule so it should be at 310K because even comparing to properties of the actual enzyme at 310K isn't a direct comparison since you aren't at the same locations on the respective phase diagrams. What could be more relevant is to make comparisons to the real world across a range of temperatures so that you can see if the trends are the same rather than just a single point.
Another less physical reason is that people have already been doing simulations at 300K, so you can't really get complaints about a simulation at 300K since everyone has already been using 300K. This is actually very common in science.

This depends on the organism you are studying. I work with Bac and human, I have simulations running at 300k and 310k. Search on the organism you are going to work on and adjust its temperament for your simulation.

You want to be able to compare your simulation with the real world, and that means you do (or look at other's) in vitro experiments, which are predominantly (at least for starters) done at room temperature because that's simplest. Further you want to be able to compare to results of earlier simulations (other MD engine, varied chemical system), and you want at least the same (nominal) temperature and pressure in your simulation. 300K is the standard, so you stick to that.
Also who says your enzymes(, ... ) stem from warm-blooded animals?

I second jheindel's suggestion that people tend to just follow the literature. It's also true that the fluctuation in temperature during your MD simulation is consistently on the range of several kelvin. Therefore, 310 and 300 K aren't that far apart.
Nonetheless, the conformational space sampled by the protein is heavily dependent on temperature (think of denaturation temperatures as an extreme example). As such, we should try our best to replicate physiological conditions. If the optimal growth temperature or physiological temperature of the biological host from which you took your protein is known, you should run your simulations at that temperature. This is especially relevant for extremophilic proteins, such as thermophiles and cryophiles.

Are your forcefields good enough to capture 10K difference? If not, it makes sense to perform most calculations at a standard condition, since it is easier to compare them with others, except if you do the study on temperature dependence (where I do not think you can rely on the absolute scale). Also, biochemistry is not dramatically different at room temperature from body temperature, and warm-blooded animals have essentially the same stuff as any other organism. Much lower or higher, sure, that makes difference, but again, we talk about 10 K.

See Table 3 here (https://pubs.acs.org/doi/pdf/10.1021/jacs.7b11926) where a 17 K difference (using pretty standard Amber12 FF and TIP3P models) led to a different mechanism of electron transfer for E. Coli.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Why is entropy favorable?
I cannot seem to grasp the logic behind it. We say that more entropy (or more disordered system) is favorable over less entropy. But why? Why is randomness preferred over proper arrangement of atoms/molecules? The only thing I can think of is more degrees of freedom. Molecules in a more random system would have more degrees of freedom, and would thus be favorable. But, again, why? Why would more degrees of freedom be favorable? I've already asked my teachers and looked up internet, but haven't found any satisfactory answer. Also, is there any way you could also explain diffusion by the same concept (diffusion is entropically favorable, right?)?
PS: if this question is better suited for physics stack exchange, I'd be more than happy to have it migrated.

Welcome to chemistry.SE! If you have any questions about the policies of our community, please visit the help center. I think your question is well-suited for this site, more so than for physics.se. One way to look at your question is thermodynamically via DG = DH - TDS. Here you can see that a large entropy change between 2 states results in a more favorable (negative) Gibbs free energy change. Anyway, best of luck with your question.

I'm not sure I understand the question (although there are answers which appear to understand them). Entropy is not favorable to me; I desperately fight it every day. When you say "favorable", do you mean that systems tend to move in the direction of more entropy, e.g. in the course of chemical reactions? Then the answer is, at least somewhat, circular: "Higher entropy" is the name for that direction ;-).

@airhuff yes I get that, what I dont get is the logic behind that equation (I mean what is really happening at the atomic level)

@peter yes that is what I meant, but I dont get what you mean by "circular". Would the system eventually return to a low entropy state? :/

No, not circular in the process but circular reasoning. "High entropy" is the name we give to the states which are more likely and therefore are those to which systems tend to evolve (It's unlikely that a system evolves to an unlikely state. You see the circular reasoning). So the question "why does the entropy of enclosed systems increase?" is answered by "because that's the name we attach to those states" ;-). The concept of entropy has been developed by observing systems and their behavior (and then, of course, understanding the principles and giving a formal description of them).

It appears you're looking for an ELI5-style answer, not an elaborate definition.
Entropy just happens – as long as the universe isn't frozen solid, things will always be moving around, and that movement tends to introduce randomness more than it tends to introduce order.
Consider a deck of cards. Shuffle it. Is it perfectly sorted? No. Why? There are 10^67 ways to arrange a deck of cards, but only one way to sort it; chances are you didn't shuffle the deck back into perfect order. The number of different available states (e.g. card sequences) shows up in the various formal definitions of entropy.
The same principle applies everywhere else in the universe: in the face of random (microscopic) state changes, the odds are stacked in favour of mishmash and against (macroscopic) "order" appearing by mere chance – and overwhelmingly so.

Thermodynamics. The second law of thermodynamics states that entropy always increases in an isolated system. This is taken as a fundamental postulate---we simply accept this statement as a fact regarding how the world works, and our justification is that no experiment has ever shown the second law incorrect. In the framework of macroscopic thermodynamics, there is no deeper principle from which we can hope to derive the second law.
Statistical mechanics. Statistical mechanics brings a microscopic perspective into thermodynamics. We talk of macrostates---macroscopic states of our system, characterized by (macroscopic) observables like pressure, temperature, volume,  etc.---which may be realized by a number of microstates---a complete microscopic description of our system, consisting of the positions and momenta of each particle. Many different microstates can correspond to the same macrostate, since a macrostate only deals with macroscopic properties.
In statistical mechanics we can go one level deeper, taking as a fundamental postulate the principle of equal a priori probabilities, the idea that the system has an equal probability of being in any given microstate. This implies that the most likely macrostate for the system is that with the greatest number of corresponding microstates. At this point, knowing also from statistical mechanics that the entropy is a monotonically increasing function of the number of microstates (specifically the logarithm of the number of microstates in an isolated system, but the functional form is unnecessary here), we can motivate the second law: the entropy of a macrostate of an isolated system is a measure of the number of its corresponding microstates; the greater this number, the more likely this macrostate will be observed. The equilibrium state of a system is simply its most probable macrostate.
Example. (Microscopic diffusion.) Let us consider a microscopic model of diffusion: a $10$-by-$10$ grid of particles, $50$ red and $50$ blue. The particles are indistinguishable except by color. In addition, we will consider two sets of macroscopic states, 'ordered' and 'disordered'. Take as initial state the configuration with the two groups of particles separated by color into two equal halves by a horizontal boundary, the blue particles on the top and the red on the bottom. The diffusion process is akin to removing this artificial boundary.
There is one initial ordered microstate and $100!/(50!)^2$ final microstates. (Note that indistinguishable microstates should be considered the same microstate.) The entropy of our system has increased, because we have relaxed a constraint and allowed more microstates into our system. Most of these final states look disordered. Diffusion is therefore an entropically favorable process that brings an ordered system into a disordered one.

Additional remark. I find statements like "entropy drives the system towards a particular state" to be somewhat misleading, because they imply that entropy behaves like a force. But entropy, as we've seen, is really just statistics.

This is an excellent answer. I have a physics degree, and had an excellent SM lecturer, and I still had an epiphany reading this.

Thanks for the excellent answer! Both you and @sebastian_k talked about somewhat the same thing, that we dont know (yet) of a more fundamental cause for entropy. Yet I am selecting the latter answer, in part because it talks with a more relatable analogy, and in part because (TBH) I didn't fully get your point on microstates (too bad, I know) :)

Do not think of entropy as 'disorder' as this is misleading, better is that it is a 'measure of disorder' but this is equally vague. It is better to think of entropy as the number of ways that 'particles' or quanta (say vibrational or rotational quanta in a molecule) can be placed among the various energy levels available.
Thus at zero energy all the particles are in their lowest levels and there is only one way of doing this and so entropy is zero. As the energy is increased then more levels can become populated and so there are many different ways of populating all the energy level this and so the entropy is increased. (Think of the calculations, usually given as examples in calculating probability, where one has to find the number of ways of placing identical balls into bags.) The connection of entropy $S$ to counting the number of ways was discovered by Boltzmann with his famous equation $S=k_B\ln(\Omega)$ where $\Omega$ measures the number of ways of distributing particles among the energy levels. From $\Omega$ you can sort of see where the 'disorder' tag  comes but using this word one misleads oneself.

If we look at why a chemical bond occurs maybe that might shed light on to this question. If we take diatomic hydrogen which has one electron from each hydrogen, the question arises why do two nuclei come together to form a bond since Coloumbic forces are increased. And the answer is because the electrons have a larger volume to exist in as they will be able to exchange between nuclei(I'm referring to the exchange integral). Thus we can summarize that a chemical bond occurs due to purely entropic reasons and as a result the energy of the system is decreased.

Everything you wrote here is wrong.


text:
How do you experimentally calculate the ionization constant of the carbonate ion?
Given sodium hydrogen carbonate, $\ce{NaHCO3}$, as well as hydrochloric acid, $\ce{HCl}$, how would you experimentally determine the ionization constant for the carbonate ion, $\ce{HCO3-}$? By experimentally, I mean by actually doing the reaction and measuring the pH and such, and ignoring the already known theoretical values.
Here's what I've got so far.
The sodium hydrogen carbonate dissociates in water.
$$\ce{NaHCO3->Na+_{(aq)} + HCO3-_{(aq)}}$$
Then, titrate the ion with the hydrochloric acid, a strong acid.
$$\ce{HCO3-_{(aq)} + HCl_{(aq)} <=> H2O_{(l)} + H2CO3_{(aq)} + Cl-_{(aq)}}$$
It's at this point that I would measure the $\mathrm{pH}$ and determine the necessary concentrations, and then plug it into the formula
$$K_\mathrm{b} = \frac{\ce{[H2CO3]}}{\ce{[HCO3-][HCl]}}$$
But I'm wondering if I need to go further with the $\ce{H2CO3}$ since it would hydrolize in water:
$$\ce{H2CO3_{(aq)} + H2O_{(l)} <=> H3O+_{(aq)} + HCO3-_{(aq)}}$$

So far, so good. You'll have to graph the $\mathrm{pH}$ of your system with the volume of titrant you've used to construct what's called a titration curve.
When you do the titration curve for this experiment, depending on the relative strengths of the bases involved, you will see one or two points of inflection points that will signal that the equilibrium conditions have changed. This will tell you if you need to take into account the second ionization. If there is a second inflection point, that's the second ionization.
Without actually going ahead and looking at the $\mathrm{p}K_\mathrm{a}$ of $\ce{H2CO3}$, I think for most computations, it could be omitted, since it is a second ionization which has a much lower constant than the first. Not unless you're looking for zwitterionic states or something like that.


text:
Final temperature after caustic dilution: Interpreting the graph
I found the below graph in the Occidental Handbook on Caustic.
I'm confused, what may be the parameter noted on the four curves? 80-100-120-140 F?
The final Temperature is on the y-axis & initial water T is fixed as 70 F.

The labels on the curves are the initial  temperatures of the 50% caustic soda solutions.

@ChesterMiller I think you can post your comment as an answer:)

The labels on the curves are the initial temperatures of the 50% caustic soda solutions.


text:
Decomposition of LiNO3
I have a doubt about decomposition of alkali metals.
Why does $\ce{LiNO3}$ decompose to give $\ce{Li2O}$ and $\ce{NO2}$ where as other alkali metals decompose to give corresponding nitrite and oxygen?

Have you checked Wikipedia?

Upon thermal decomposition, $\ce{LiNO3}$ gives lithium oxide ($\ce{Li2O}$),
  nitrogen dioxide, and oxygen:
$$\ce{4 LiNO3 → 2 Li2O + 4 NO2 + O2}$$
Other group I nitrates decompose differently, forming the nitrite salt and oxygen. Because of its relatively small size, the lithium
  cation is very polarizing, which favors the formation of the oxide.

You actually can get lithium nitrite from the nitrate by controlling the decomposition temperature around 500°C.  But yes, as Nilay points out the nitrite is more prone to decomposing further than sodium nitrite because of the polarizing power of the lithium cation.


text:
USNCO question - Finding pH of salt solution from pKa of its parent acid
The question from 2017 USNCO is as below:

What is the $\mathrm{pH}$ of a $\pu{0.25 M}$ solution of $\ce{NaCN}$? (The $\mathrm{p}K_\mathrm{a}$ of $\ce{HCN}$ is $9.21$.)
(A) $4.91$
(B) $8.61$
(C) $11.30$
(D) $13.40$

The answer sheet states (C).
My understandings/reasoning:

salt hydrolysis: solution of salt derived from strong base and weak acid is basic;
$\ce{NaCN}$ is derived from the base $\ce{NaOH}$ (strong) and acid $\ce{HCN}$ (weak according to given $\mathrm{p}K_\mathrm{a}$);
Solution is basic, so (A) is incorrect.

However, I am unable to proceed from here.

Sodium cyanide is a salt formed by a strong base and a weak acid, so the reduced ionic equation of hydrolysis is
$$\ce{CN- + H2O <=>[$K_\mathrm{h}$] HCN + OH-}$$
$$
\begin{align}
K_\mathrm{h} &= \left.\frac{[\ce{HCN}][\ce{OH-}]}{[\ce{CN-}]} \qquad \right|\cdot \frac{[\ce{H+}]}{[\ce{H+}]}\\
&= \frac{[\ce{HCN}][\ce{OH-}][\ce{H+}]}{[\ce{CN-}][\ce{H+}]}\\
&=\frac{K_\mathrm{w}}{K_\mathrm{a}} \label{eq:1}\tag{1}
\end{align}
$$
Sodium cyanide is highly soluble salt, so one can neglect auto-protolysis of water, and also express hydrolysis constant (assuming $[\ce{OH-}] = [\ce{HCN}]$ and $C$ – concentration of sodium cyanide) as follows:
$$K_\mathrm{h} = \frac{[\ce{OH-}]^2}{C-[\ce{OH-}]} \label{eq:2}\tag{2}$$
Equating \eqref{eq:1} and \eqref{eq:2}:
$$\frac{10^{-14}}{10^{-9.21}} = \frac{[\ce{OH-}]^2}{0.25-[\ce{OH-}]}\, ,\label{eq:3}\tag{3}$$
and solving the quadratic equation \eqref{eq:3} for $[\ce{OH-}]$, one can find that $[\ce{OH-}] \approx \pu{2e-3 M}$, $\mathrm{pOH} = 2.70$, hence
$$\mathrm{pH} = 14.00 - \mathrm{pOH} = 14.00 - 2.70 = 11.30$$

Thank you. I noticed that you used water's dissociation constant at 25°C. However, many questions specify the temperature, while this doesn't, leading me to believe that a temperature-independent solution would be preferred.

@GeorgeTian There is none unless the problem itself explicitly tells you the temperature or $K_\mathrm{w}$. In the absence of such info, it's safe to assume that $K_\mathrm{w} = 10^{-14}$ and move along.


text:
Is it possible to self-teach yourself most of organic chemistry and biochemistry without access to a lab or university resources?
I'm stationed on a US Navy warship and I often wonder if its possible to master these subjects without access to a college instructor. I've been reading introductory books on organic chemistry and while I've gained a lot of knowledge over the past year, I still cannot carry out any experiments.
I have three more years left on my contract before I can ever step foot in a chemistry classroom. I would like to eventually know everything about organic chemistry and biochemistry from reading IR spectroscopy / NMR signals to knowing everything about the properties of various functional groups along with their medicinal applications. I would like to get to the stage where I can propose what reagents to use for various reactions, but I realize that is made more difficult by the fact that I don't have direct experience in the lab.

I think you might find this answer relevant (from the question House-hold chemistry experiments for a high-school student).

Teach yourself,  yes, master,  no. Mastery will come with the (simplified)  cycle of hypothesis -> test -> conclude or re-hypothesise,  as appropriate. Learning reagents out of the lab won't give appreciation of ease of handling, scale up, safety, by-product formation or downstream processing ("work-up" in the lab), nor the impact of heat dissipation or retention during such reactions, gas evolution or incomplete phase splits (emulsions); simply scratching the surface with these examples

If you have a general background you are clever and study good book you can surely achieve better knowledge than not so talented university students. This is possible. Forget mastering. But there is no need to. Most of working chemist and even researcher can spend their life about a couple if not single reaction/ mechanism/ class of compounds.....etc

As much as I'm intrigued by this question, frankly speaking, it doesn't really fit on the site. As of posting this comment, 3 close-votes have been cast under "Primarily Opinion Based" (which is what this is). I really admire your zeal to self-learn this (I've a similar tryst with Org.Chem. and BioChem. , so I think I understand your plight). You're more than welcome to pop over to the site chat room for help. I'd recommend you talk to Jan, orthocresol, andselisk, Loong, Martin about those topics. Cheers, and good luck to you!


text:
Why are 3 equivalents of organolithium needed in synthesis of ketone from carboxylic acid?
For the reaction of a carboxylic acid and organolithium reagents to synthesise a ketone, Clayden's, on p 219, mentions:

Notice that three equivalents of organolithium are needed in this reaction: one to deprotonate the acid, one to deprotonate the hydroxyl group, and one to react with the lithium carboxylate.

It seems to me that "the acid" and "the hydroxyl group" refer to the same thing. A carboxylic acid molecule only has one acidic proton and it belongs to a hydroxyl group. Thus, shouldn't these two terms refer to the same thing and only two equivalences of the organolithium reagent be necessary?
Reference
Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

The text box you are quoting from (p 219) refers to this specific reaction. The compound in question possesses an α-hydroxyl group, which needs to be deprotonated too.

Notice that three equivalents of organolithium are needed in this reaction: one to deprotonate the acid, one to deprotonate the hydroxyl group, and one to react with the lithium carboxylate. These chemists added a further 0.5 for good measure.

However, the scheme directly above this incorrectly indicates that 3 equivalents of MeLi are required for a generic carboxylic acid:

In the absence of any other acidic groups, only 2 equivalents of organolithium are needed. In practice, however, more than 2 may be added.
Note that the 3 protons are still needed to get to the protonated species shown directly afterwards. In practice the proton source (e.g. saturated aqueous $\ce{NH4Cl}$, or just plain water) is usually added in excess.

In practice, ‘more than 2’ is usually on the order of 2.1 or so.


text:
Why is the cyclopropenium ion aromatic?

I don't understand how the 'cyclopropenium' ion is aromatic. According to my understanding, the carbons within the central 3 carbon ring must each be sp2 hybridised with a single electron in a p orbital - since each carbon is bonded to 3 other atoms and has 4 valence electrons. How does that lead to aromaticity, since this doesn't fullfil Huckel's rule?

https://en.wikipedia.org/wiki/Cyclopropenium_ion#Bonding

Related: What is the reason for the exceptional stability of tricyclopropyl-cyclopropenyl carbocation?

https://en.wikipedia.org/wiki/Aromaticity.

It's because the of the small size of the molecule. The hydrogen on the cation does not interfere with pi electrons orbiting around the cyclic molecule.

It does meet Huckel's rule.  When you have $m$ carbon atoms in a conjugated ring with a positive charge of $q$, there are $m-q$ pi electrons.  Here $m=3, q=1$ therefore the number of pi electrons is $m-q=2=4×0+2$.
A Huckel model calculation indicates that cyclopropenyl cation has roughly as much stabilization as benzene, but it's distributed over three instead of six carbon atoms.  Thus aromaticity is especially powerful in a cyclopropenyl cation ring.   This explains how a positive (formal) charge can be stabilized on as few as three carbon atoms and the ring holds together despite steric strain.

Thanks very much Oscar, I supposed i overlooked the positive charge. You mention how stabilization is distributed over three carbons instead of six, are you referring to the positive charge?  Am i right in thinking that in general a larger distribution of charge lead to greater stability?

I am comparing the pi-electron stability based on the "Frost circle" which is a picture of the Huckel model) versus ordinary valence bond structures.  Benzene gives a difference equal to one "extra" pi bond in this model.  Cycylopropenyl cation gives that amount of stabilization too, but each ring atom gets one third of it not one sixth.  That bigger per-atom stabilization helps stabilize the charge and fight steric strain.

the carbons within the central 3 carbon ring must each be sp2 hybridised with a single electron in a p orbital

This is incorrect.
The requirements for Huckel aromaticity are:

There must be a continuous, planer ring of overlapping $p$ orbitals.
There must be $4n+2$ $\pi$ electrons in the system.

The cyclopropenyl fulfills these requirements because it is planar, has overlapping $p$ orbitals, and has 2 $\pi$ electrons ($n = 0$).

The cyclopropenyl cation is aromatic because it is meeting all the definitions of Huckel's rule of aromaticity:

All carbons are sp2 hybridised.
There are 3 carbon atoms which form a conjugated system and moreover it has a positive charge therefore it has $3$-$1$=$2$ pi electrons.
If we put $n=0$ in Huckel's expression ($4n+2$) pi electrons we are getting 2 as the result.
The cation also has overlapping planar and parallel p orbitals. You may be knowing that for maximum overlapping, orbitals should be parallel to one another and planar.

Since the required criteria are fulfilled, it is aromatic.


text:
Dissolution of hydrogen sulphide gas in water, and its reaction with cement
INChO 2016 problem 1.8
Each Problem in this question paper is like a comprehension, whose each subpart picks up where the previous left. This makes reading the previous subparts nearly mandatory for solving Problem 1.8, in order to gain proper context, sorry for that! I've copied the most relevant parts here for this subpart though
Problem 1 in question paper, Answer key
Subpart 8 of Problem 1:

Water vapour in this air condenses on the tank walls. H2S in the air then slowly dissolves in
  these droplets. Since oxygen is available in immediate vicinity of the droplets (because of
  high surface area to volume ratio as compared to sewage below), sulphur oxidizing bacteria
  (SOB) start accumulating in these droplets. In this new aqueous system, SOB oxidize
  sulphide species and produce sulphuric acid. The cement part in concrete walls can be
  represented as $\ce{(CaO)3(Al2O3)}$, which is attacked by sulphuric acid. Write the balanced
  reaction(s) responsible for sulphuric acid production in the water droplets and its possible
  reaction/s with the cement in concrete.

My attempt:
For the formation of sulphuric acid, I thought this:
$$\ce{H2S}+\ce{2O2->}\ce{H2SO4}$$
which seems a good enough redox reaction for me. (Sulphide oxidised to sulphate, whereas the oxygen got reduced).
This acid then formed should react with both $\ce{CaO}$ and $\ce{Al2O3}$ in the cement, because the former's a good base, while the latter's weakly basic and amphoteric anyway. Thus the second reaction should be:
$$\ce{6H2SO4\+(CaO)3(Al2O3)->3CaSO4\+Al2(SO4)3\+6H2O}$$
My query:
Neither of my answer's matches with the given answer, which is:
$$\ce{HS-(aq) + 2O2 + H2O -> SO4^{-2}
(aq) + H3O+}$$
and
$$\ce{CaO(s) + H2SO4(aq) -> CaSO4(s) + H2O}$$
I hope I've been clear enough, please help me understand where I went wrong. Thank you!

$\ce{H2S}$ on dissolving in water would give the bisulphide ion and hydronium ion. This is what makes aq. $\ce{H2S}$ slightly acidic.
So, you just need to balance $\ce{HS^-\to SO_4^{2-}}$. Again, the sulphate ion as we are dealing with a dilute aqueous medium.

As for the further reactions with the concrete, I suppose the $\ce{CaO}$ would react "faster" as it is basic compared to the amphoteric $\ce{Al2O3}$. The $\ce{H2SO4}$ concentrations due to the sulphur-oxidising bacteria would be quite low. Maybe the sulphuric acid is not enough to quench the $\ce{CaO}$.

@EashaanGodbole Yes, that could be a possibility.


text:
Determination of hardness of water using EDTA
Finding hardness of water by using EDTA. For determining permanent hardness we take the sample of about $\pu{50 mL}$ and then add Eriochrome Black T (EBT) indicator that results in wine red color due to calcium complex made by and also ammonia buffer. Then titrating it with ethylene diamine tetra acetic acid (EDTA) where change in color (i.e. sky blue) marks the end point of titration.
While carrying out experiment we add EBT as an indicator. Is there any criteria to add so in a fixed particular amount?

Thank you for your edit, this clarified a lot. I retracted my close vote and fixed few typos.

There really is no need to add EBT in a fixed amount. Calcium EDTA complex is far more stable at $\mathrm{pH} \approx 10$ (ammonia buffer) and will be the dominant Ca-compound in the solution even if there is a huge excess of EBT.
However, the amount of EBT is important to track the color change. If there is too much EBT, the color will be too dark, and one can easily miss the red-blue transition. If there are only trace amounts of EBT, the colors will be pale and again the transition is going to be barely visible.
The amounts of EBT given in the lab manuals are typically approximate and are estimated for the convenience of the personnel and to reduce the error in determining equivalence point due to unclear reading of the solution's color.


text:
Justifying the tremendous −I effect of trimethylammonium group by molecular orbital theory
The $\ce{-[NMe3]+}$ group is ranked almost at the first position in the −I effect series, which is often given at the very beginning of the teaching of general organic chemistry.
I know this can be justified by hyperconjugation, but I am interested in a rigorous justification by molecular orbital theory.

This group has a positive charge, any molecular orbital effect will be negligible compared to that.

https://chemistry.stackexchange.com/questions/88185/justifying-the-tremendous-%E2%88%92i-effect-of-trimethylammonium-group-by-molecular-orbi/135045

Powerful electron withdrawing substituents, like $\ce{-NO2}$ can be interpreted as having a large effect on the $\pu{\pi}$-system of an aromatic ring. Resonance structures can be drawn that put a positive charge at the ortho and para positions and put the electron out in the nitro group, on an oxygen.

This must be adjusted for $\ce{-N(CH3)3+}$, because it has no $\pu{\pi}$ electrons, and we can't draw structures with a double bond between the nitrogen and the ring carbon. It works thru a very strong $\ce{-I}$ effect. However, if we use the concept of hyperconjugation, and donate a $\pu{\pi}$ electron from the ring to the nitrogen and then a methyl group (making it $\ce{-CH3}$), unbounded to any other atom, we can imagine the oxygens in the picture replaced with methyls and will get the same distribution of charge in the ring: a positive charge at the ortho and para positions. This will result in meta substitution if any reaction does occur.
For $\ce{-N(CH3)3}$, it's like hyperconjugation in reverse. Interestingly, while $\ce{-N(CH3)3}$ makes it very difficult to substitute, and always meta, $\ce{-NH3+}$ is only a weak meta director.
Reference

Advanced Organic Chemistry, L.F. Fieser, 1961, p628

The question is so good that maybe it deserves another correct answer. While my answer above attempts to answer using the criterion mentioned by user57048 (hyperconjugation), hyperconjugation is generally a small effect, but the $-I$ effect of $-N(CH3)3+$ is so large that maybe an entirely different approach could be appealing - not that you could necessarily distinguish between them.
The inductive effect is not in question, and does not need hyperconjugation to explain it. But the $meta$ directing effect, without the possibility of a double bond between carbon #1 on the ring and the nitrogen presents a problem. The reference I used in my first answer also discusses this at length, but I'd like to expand it a bit.

The question mentions the $-N(CH3)3)+$ group, but we need one more thing: an aryl group attached to the nitrogen. And we need one more thing for explaining the $meta$ direction: the transition state. This is what is brought up in the discussion by Fieser, but without using the specific words $transition state$. While in the example of the $-NO2$ group, a charge separation is generated such that it can minimize the $pi$ electron density on the $ortho$ and $para$ positions, with $-N(CH3)3+$, we only have to consider the inductive effect as far as carbon #1. The incoming electrophile carries the positive charge that resonates around the ring and directs it to the $meta$ positions because attaching to the other positions puts a positive charge on the #1 carbon, which would be highly unfavored because it places two positive charges next to each other.
The question seems to have been of long standing. The idea of aronium ion intermediates was put forth by Pfeiffer and Wizinger. The structures shown as aronium ions $a$ and $b$ in the picture are the ones that do not lead to product; $c$ leads to product, and would seem to be a transition state (or close to it) rather than a true intermediate.
Reference:
P. Pfeiffer and R. Wizinger, Ann., 461, 132 (1928)

$\ce{Me}$ is a pushing group which pushes electron towards the nitrogen atom. As a result electron concentration is towards the nitrogen atom.
The anti bonding electrons of nitrogen are shifted to the bonding orbital of nitrogen. This is because the bonding orbitals are closer to the more electronegative atom which in this case is nitrogen.
So nitrogen becomes more "electron hungry" and in order to stabilize itself tries to pull electron towards itself and thereby renders a -I effect.

Will somebody explain what i have done wrong in this answer.

This is super hand wavy and probably not what the OP wanted.

Relative to what you did wrong: methyl does indeed push; there are two methyls pushing toward the N in -NMe2. However, when you add the third methyl (Me+) to make NMe3+, it becomes an electron withdrawer, and this FAR outweighs the two little electron pusher methyls. They all average out, pulling on the N, which then is a big puller from whatever it is connected to.

@JamesGaidis could you please write an answer for this.


text:
Why is this β hydrogen in trans position to the leaving group?
Problem:

One of the most widely used methods for the formation of $\ce{C=C}$ is β - elimination reaction of the following type:

The spatial arrangement for a bimolecular elimination (E2) of the type shown above requires a trans geometry of the β - hydrogen and the leaving group (X). In most of the cases, the most substituted olefin is the favored product. However, in the case of ammonium and sulphonate salts, the least substituted olefin is formed as a major product
Identify the major product in the following reaction:

My query:
As you can see in the diagram they've drawn, the $\ce{NR3+}$ is in the plane of the paper, while $\ce{H}$ is NOT in the plane of the paper. Thus, I would not consider them as trans, since they are not exactly opposite to each other, but are in fact, inclined to each other at certain angle. In the current diagram, I would only consider (1) the two H atoms on a wedge and a dash position, (2) the left phenyl and right $\ce{NR3+}$, (3) the right phenyl and the methyl - as trans to each other, as then they are exactly opposite.
As is obvious in the answer key, they've considered the given H atom and the $\ce{NR3+}$ as trans and have eliminated them to yield:

Thus I wish to know exactly where my reasoning is wrong? Thanks!

Complete problem 2 and its answer key for reference

Your assessment of which groups are trans to each other is correct but you have not taken into account that the molecule can adopt other conformers.
The classic confomation for E2 elimination is when the hydrogen and the leaving group are trans. This is because it maximises the orbital overlap between the $\ce{C-X \sigma ^*}$ and the $\ce{C-H \sigma}$ orbitals which are key to the reaction.

As shown in the question, the molecule will not undergo elimination without rotating into the appropriate conformation where the hydrogen and the leaving group are trans. This leads to the product shown.
Other points to note:

However, in the case of ammonium and sulphonate salts, the least substituted olefin is formed as a major product

This seems to be a complete red herring because there is no option for variably substituted products in this case (or in any other case of E2 elimination that I can think of?).
E2 elimination is also possible when the hydrogen and leaving group are syn, although it is usually less favourable. However, in this case the positive charge on the leaving group and the negative charge on the base might promote this pathway, leading to a different product with the two phenyl groups cis.


text:
Determining the rate equation from a reaction where one reactant is in excess
Lets say
rate = k[A][B]n
(A is in excess, lets say at a concentration of 50mM)
My data includes the initial concentration of B, and the initial rate of reaction at that concentration
I can rewrite the rate equation as as:
rate = kapp[B]n
I then take the natural logarithm of this equation and plot a graph of ln(rate) against ln[B], the gradient will give me the value of n.
Am I right in thinking that the y intercept will give a value of kapp? And if it does, to determine the true rate constant do I just divide kapp by 50? If not, how do you determine the true rate constant?

There is no real need to invoke a "$k^{app}$".
Let the rate be $r$, so:
$$r=k[\ce{A}][\ce{B}]^n$$
Since as $\ce{A}$ is in large excess:
$$[\ce{A}]=C(A)\gg[\ce{B}] \implies r=kC(A)[\ce{B}]^n$$
So $[\ce{A}]$ doesn't vary perceptibly during the rate measurement interval.
Plot the value of $\ln r$ versus $\ln [\ce{B}]$ in the knowledge that:
$$\ln r=\ln k + \ln C(A) + n\ln [\ce{B}]$$
So the gradient of that graph is $n$.
And the intercept of that graph is $\ln k + \ln C(A)$. Since as $C(A)$ is known, $\ln k$, and thus $k$, is also known.

Similar to the pseudo-first order approximation in case of Fischer esterification?

Don't know that, sorry!


text:
Greater bond distance for silicon diamond structure than for carbon diamond structure?
In the silicon diamond structure the bond distance is about two times the bond distance of the carbon diamond structure. Is this caused only by the greater dimension of $\mathrm{Si}$ atoms with respect to $\mathrm{C}$ atoms? Or are there other reasons for a greater bond distance in the case of silicon?

How could you conclude that is not?  It is tricky to isolate bond length from atom size.

Yes, both structures belong to the same diamond cubic type, and the interatomic distances for diamond and silicon (you can use "Measurements" tool from the JSmol menu)
$$
\begin{align}
d(\ce{C-C}) &= \pu{1.54 Å} \\
d(\ce{Si-Si}) &= \pu{2.35 Å} \\
\frac{d(\ce{Si-Si})}{d(\ce{C-C})} &\approx 1.53
\end{align}
$$
are in a good agreement with the corresponding covalent radii
$$
\begin{align}
r_\mathrm{cov}(\ce{C}) &= \pu{0.76 Å} \\
r_\mathrm{cov}(\ce{Si}) &= \pu{1.11 Å} \\
\frac{r_\mathrm{cov}(\ce{Si})}{r_\mathrm{cov}(\ce{C})} &\approx 1.46
\end{align}
$$
which would set the silicon atoms further apart from each other.


text:
Change in Free Energy in Gas Reaction
I try to estimate whether particles form pairs or not.
What I know is the difference in the potential Energy $\Delta U_\mathrm{pot}$ which occurs when both particles form a pair.
When I have a system with $N$ particles, at constant volume and constant temperature the formation of pairs, should reduce the entropy of the system. So to see whether the pair formation occurs spontatnous I want to calculate the Gibbs Free Energy.
I assume that the particles behave like an ideal gas, but each particle consists of $n$ atoms has $(6n - 6)$ degrees of freedom.
The change in the gibbs free energy from system 1 ($N$ single particles, each consists of $n$ atoms) to system 2 (particles from $0.5N$ pairs, each pair consists of $2n$ atoms) I have:
\begin{equation}
\Delta G = \Delta U + V\Delta p - T\Delta S
\end{equation}
The internal energy I can calculate from the equipartition theorem:
\begin{align}
\Delta U &= U_2 - U_1\\
 &= U_{\mathrm{kin}, 2} - U_{\mathrm{kin}, 1} + \Delta U_\mathrm{pot}\\
 &= 0.5 N \cdot (6n - 3)kT - N \cdot (3n - 3)kT + \Delta U_\mathrm{pot}\\
 &= N(3n - 1.5)kT - N(3n - 3)kT + \Delta U_\mathrm{pot}\\
 &= 1.5 N kT + \Delta U_\mathrm{pot}
\end{align}
$V\Delta p$ I get from the ideal gas law
\begin{align}
V\Delta p &= \Delta N k T\\
V\Delta p &= 0.5 N k T
\end{align}
Now my problem is $T \Delta S$. All I find is the the Sackur–Tetrode equation
\begin{equation}
\frac{S}{kN} =  \ln\left(\frac{V}{N\Lambda^3}\right)+\frac{5}{2}
\end{equation}
which describes only the change the entropy of a monatomic ideal gas ($\Lambda$ is the thermal wavelength).
How can I calculate the entropy difference for the case of multiatomic ideal gas?

Are you saying that you are assuming that the reaction goes to completion?  Are you familiar with the concept of free energies of formation?  Of chemical potential?  Do you know how to determine the partial molar free energy of a species in an ideal gas mixture?

Yes, I want to compare both extreme cases, and therefore am assuming completion. I am aware of free energies of formation or the chemical potential, but these are hypothetical species, and therefore there exist no tabulated standard values for the compounds.

If I would know the free energy of an multiatomic ideal gas (as a function of the number of degrees of freedom), I could of course directly calculate the change in the free energy.
How does the molar free energy of a mixture help me, when I am considering the cases of full conversion?


text:
Why in SN1 reaction the order of reactivity of alkyl halides increases from fluorine to iodine?
In $\mathrm{S_N1}$ mechanism alkyl halides need to be ionized, so wouldn't it make more sense if alkyl fluoride reactivity were the most, since it is the most electronegative element?

https://chemistry.stackexchange.com/questions/6808/how-would-substitution-reaction-results-differ-between-reacting-halides-with-a-t https://chemistry.stackexchange.com/questions/55829/why-are-more-electronegative-atoms-better-leaving-groups

https://chemistry.stackexchange.com/questions/76631/must-a-good-leaving-group-be-a-weak-base-or-a-weak-nucleophile

Properties of a neutral fluorine atom do not necessarily extend to a fluoride anion.

I can see why on first thoughts you would think fluorine is going to be the most reactive of the halogens, this may be due to the fact you always think of fluorine as the most electronegative; and therefore the most comfortable carrying a negative charge.
However, electronegativity is only sometimes a good indicator of the stability of the leaving group/conjugate base (i.e the F-, Cl-, Br-, I- leaving groups in a typical SN1 reaction). The reason why iodine functional groups are more reactive is purely because it's more stable when carrying that negative charge after it has left. This is because iodine is a much much larger anion and can delocalise that negative charge over a much larger area.

So normally the rule for good leaving groups, or when you are looking at acidity (HF,HCl,HBr,HI), is that the more stable the leaving group/conjugate base, the more reactive that functional group is going to be. There are other factors to consider when looking at different leaving groups, but in the case of the halogens, the increasing size of the anion, and therefore its overall stability is the determining factor in its leaving group ability..


text:
Is there a sure shot way of telling whether a reaction is cis addition or trans addition?
For example we say Hydrogenation in cis addition , by looking at a reaction and analyzing can we say whether it’s cis/trans addition? Or Is it purely on the basis of experiments we say it’s cis/trans

What do you mean by looking at the reaction?

Like I give you a reaction and you can tell me whether it’s cis or trans addition

That's exactly the problem. What does that mean? You provide the mechanism? The products? The name of the reaction?

there cannot be a way in which you can always say whether a reaction is cis- addition or trans- addition just by looking at the reaction unless you know its mechanism or the fact itself and mechanism is proposed using the experimental facts. So experiment is the ultimate source.


text:
Are gem-diols on alkenes stable?
Geminal diols are unstable, except for a few exceptions like chloral hydrate.
I wish to ask if a gem-diol alkene is also stable. Their structure:

suggests to me that the intramolecular hydrogen bond should stabilize the gem-diol.

No enough. This is the enol tautomer of a carboxylic acid...

@Zhe Oh! Sorry but I didn't know that the carbonyl group on the carboxylic acid can tautomerize. Please post it as an answer. Thank you!

That's a bit short for an answer for me, but you're welcome to answer your own. Nevertheless, the main point stands. Enols are generally not stable. Doesn't really matter if it's a geminal diol.

I’m sure I answered on a four-membered ring in hydrogen bonds somewhere … https://chemistry.stackexchange.com/questions/60640/why-cant-alcohols-form-hydrogen-bonded-dimers-like-carboxylic-acids

@Jan  Oh, so interpreting your answer, it means that the configuration I drew above is highly unstable because there are four atoms in one strained ring of angle 90deg?

Well, the $\ce{O-C-O}$ angle will be close to 120°. There is no way hydrogen will be aligned well between the two oxygens for an overall stabilising effect. I am hesitant to talk about ring strain, because there is no ring that would be formed and that could include strain.

Anyway the point isn't really about hydrogen bonding. Even if the stabilisation from the hydrogen bonding were significant it's highly unlikely that an enol of a carboxylic acid would be stable.

What you call a "gem-diol alkene" might also be called an "ene-diol" or "endiol" by analogy to the name enol.
Enols are a tautomeric form of aldehydes or ketones.  They have one double bond and one C-O single bond.  Since "ene-diols" have one double bond and two C-O bonds, the oxidation number of the O-bonded carbon is 2 higher than with enols.  This corresponds to the oxidation number of carboxylic acids.  Thus, "ene-diols" are (theoretically) tautomers of carboxylic acids.
Ene-diols are not stable because they would rapidly tautomerize the more stable carboxylic acid form.
Hydrogen bonds like the one you have drawn are unlikely to stabilize enols very much, for reasons that Jan noted in a related question here on chem.SE.


text:
Order of Reactivity of Halogens in Electrophilic Addition

What is the order of reactivity of halogens in electrophilic addition reaction via formation of cyclohalonium intermediate?
(A) $\ce{F2}$> $\ce{Cl2}$> $\ce{Br2}$> $\ce{I2}$
(B) $\ce{I2}$> $\ce{Br2}$> $\ce{Cl2}$> $\ce{F2}$
(C) $\ce{I2}$> $\ce{Br2}$> $\ce{F2}$> $\ce{Cl2}$
(D) $\ce{Br2}$> $\ce{Cl2}$> $\ce{I2}$> $\ce{F2}$

I tried to predict it using the mechanism, but couldn't reach a conclusion. Is this based on thermodynamic and kinetic data?
Here's the mechanism involving bromine, for reference:

Could someone provide a detailed explanation for the above problem, with data if possible?

related https://chemistry.stackexchange.com/questions/41130/does-the-addition-of-chlorine-to-carbon-carbon-double-bonds-proceed-through-a-ha

Good to learn something new, but it didn't answer my question.

Edit your question to remove salutation/thanks etc. Also add what you think about possible answer. This is rather important question so it should be rather improved then tempt to be closed as homework.

The rate-determining step of halogenation of alkenes is the formation of cyclic intermediate.
The cyclohalonium intermediate of bromine will be more stable than that of chlorine owing to its lower electronegativity.
The following paragraph is taken from Peter Sykes [$1$, p.$181$-$182$]:

It is not normally possible to add fluorine directly to alkenes as the reaction is so exothermic that bond fission occurs. Many alkenes will not add iodine directly either and when the reaction does occur it is usually readily reversible.

As is clear $\ce{I2}$ will be more reactive than $\ce{F2}$.
Hence, the correct answer is D.
References:

Peter Sykes; A Guidebook to Mechanism in Organic Chemistry 6th ed.; 2003 ISBN-10:8177584332

Ans is (D)
Since F+ is not possible and
in case. of I2 it will be reversible
so In I2 &F2 the order is I2>F2
So ans is (D)

So how does Fluorine react ?

I would say (B).
The mechanism involves the breaking of the Halogen–Halogen bond. The strength of this bond decreases down the group because atomic radius increases. The attraction of the halogen nuclei on the electrons in the covalent bond decreases as the valence electrons are further away from the nucleus.
Therefore, the I–I bond is weakest while F–F is strongest. Hence, I2 is most reactive and F2 is least.

ans is d               Normally not possible to F2 because it is highly exothermic with I2 reversible reaction takes place
due to more E.N of Cl2 compare to Br2. It does not give e pair to carbocation


text:
How dried up detergent water led to precipitation of snowflake like shapes?
I forgot to drain the water off the bucket (after cleaning stain from a piece of clothing with detergent) and left it for few days. After few days when I needed it back, I found these glittery very small snowflake like shapes at the bottom. Pardon if it's very basic thing, I haven't studied chemistry after school. Thanks for the help.

There are many water soluble molecules in soap. I believe that as the water evaporated, the concentration of these molecules became great enough for them to start crystallizing. This process is used in chemistry as a means of purification and is known as recrystallization.
Are you considering the X type shape of the crystals to be snowflake like? I am not quite sure why they all seem to have similar geometries, I want to guess that they are crystals of citric acid.

Yes X ones, But why X shapes?


text:
Electrolysis of diluted sulfuric acid
My teacher told me that electrolysis of dil. $\ce{H2SO4}$ (concentration below $50\%$) using inert electrodes results in gradual increase of the concentration of $\ce{H2SO4}$.
I can't seem to figure out why. I know that the sulfate ions will remain in the solution, but aren't the $\ce{H+}$ ions of $\ce{H2SO4}$ reduced at the cathode as well?
Another doubt I have is that I've seen different equations for the oxidation at the anode for this electrolysis in different books. Some of them show that $\ce{OH-}$ will be reduced, and some show that $\ce{H2O}$ will be reduced! I'm really confused about all this.

In a dilute solution of sulfuric acid, there are the following species present: $\ce {H_2O}$, $\ce {H^+}$, $\ce {OH^-}$, $ \ce {HSO_4^-}$, $\ce {SO_4^2-}$. Of course, the water molecules are present in the highest concentration, much higher than the other species, since it is a dilute solution. For simplicity, we would also represent the proton and hydroxide ion as unsolvated. However, do take note that they are in fact heavily solvated in aqueous solution.
In some textbooks, it may be said that for the electrolysis of dilute $ \ce H_{2}\ce S\ce O_{4}$, the oxidation half-equation is written as $\ce {2H^+ + 2e^- -> H_2}$ and the reduction half-equation is written as $\ce {4OH^- -> 2H_2O + O_2 + 4e^-}$. However, the concentrations of these ions in solution are rather small compared to the concentration of the water molecule in solution. Albeit these reactions may be taking place, they should not be the main reactions and are unlikely to be representative of what is actually happening in solution.
In fact, since water molecules are the dominant species in solution, it is more likely that they undergo the redox reactions. The following half-equations represent the reactions happening in the solution more accurately:
Reduction: $\ce {2H_2O + 2e^- -> H_2 + 2OH^-}$
Oxidation: $\ce {2H_2O -> 4H^+ + O_2 + 4e^-}$
Since water molecules are being used up in the electrolysis process, the concentration of the remaining ions increase as the solution is electrolysed. Thus, the concentration of the sulfuric acid solution should increase.

Then what is the purpose of using an electrolyte (sulfuric acid)?

$\ce{OH-}$ can't be reduced. It may only be oxidized by giving an electron at the anode, while $\ce{H+}$ is reduced instead of $\ce{OH-}$.
Secondly, the reducing property of $\ce{H2O}$ is greater than that of $\ce{H+}$ ion. Moreover, it is greater in number than other ions like $\ce{H+}$ ions etc. So it will be preferentially discharged although it is already discharged but preferentially reduced at the cathode.


text:
What pressure units should be used when calculating electrode potential with Nernst equation?

Example 17: Calculate the electrode potential of given electrode
  $$\ce{Pt,Cl2 (\pu{1.5 bar}) | 2 Cl- (\pu{0.01 M})}\,; \quad E^\circ_\ce{Cl2/2Cl-} = \pu{1.36 V}$$
Solution: The reaction of electrode is
  $$\ce{\underset{\pu{1.5 bar}}{Cl2 (g)} + 2 e- -> \underset{\pu{0.01 M}}{2 Cl-}}$$
  $$E = E^\circ - \frac{0.0591}{n}\log{\frac{[\ce{Cl-}]^2}{P_\ce{Cl2}}} = 1.36 - \frac{0.0591}{2}\log{\frac{(0.01)^2}{1.5}} = \pu{1.483 V}$$

In the solution, the author has directly used the pressure (which is in bars) in the Nernst equation without first converting it to its SI units. How can this be right?

Just as you are supposed to use activity instead of concentration for dissolved components when applying Nernst equation, you are also supposed to use fugacity instead of the partial pressure for the gaseous components. The reason for that will become transparent if you follow the derivation of Nernst equation using chemical potential.
Fugacity is defined with a reference to the standard state with the pressure of $\pu{10^5 Pa}$ or $\pu{1 bar}$. At ordinary pressures fugacity is numerically approximately equal to the pressure expressed in bars or atmospheres ($\pu{1 bar} \approx \pu{0.987 atm}$), so the author of the book is right.

Oh…I didn't know that. We didn't drive Nernst equation…I'm in class 12, and so I guess that's above my level... Well, thanks man! Also, what's the difference between activity and concentration?

@SenthilArihant No prob; I'd suggest to read https://en.wikipedia.org/wiki/Thermodynamic_activity


text:
Chemistry-bond energy
Considering the hyperconjucation effects with C-T,C-D and C-H bonds, the maximum efficiency is given by C-H bond.This should mean that C-H must be the weakest bond and must hve the least bond energy compared to others.While all are isotopes of the same element with completely identical outer electronic configuration,then with is the difference in the bond energy?

Last line-why is the difference in the bond energy?

Related: Why does hyperconjugation help for ring cleavage?; Why is a C–D bond stronger than a C–H bond?

Why is C-D bond stronger than C-H bond

In a nutshell: The isotope is only irrelevant for the electronic configuration of a free atom. In any molecule, the vibrational ground state is not at zero, and it should be obvious the mass makes a difference if there is relative motion. See also https://en.wikipedia.org/wiki/Kinetic_isotope_effect

Heavier isotopes make stronger bonds, therefore C-D and C-T bonds are stronger than C-H bonds. The difference between the bond strengths is however not very significant and can be ignored for most practical purposes.
Therefore, the extent of hyperconjugation can be given as: C-H > C-D > C-T (the difference however is not very significant).
The exact reason for this little difference involves a quantum mechanical explanation, which you should not bother about if you are studying high school chemsitry. If you want to read that, you can refer the links in the comment section.


text:
Can a lone pair be involved in cis/trans isomerism?
I was trying to find whether a molecule showed isomerism, and this is what I came up with.
I thought this was the case as I took the nitrogen to be sp2 hybridised, thus trigonal planar (with the lone pair not shown on the diagram). The molecule can't rotate about any single bonds because resonance makes the whole molecule planar. Am I right in thinking this shows cis trans isomerism? If not, what sort of isomerism might it show?

The single bond can rotate. The barrier to rotation might be slightly higher than usual, but it's not large enough to entirely prevent rotation. You are correct about the double bond displaying cis-trans isomerism.

I've heard this being referred to as "syn-" and "anti-", rather than "cis-" and "trans-".

Yes, due to the lack of rotation about the C=N bond this will display the desired isomerism. For a much simpler example see $ \ce{N2F2} $

(taken from https://en.wikipedia.org/wiki/Dinitrogen_difluoride )


text:
Can benzaldehyde undergo chlorination (aka halogenation)?
I've read articles online and they rarely mention about chlorination of benzaldehyde but more on bromination.
Also, I would like to know the catalyst used for this is FeCl3 or AlCl3?
Thanks for your help!:)

Homework? What does the literature say about the bromination? What does the literature say about using iron or aluminium chloride for other substrates? Please elaborate.

In my syllabus we only learnt about halogenation for benzene and toluene and its FeCl3 as the catalyst. However, I’m unsure if benzaldehyde is able to halogenate the same way as benzene and toluene. So thats my qns.

Please edit your question accordingly, and make clear that you want the aromatic substitution, not the chloraldehyde. And a hint: FC-alkylation works with catalytic amounts of AlCl3, FC-acylation not.

The experimental process of chlorination of benzaldehyde to form benzoyl chloride can be found here:

A round bottom flask is fitted with a reflux condenser and a gas inlet
  tube. 100 grams of cold and dry benzaldehyde are added to the flask
  and saturated with the current of dry chlorine. The chlorine gas is
  readily absorbed and the evolution hydrogen chloride takes place. When
  the reaction has moderated, the external heat is applied by
  keeping gentle boiling. The stream of chlorine is continued until no
  more hydrogen chloride is evolved. A stream of dry air or carbon
  dioxide is then passed through the apparatus in order to remove
  an excess of chlorine. Benzoyl chloride is obtained by distillation as
  a colorless fuming liquid with a very irritating smell. Yield almost
  quantitative.
Preparation of organic compounds, E. de. Barry Barnett, 52, 1912

There is also a German paper written by Freidrich Wohler and Justus von Leibig which mentions an early preparation of benzoyl chloride by the action of chlorine over benzaldehyde.

I think the OP wanted to know about halogenation of benzene nucleus.

Bromine Chloride in the presence of AlCl3 reacts with benzaldehyde to give 3-Bromobenzaldehyde - see US Patent 4036887 A


text:
Changing the suffix of the name of a compound loses one hydrogen atom?
I just chanced upon these intriguing pubchem entries: Ethyl Carbylamine and Ethyl Isocyanide. Their structures - as given on that site - are respectively:
 and .
Notice that the only difference is the H atom on ethyl carbylamine, which is missing in ethyl isocyanide!
I always understood carbylamine and isocyanide being interchangeable suffixes for an IUPAC name, but it seems like I've been proved wrong.

What is the reason for such an anomaly in the structures of these two compounds?

No, you weren't wrong. PubChem isn't as good as one could hope.

@Mithoron  Could you please post an answer so I could accept it? Thank you!

Both carbylamine and isocyanide are same. It is only that carbylamine has become obsolete. See: https://goldbook.iupac.org/terms/view/C00853


text:
Calculate the volume of ammonium hydroxide needed to get desired pH

I have an $29\%$ aqueous solution of ammonium hydroxide. How much do I need to add to $\pu{20 ml}$ of water to get $\mathrm{pH}$ $12.8$. Ammonium hydroxide $K_\mathrm{b}$ is $1.774\cdot 10^{-5}$.

What I tried:
$$K_\mathrm{b} = \frac{[\ce{OH-}][\ce{NH4+}]}{[\ce{NH4OH}]} \approx \frac{[\ce{OH-}]^2}{[\ce{NH4OH}]}$$
Could someone explain the rational for this approximation if it is correct?
Then
$$
\begin{align}
-\log{K_\mathrm{b}} &= -2\log{[\ce{OH-}]} + \log{[\ce{NH4OH}]} \\
\mathrm{p}K_\mathrm{b} &= 2\mathrm{pOH} + \log{[\ce{NH4OH}]} \\
[\ce{NH4OH}] &= 10^{\mathrm{p}K_\mathrm{b} - 2\mathrm{pOH}} \\
&= 10^{\mathrm{p}K_\mathrm{b}-2(\mathrm{p}K_\mathrm{w}-\mathrm{pH})} \\
\mathrm{p}K_\mathrm{b} &= -\log{(1.774\cdot 10^{-5})} = 4.75 \\
\mathrm{p}K_\mathrm{w} &= 14 \\
\mathrm{pH} &= 12.8
\end{align}
$$
Plugging in:
$$[\ce{NH4OH}] = 10^{4.75-2(14-12.8)} = 223.9$$
This is much too high and does not make sense.

Please explain/show what attempts you have made to solve this problem on your own; these types of open questions do not to adhere to the posting policy of the site.

Sure I will edit the post with what I tried. This is not a homework problem, I am a researcher trying to do chemistry in the lab, but my background is in physics, so I am struggling with some of the elementary concepts.

Technically, you calculated the concentration correctly (math-wise). However, your initial hypothesis is wrong.
First, you cannot get ammonia hydroxide solution of $\mathrm{pH}$ $12.8$ at the room temperature to work with. At $\pu{20^\circ C}$ concentrated solution would be approx. $35\%$ with $\mathrm{pH} \approx 12.0$. The higher the temperature, the lower the solubility and the $\mathrm{pH}$ will be. Having $29\%$ stock ammonia solution, there is no way to reach $\mathrm{pH}$ $12.8$.
Second, when you work with borderline/concentrated solutions, you are no longer allowed to use concentrations when describing the equilibrium, you need to use activities instead. The formula you derived would give a reasonable result for the moderately concentrated solutions only. You can adapt it to establish the relationship between the mass fraction $\omega$ and $\mathrm{pH}$, knowing the density of the solution $\rho$ ($M$ is molar mass of ammonia)
$$\mathrm{pH} = 14 + 0.5\lg{\frac{\omega\rho K_\mathrm{b}}{M}}\, ,$$
but that's pretty much all to it.

Firstly, "$\ce{NH4OH}$" doesn't really exist. Ammonia solutions generate hydroxide ions via:
$$\ce{NH3(aq) + H2O(l) -> NH4+(aq) + OH-(aq)}$$
The equilibrium constant $K_b$ is given by:
$$K_b=\frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}$$
Now let the nominal concentration of ammonia be $C(\ce{NH3})$, then because $\ce{NH3}$ is such a weak base:
$$C(\ce{NH3})\approx [\ce{NH3}]$$
So that we have:
$$K_b\approx\frac{[\ce{OH-}]^2}{C(\ce{NH3})}$$
And with $pH+pOH=14$ and $[\ce{OH-}]=10^{-pOH}$, you can then calculate $C(\ce{NH3})$, in the $20\ \mathrm{ml}$.
Then calculate the number of moles of $\ce{NH3}$ contained in that volume and calculate how much of the strong $\ce{NH3}$ solution you need to add to the $20\ \mathrm{ml}$ to reach $pH=12.8$.
Use the rule $C_1(\ce{NH3})\times V_1=C_2(\ce{NH3})\times V_2$ for the above.

This is pretty much what OP has already shown in the question. Have you actually tried to calculate the concentration? It's still going to be two-hundred-something mole per liter; it's not the math that it wrong, it's the initial assumption.


text:
Why does saline have a pH of 5.5?
I am a physician and administer 0.9% NaCl in water ("normal saline") to patients frequently. The bag says it has a pH of 5.5 (compared to the human body of 7.4).
I am asking why the pH is so low. It is because NaCl is the salt of HCl? I read on wikipedia that the pH of NaCl aqueous solutions is 7. I teach medical students and want to give an understandable explanation why the pH on a bag of "normal saline" reads pH 5.5 and when we give lots of saline to patients, they develop acidosis in the blood.

There is an entire article dedicated to this topic, actually:)

Wow. Thank you for the super fast replies!!! That is awesome. I will go through the article you linked and see what it has to say. Much appreciated.

It would seem that a normal saline solution isn't buffered at pH 5.5. Thus there is really very little acid in an absolute sense. A typical carbonated beverage would have thousands of times more acid.

Check this article: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3638298/ Saline (0.9% salt solution) seems to be a beast to measure its pH, but harmless to the patients.

The problem with acidic saline is the activation of bone resorption pathway to release calcium to normalize the blood pH to 7.4, putting old patients at risk of bone fracturs in cases at which high volume of acidic saline is administered.

Pure water in contact with air (like rainwater) also is slightly acidic (approximately pH 5.6).

This is an unfortunate example of a known problem with a known solution that for some reason is not implemented. Ringer's solution is an attempt to solve the problem. A lot is known about blood tonicity and an ideal ionic mix could be ascertained, but for some reasons it has not been done. A similar situation existed for the relationship between scurvy and the cure, fresh fruits and veggies. the problem waxed and waned for centuries. Similar problems are still possible with attitudes towards vaccines and vitamin supplements.

Medical saline has a $\mathrm{pH}$ of 5.5 do to the dissolved $\ce{CO2}$ within the solution as well as factoring in the degradation of the PVC packaging.
I reference the attached article from the International Journal of Medical Sciences from 2013:
Benjamin AJ Reddi, “Why Is Saline So Acidic (and Does It Really Matter?),” Int. J. Med. Sci. 2013, 10(6), 747-750 (doi:10.7150/ijms.5868)(PDF).


text:
How exactly can an increase temperature be used to break the chemical bonds of a gas [SO2, for example]?
So I understand the ideal gas law at a basic level, and that a rise in temperature can break bonds, however, I am unclear of how this works in a practical sense.
Say I have a container full of $\ce{SO2(g)}$ and I would like to break $\ce{SO2}$ into Sulfur ($\ce{S}$) and molecular oxygen ($\ce{O2}$). Would simply increasing the heat of the container to $n~\pu{K}$ be sufficient to break the bonds to get $\ce{S}$ and $\ce{O2}$ by themselves?

Imagine the molecules in motion. As temperature increases, the motion becomes more violent.  What do you think happens when the collisions between molecules or vibrations within a molecule increase?

Right, they break apart. I think I got it, thanks!

As we provide heat to any molecule, the energy of the molecule increases. Specifically, the energy of atoms increases. Using the fact that stability is inversely proportional to energy, we can conclude that as energy increases, the bonds between the atoms become more unstable and eventually break and constituent atoms disassociate.
This heat provided is known as bond disassociation enthalpy, or bond disassociation energy [BDE].


text:
Is it right to say that fluorescent lamps are actually fluorescent?
Fluorescence occurs rapidly ($\approx 10^{-6}$ to $10^{-9}$ s) but phosphorescence is slow, and is therefore observable ($> 10^{-4}$ s). So, is it right to say that fluorescent lamps actually fluoresce when they actually do is to phosphoresce? Or am I wrong in saying that?

Fluorescent lamps do primarily work by fluorescence. According to this Wikipedia article:

"The inner surface of the lamp is coated with a fluorescent (and often slightly phosphorescent) coating..."

First off, an electric current excites the outer electrons of gas atoms (typically mercury) within the lamp to a higher energy state. When the atoms relax back to their stable state, UV light is emitted. This UV light is then absorbed by a fluorescent material coating the inside of the lamp, which in turn fluoresces in the visible region of the EM spectrum. Some confusion could arise here as this coating is called a phosphor, even though the type of luminescence is fluorescence, not phosphorescence.
The rate at which the luminescence occurs is not important because there is a continuous supply of electric current. If it were necessary to provide light for any significant period of time after the power source was removed, then some type of phosphorescence would be necessary due to it's much longer lifetime as compared to fluorescence, as you pointed out. However, this is clearly not a requirement for a standard fluorescent lamp.

I wonder if the phosphors are solely fluorescent. Fluorescence would happen within nanoseconds. A phosphorescence  half-life of about 20 milliseconds would greatly help reduce the 120 Hz flicker of the fluorescent lamp.

@MaxW Your comment got me reading more about this and sure enough the phosphors used to coat the lamps can be slightly phosphorescent. Your point makes sense and I suspect it's a matter of cost that drives the use of fluorescent materials vs phosphorescent. That's my best guess anyway.

Ahh right! There's a continuous supply of energy that's why we can observe it. And the amount of outer electrons excited relates to the intensity of light emitted by the lamp that's why they can get pretty bright!


text:
Validity of the equation K_eq = ratio of rate constants of forward and reverse reactions
If we have a multi-step reaction (which involves more than one step to reach the products; a complex reaction, one may say), then is the following equation valid?
$$K_\mathrm{eq} = \frac{K_\mathrm{fwd}}{K_\mathrm{rev}}$$
$K_\mathrm{eq}$ – equilibrium constant of the reaction; $K_\mathrm{fwd}$ – rate constant of the forward reaction; $K_\mathrm{rev}$ – rate constant of the reverse reaction.
I know that it is valid for an elementary or single step reaction, where the expression for the rate involves raising the concentrations of the reactants to their coefficients in the balanced equation.
But is it so for a multi step reaction too? (Where the law of mass action is no longer valid?)

The case for non-elementary reactions is interesting. In fact, equilibrium constants arise from thermodynamics, with no reference to kinetics whatsoever. In the case of an elementary reaction, the strategy of equalling the forward and reverse produces the same equation, but why does or why doesn't that strategy extend to non-elementary reactions? Does microscopic reversibility guarantee the kinetics of even a very complex network of reactions generate a simple equilibrium constant between the stating materials and final products? I wonder if anyone can provide an answer touching on that.

If each 'step' in a sequence of N steps is an equilibrium and all steps are at equilibrium, there are N equilibrium constants and 2 N rate constants. Write out the differential rate law for each step (e.g. for $\ce{A<=>B}$, where $A$ does not appear in any other reaction, $\frac {d[A]}{dt}=-k_1 \cdot [A] + k_2 \cdot [B]$), and equate each to $0$. Depending on where reactants and products appear in each step, you will find whether the equation you mentioned holds or not, and in some cases you may be able to find a single expression linking the initial reactant and final product.

@user6376297 I don’t quite get it. Anyway, what I asked wouldn’t be valid in that case, yes?

I edited your question a bit for clarity. Note that notation $K_\mathrm{b}$ have another meaning, where b stands for "base"; also, the term "reverse" is preferred to "backward". Notations are updated according to IUPAC recommendations.

Related: How is it that the equilibrium constant does not depend on the mechanism?; Relationship between rate equation and equilibrium constant

@SenthilArihant : I don't think your question can be answered as such. As I wrote, you need to specify what the individual steps are, to know which concentrations are linked at equilibrium, and how. And BTW, if your overall transformation is composed by multiple reversible steps, what do the single forward and backward rates you mentioned in your post mean? Which reaction do they refer to? You should at least have the situation completely clear in you mind before you question its details.

I meant the ‘K’ of the overall reaction, which occurs when I write the rate law expression for the overall reaction.

Here’s an example: https://chemistry.stackexchange.com/q/88280/56916  Will my equation be valid here?

@andselisk So, according to the 2nd post, unless, for a reaction, the law of mass action is valid, my equation won’t be valid, correct?

@SenthilArihant Still, depends on what reaction you apply it to. It works fine for the brutto-reaction if you use corresponding constants for the forward and reverse brutto reactions.

@andselisk So, what you mean is that if, by chance, the rate law expression (determined using the slowest step of the reaction mechanism) has the concentrations of the reactants raised to the powers of their stoichiometric coefficients in the balanced equation, then my equation is valid?

Yes, I agree with that. But is that what you’re saying?

@SenthilArihant Kinetics of that slow step is already "imprinted" in the $K_\mathrm{fwd}$ and $K_\mathrm{rev}$ values for the brutto reaction.

@andselisk I’m not getting what you are saying. And what’s the brutto reaction?

@SenthilArihant That's adaptation of the German word "die Bruttoreaktion", which means "overall reaction" - the one I suppose you've written the ratio for.

Yeah, so are you saying what I think you’re saying in my 2nd comment before this one?

And so, in this example: https://chemistry.stackexchange.com/q/88280/56916.   The equation won’t be valid yes?


text:
Calculating rate constant and order of a multi-step reaction
If I’m given a multi-step reaction, say:
$$\ce{2 NO + O2 -> 2 NO2} \label{rxn:R1}\tag{R1}$$
with the following mechanism:
$$
\begin{align}
\ce{NO + O2 &<=>[$K$] NO3} &\quad &\text{(fast)} \label{rxn:R1.1}\tag{R1.1}\\
\ce{NO3 + NO &->[$K_1$] NO2 + NO2} &\quad &\text{(slow)} \, , \label{rxn:R1.2}\tag{R1.2}
\end{align}
$$
then will the rate constant of the whole reaction be equal to the product of the equilibrium constant of the fast reaction and the rate constant of the slower reaction?
Because this is how they made us solve it at my institute:
$$K_\mathrm{eq} = \frac{[\ce{NO3}]}{[\ce{NO}][\ce{O2}]}$$
$$
\begin{align}
\text{Rate of slow step \eqref{rxn:R1.2}} &= K[\ce{NO3}][\ce{NO}] \\
&= KK_\mathrm{eq}[\ce{NO}][\ce{O2}][\ce{NO}] \\
\Rightarrow\text{Rate of reaction} &= K'[\ce{NO}]^2[\ce{O2}]
\end{align}
$$
where $K' = KK_\mathrm{eq}$, and the order is $2 + 1 = 3$.
So, is $K'$ the rate constant for the overall reaction? And is the reaction a 3rd order reaction?

The method you've indicated in your solution is correct. Regarding:

Then will the rate constant of the whole reaction be equal to the product of the equilibrium constant of the fast reaction and the rate constant of the slower reaction?

In your question, the final $K'=K_1\cdot K_{eq}$, but this is not always true.
Remember that you've to adjust the final rate law expression in such a way such that it is free from all intermediates. In doing so, you might need to multiply the $K$ of another reaction, as you did above. But you may also need to divide or multiply the square/cube/etc. of another reaction.
Hence, $K_{final}$ is not always equal to $K_1\cdot K_2\cdot...\cdot K_n$. So, your assumption is incorrect.

Yeah, I get that. So whatever K’ I get in the final expression will be the rate constant of the overall reaction always?

Also, in the initial rate law expression (with the unstable NO3 concentration) will I always have to replace the most unstable species (i.e. NO3 here) to get be final rate law expression from which I’ll deduce the order of the reaction? Is that some kind of rule?

@SenthilArihant "Is that some kind of rule? " Yes, the rule is that the final rate law expression for a multi-step reaction should not contain any term for the intermediates in it.

@SenthilArihant "I get in the final expression will be the rate constant of the overall reaction always?" Yes, of course