text:
What does it mean that the standard molar entropy value is the amount of energy that a substance must have to exist at a certain temperature?
What does it mean that the standard molar entropy value is the amount of energy that a substance must have to exist at a certain temperature?
According to this reference: http://khimiya.org/volume15/Entropia.pdf the meaning of standard molar entropy is the relative amount of energy that the substance must have to exist at for example 298 Kelvin.
An example is given of diamond and graphite, which has 2.4 J/K and 5.7 J/K respectively. I see that graphite has less strong bonds than diamond, i.e. diamond has higher potential energy stored inside its bonds. But I don't understand how that translates into a lower amount of energy needed to exist at a certain temperature? Is this the energy that needs to be supplied to the diamoned to raise its temperature to 298 kelvin? But isn't it than just a kind of heat capacity? What happens if you did not transfer enough energy to make it exist stably at that temperature?

The meaning of standard molar entropy values
When seen in their relation to the energy content of a substance, standard molar entropy
values, $S^0_{298}$, give useful insight that is absent when - as is usual -
those J/K values are treated as merely abstract numbers to be added
or subtracted in determining a $dS_0$ reaction . A $S^0_{298}$ value for a
substance is the number of joules of energy/T transferred
incrementally (reversibly, from the surroundings at each T) to a mole
of substance from 0 K to 298 K. Thus, this number is a rough indicator
or approximate index (not the joules dispersed at 298 K, nor the total
joules dispersed from 0 K to 298 K!) of the relative amount of energy
that the substance must have to be exist stably at 298 K. This is why
$S^0_{298}$ values illuminate e.g., the difference in rigidity of bonding:
the more rigid bonds of diamond (2.4 J/K) vs. the looser interatomic
bonds in graphite (5.7 J/K) ,

https://chemistry.stackexchange.com/questions/126803/what-is-the-precise-definition-of-standard-molar-entropy

What you say in the title is simple incorrect.

I copied that from a well known author. I provided the link. Could you help me understand it?

Then you should cite properly, like whole paragraph, with revenant bit in bold. Even then it looks like mistaking a metaphor for definition.

I have added the citation, thank you for the suggestion

Chemistry, Vol. 15, Iss. 1 (2006): A MODERN VIEW OF ENTROPY, Frank L. LAMBERT, Occidental College, Los Angeles, USA. No DOI that i could see. Here is an archive version: https://web.archive.org/web/20201111235310/http://khimiya.org/volume15/Entropia.pdf

The wording of the cited paragraph is confusing, but it does not say that "the standard molar entropy value is the amount of energy that a substance must have to exist at a certain temperature" (as claimed in the question). In fact the author says specifically that this is not the case. Rather, it is "is a rough indicator or approximate index" of the amount of energy.
Mathematically, the change in internal energy of a substance as it is heated (at constant volume to avoid PV work) from absolute zero to a temperature T would be $$\Delta U = \int_0^T C_V dT,$$ where $C_V$ is the heat capacity at constant volume and is a function of T. For one mole of substance, we would use the molar heat capacity $\overline{C}_V$.
This is similar to but not the same as the expression for absolute entropy, which is $$S^\circ_T = \int_0^T \dfrac{C_P}{T} dT,$$
with the crucial difference being the T in the denominator. (The difference between $C_P$ and $C_V$ is of secondary importance.)
As long as the change in the heat capacity as a function of temperature is similar between two substances, then the relative ranking based on absolute entropy will be similar to a ranking of the internal energy, which is what the author means when he says that the former can be used as "a rough indicator or approximate index" of the latter. But clearly they are by no means the same and I would argue that it's somewhat misleading to suggest that the entropy should be used in this way.

Thank you for your answer, however I still don't understand how it illuminates the difference between the bonds of graphite and diamond. How are the bonds related to the heat capacity? Also why is it a rough indicator and not to be taken too literal?

The heat capacity gives us a measure of how much energy is distributed into vibrational modes as we increase the T. Conceptually, when there are looser bonds, the structure can vibrate more at lower temperature, so the heat capacity is greater for a given T. So the total energy that is needed to get the substance up to a given temperature is greater. The reason it is a rough indicator rather than a literal measure is because the first integral (the energy) is obviously much greater in magnitude (and in different units) from the second, which is the entropy.


text:
Are there any chemistry research classification systems?
There are research classification systems for mathematics (MSC) and physics (PACS and subsequently PhySH) which help the researchers to navigate through codes composed by letters/numbers made by associations that specialise in those pieces of human knowledge. Are there any classification systems of research in chemistry?

I guess I always thought there would be… Although, frankly, PACS is not very useful…

I'm not sure if it counts as a classification system, but SciFinder can be used to filter research papers based on many criteria. Each chemical has a CAS number as well.

source: Stack Exchange

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

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Does E1 mechanism always imply first order reaction?
Oxford University Press, Okuyama & Maskill: Organic Chemistry — Chapter 13: Multiple Choice Questions, Question 1:

Which of the following statements regarding the E1 mechanism is wrong?
a) Reactions by the E1 mechanism are unimolecular in the rate-determining step.
b) Reactions by the E1 mechanism are generally first order.
c) Reactions by the E1 mechanism usually occur in one step.
d) Reactions by the E1 mechanism are multi-step reactions.

Option c is the correct answer, but I have a few questions regarding option b being true:

Is the reaction which follows E1 mechanism always a first order reaction? Are there any examples with the second or higher orders?

Dehydration of secondary alcohol in the presence of concentrated sulfuric acid follows E1. What is the order for this reaction?

Does the number one in E1 represent molecularity or order of the reaction? According to me, it is usually molecularity of RDS, not the order, because option b says E1 reactions are generally 1st order, not always.

For the second question, hydrogen ion acts as a catalyst so I think that the rate should be proportional to it but it generally has fairly constant concentration so I think that it should be pseudo first order reaction. However I am not sure about it. Similar case also happens with reaction of tertiary alcohol with hydrohalic acid which follows SN1 mechanism. Here also according to your logic, it could be 2nd order reaction

I think the problem lies in definition of terminologies. There are various kinds of reaction which occur in multiple steps, but just because one of the elementary step reaction follows SN1 or E1, it does not mean the whole reaction will have the first order.
Same thing here. The cation formed when alcohol attacks on $\ce{H+}$ undergoes SN1 and the rate will be only proportional to concentration of this cation. Further, the concentration of this cation is proportional to both hydrogen ion concentration and alcohol concentration.
In conclusion, E1 is always the first order reaction, but the dehydration of alcohols is the second order reaction (or pseudo first order if proton is abundant) because water expulsion is just one of the elementary steps happening in the reaction.

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How to rewrite kinetic mass balance to get to proper SI units?
Suppose from kinetic mass balance I get the following differential equation of the molarity substances $A$,$B$, and $C$ in a reactor, with molar in- or outflow rate $\phi$, and rate constant $k$:
$$
\frac{\mathrm{d}}{\mathrm{d}t}[C]
= k[A]^2[B]^3 + \phi{C_{\mathrm{in}}} - \phi{C_{\mathrm{out}}}
$$
Where the dimensions convert to:
$$
[\pu{mol L-1 s-1}]
= [\pu{s-1}] [\pu{mol5 L-5}] + [\pu{mol s-1}] - [\pu{mol s-1}]
$$
Which can be simplified by adding the flow rate terms to:
$$
[\pu{mol L-1 s-1}]
= [\pu{s-1}] [\pu{mol5 L-5}] \pm [\pu{mol s-1}]
$$
By cancelling similar terms mole,seconds on both sides to:
$$[\pu{L-1}] = [\pu{mol4 L-5}] \pm 1$$
This seems that something is off about the rate constant, but also that there is no $\pu{L-1}$ term in the molar flow rate.
I don't expect exact answers to this, but could someone point me the right way, or send some clear explanation on either reaction rate constants or on kinetic mass balances? It seems I cannot find clear explanations, not even in textbooks.

Consider Formatting of chem expressions by using mhchem MathJax package, available in CH SE by default, involving MJ syntax \ce{} and \pu{}
E.g. write $\ce{H2SO4}$ or $\ce{a A <=> p P}$ or $\pu{6.022E23 mol-1}$  to get $\ce{H2SO4}$ or $\ce{a A <=> p P}$ or $\pu{6.022E23 mol-1}$ (all eventually with double dollars in the display mode like $$\ce{H2SO4}$$. // E.g. typographic rules say unit symbols are never in italic.

Okay will do, but do you have an answer to my question?

I know nothing about this, but looking at the dimensions you firstly seemed to have used for concentration $mol    L^{-1}$ on the LHS and simply $mol$ on the RHS, fixing that solves your second problem, and secondly by dimensionality arguments your assertion that the dimensions for k are $s^{-1}$ must be wrong. But don't ask me how to justify what the correct dimensions must be ($mol^{-4} L^4$).

Yes the units of k should be mol$^{-4}$L$^4$

@IanBush I do not understand what you mean, the concentration is the same on both sides, except that on the LHS it is the derivative, so it is over time.

@Andrew but that is not a possible dimension for $k$ right? The general formula for the dimension of $k$ is $[mol^{1-n}s^{-1}]$ which would translate for a reaction of $n=5$ (because sum of degrees $2+3=5$) to $[mol^{-4}s^{-1}]$. How would you explain this?

Sorry forgot the s-1. I was intending to say that you needed to add those units to the s-1 that you already had

Purpose of comments is not providing answers. There are 2 simple rules for dimensions in equations: Every additive/subtractive term on both eq. sides must have the same dimension. Arguments of elementary functions must be dimensionless.Otherwise, something is wrong.

It seems I took the general formula for $k$ wrong. The general formula is $[\pu{mol^{1-n}L^{n-1} t-1}]$ so indeed the $k$ can be $[\pu{mol-4L4t-1}]$. The flow rate would have to be divided by a volume $V$ to get to the same dimension.


text:
Is a compound optically active if plane polarised light is deflected by an angle of n*(2π) angles?
Is a compound optically active if plane polarised light is deflected by an angle of  n*(2π) angles(like 360°); as there won't be any deflection in the analyser of the polarimeter?

yes for $n\neq 0$, because compound does interact with light and causes optical rotation. The way to measure this would be to take several measurements at different concentrations.

Don't this mean that actual angle also depends upon concentration like specific rotation?

On concentration and path. It is analogical to Lambert-Beer law ( A=eps.c.l ).

@Poutnik Okay, I got it.Should I delete the question now?

@Shiva I would prefer that you don't delete it since I also didn't knew about it earlier. Let others also know about it.

I don't have time at the moment, but there's worth in an answer differentiating between observed and specific rotation ( https://pediaa.com/difference-between-optical-rotation-and-specific-rotation/ ) and throwing in a few formulae connecting them

related question https://chemistry.stackexchange.com/q/166327/124759

@IanBush Thanks, It wasn't related to my question but I got new things to learn.

The answer to your question was already provided as a comment to the answer in the link Ritil provided above.

@ron Yes, please tell what should I do. Feel free to delete it if required.

Yes
The actual rotation you observe in a polarimeter depends on several factors: the path length, the concentration (if in solution) and the specific amount of optical rotation caused by the compound.
If you observe the actual rotation at a single concentration you might observe no apparent effect because the rotation was 360°. But at a different concentration or a different path length with the same substance you would observe a rotation.
So it is pretty easy to avoid the mistake of assuming no rotation by simply taking more than one measurement while varying the other factors.


text:
Does no optical rotation always implies optical inactivity?
This question popped up in my mind in reference to this question,
Is a compound optically active if plane polarised light is deflected by an angle of n*(2π) angles?
Suppose that I give a chemistry exam in future in which a question is asked like this,

A pure organic compound(not a racemic mixture) was analysed once and it was found that that there was absolutely zero rotation of PPL when passed through it(not even slight rotation as in cryptochirality). This means that it must be an optically active compound.

Now there are two options whether the above statement is true or false. What will be the answer? Assume that the organic compound is pure ie it is not a mixture (not even a racemic mixture). Also assume that the enantiomers (if possible) are separable which means that there is nothing like amine inversion or any other similar mechanism through which enantiomers can interconvert into each other.
Though it might be quite intuitive that the answer should be false but think about the person who set this question. (It has quite often happened with me that even after having the correct logic, my answer is consider wrong by majority.)

The statement is false. This does not mean that. Maybe it does not mean the opposite either, but that's irrelevant.

Be reminded that Emil Fischer observed no rotation from arabitol derived by reduction  of (+)-arabinose but yet its borax complex did.

What is there to support? See Zhe's comment in @Mithoron link. For a reference, try: Emil Fischer's Discovery of the Configuration of Glucose. A Semicentennial Retrospect, C. S. Hudson, J. Chem. Educ. 1941, 18(8), 353-356; https://doi.org/10.1021/ed018p353

Now there are two options whether the above statement is true or false. What will be the answer?

I choose to false. We have two cases for no rotation of plane polarized light. One of the case is that the compound is optically inactive. And the other case is when the analyte is a racemic mixture.
Wikipedia states that,

A racemic mixture, or racemate, is one that has equal amounts of left- and right-handed enantiomers of a chiral molecule.

So the rotation of plane polarized light is a combined effect of nature of compound and nature of analyte. So no rotation doesn't implies that the compound is optically active.

Suppose that it is given that it is a pure compound and not a racemic mixture. Now what?

Your tag doesn't seem to fit with the content in the question. It might be a possibile duplicate of this one
https://chemistry.stackexchange.com/questions/166324/is-a-compound-optically-active-if-plane-polarised-light-is-deflected-by-an-angle

I have already referred to this question in my question, since the question came after reading that only, but the difference from that post is that my question is regarding exam. What will I do if such a question comes in an exam?

You have not considered the case where the specific rotation of the optically active, non-racemic compound is 360° or a multiple thereof.  That is why the rotation must be measured at at least 2 concentrations (and\or path lengths).  If you experimentally measure a 0° rotation and then remeasure with the concentration reduced to 75%, you will experimentally observe a rotation of 270° if the specific rotation is actually 360°.  If the compound is optically inactive, then the rotation will remain 0°.

@Ritil Stack Exchange is a place to ask questions about chemistry. It's not a place to ask about answering techniques for exams. We can tell you facts. What you do with those facts in your exam is your business.


text:
Vinylic Nucleophilic Substitution
While reading "Francis A. Carey, Richard J. Sundberg - Advanced Organic Chemistry Part A. Structure and Mechanisms-Springer (2007)", I came across the following schematic for addition of electrophiles to vinyl ethers:

I understand that such a reaction has application in polymer chemistry. For example, under the right acidic conditions, methylvinylether will form polyvinylmethylether:

Under what conditions will the electrophile react with the lone-pair of oxygen instead of the $\beta$-Carbon?
In another section, I found that hard-hard and soft-soft interactions are preferred over hard-soft interactions:

Many questions arise from reading this. Is the lone pair on the oxygen atom hard or soft? And accordingly, what kind of electrophile should be used? Do stearic factors play a role?
It has been pointed out in the comments that addition of electrophile to oxygen is a reversible and non-productive. What I was thinking was something like this:

I guess the reaction could be called a vinylic nucleophilic substitution.

Addition of an electrophile such as a proton to the oxygen of methyl vinyl ether (MVE) is a reversible, non-productive event. MVE in the presence of $\ce{H3O+}$ gives hydrolysis to acetaldehyde. With methanol/$\ce{H+}$ the dimethyl acetal of acetaldehyde is produced.


text:
Is Bent's Rule an alternative explanation for the Inductive Effect?
Today I read that Bent's Rule can be used to explain the inductive effect (I was reading the Consequences section of this Wikipedia article). If I understand it correctly, for a molecule CL3R, the more electronegative the substituents (denoted L) are, the more s character can be saved for the hybrid orbital that the C uses for the C-R bond. Higher s character for the hybrid orbital means lower energy, so when the C-R bond forms, the resulting molecular orbital has more electron density around the carbon. This means that the carbon withdraws electrons better, so the electron-withdrawing property of the substituents has been transmitted to the carbon.
However, when I originally learned about the inductive effect, I understood it through a Coulombic argument: the three electronegative substituents withdraw electrons from the carbon, creating a partial positive charge on the carbon, which gives it a better pull on the electrons in the C-R bond (in other words a higher electronegativity).
Are Bent's Rule and Coulomb's Law simply two separate ways of justifying the inductive effect, or should they be seen as working together to produce the inductive effect?
And of course, are these both just ways of simplifying the molecular orbital explanation of the inductive effect?


text:
Why is reported effective pore size smaller that the one declared in the name of the molecular sieve type X?
The molecular sieve type is often coded with <number><letter> combination, where <number> specifies the pore size in ångströms and <letter> refers to the Linde type of zeolite. For sieve types 3A, 4A and 5A the reported pore sizes ubiquitously match the code, i.e. being 3 Å, 4 Å and 5 Å, respectively, with virtually no discrepancy neither on the manufactures' websites nor in the literature.
However, the majority of the companies selling 13X type sieves across both EU and US (ADCOA, VWR, GeeJay Chemicals Ltd., Lawrence Industries, MTE Group, …) report nominal pore size 10 Å instead of 13 Å. Analogously, the spec sheets for the mol sieves type 10X report the pore diameter 8 Å instead of expected 10 Å.
Surprisingly, there wasn't any mismatch in older literature and patents [1, 2], and the 13X sieves were claimed to have 13 Å pores, which seems logical. An editorial remark of a similar nature has been made for the 2006 paper [3] pointing out the inconsistency between expected pore size for 13X zeolite and the pore size reported by Sigma-Aldrich which authors used in their work. Based on discrepancy of the data presented in the literature at that time, the authors suggested that

microscopic structure of commercially produced 13X may not be fully independent of the sample preparation procedure

and from the rest I draw a conclusion that they infer that commercially available sieves of 10X and 13X types could be faujasite, zeolite of Linde type X and synthetic zeolites of similar composition, which could deviate from one manufacturer to another. I am not finding this completely convincing since the reported pore size doesn't fluctuate around one number, it's consistently understated by 2 Å and 3 Å for 10X and 13X sieves, respectively.
What exactly is going on here and where can I get a refund for my missing 3 ångströms? Were the initial structures of the X type zeolites and, accordingly, the pore dimensions, incorrectly determined? Or did the manufacturers changed their product and continued to ship it using the former name?
References

Meyer, J. R. Method for Producing Exceptionally Pure Hydrogen. US3011589A, December 5, 1961.

Read, P. L. Sorption Vacuum Pump. US3264803A, August 9, 1966.

Swenson, J.; Jansson, H.; Howells, W. S.; Longeville, S. Reply to “Comment on ‘Dynamics of Water in a Molecular Sieve by Quasielastic Neutron Scattering’ ” [J. Chem. Phys. 125, 077101 (2006)]. J. Chem. Phys. 2006, 125 (7), 077102. DOI: 10.1063/1.2272835.

The "pore size" depends on what the analysis filled into the pores. Perhaps the standard gas used was changed at some time in the past? Nitrogen, argon, CO2, ...


text:
What is the necessary and sufficient condition for a mixture to be ideal?
I am trying to understand some concepts from solution thermodynamics related to ideal mixtures and fugacity. My main question is what is the necessary and sufficient condition for a mixture to be an ideal mixture? Below is my attempt to answer this and some related follow up questions.
The fugacity of component $i$ in a mixture is typically defined by two requirements:

$f_{i}(T,p,\{x\})=p_{i}\exp\left(\frac{1}{RT}\left[\mu_{i}(T,p,\{x\})-\mu_{i}^{IGM}(T,p,\{x\})\right]\right)$
$\lim_\limits{p \to 0}\left(\frac{f_{i}(T,p,\{x\})}{p_{i}}\right)=1$

Where IGM is an ideal gas mixture. Using the first part of the definition, we can rewrite the second condition as:
$$\lim_\limits{p \to 0}\left( \mu_{i}(T,p,\{x\})-\mu_{i}^{IGM}(T,p,\{x\}) \right)=0$$
From this we can derive an expression for fugacity that we can calculate explicitly:
$$f_{i}(T,p,\{x\})=p_{i}\exp\left(\frac{1}{RT} \lim_\limits{p’ \to 0} \int_{p’}^{p} \left[ \bar{V}_{i}(T,p,\{x\}) - \frac{RT}{p} \right] dp \right)$$
And from this we can show:
$$\frac{f_{i}(T,p,\{x\})}{x_{i} f_{i}(T,p)}=\exp\left(\frac{1}{RT} \lim_\limits{p’ \to 0} \int_{p’}^{p} \left[ \bar{V}_{i}(T,p,\{x\}) - \bar{V}_{i}(T,p) \right] dp \right)$$
An ideal mixture is defined as a mixture where the chemical potential for any component $i$ is given by:
$$\mu_{i}^{IM}(T,p,\{x\})=\mu_{i}(T,p)+RT\ln x_{i}$$
All other properties of an ideal mixture follow from this definition, since the chemical potential is equal to the partial molar Gibbs free energy. One key property is that $\bar{V}_{i}(T,p,\{x\}) = \bar{V}_{i}(T,p)$ at all conditions in an ideal mixture. That is, the volume change of mixing is zero. In turn this means the integrand above is always zero and the expression reduces to the Lewis-Randall rule:
$$f_{i}^{IM}(T,p,\{x\})=x_{i}f_{i}(T,p)$$
From the first part of the definition of fugacity we can also express the chemical potential as:
$$\mu_{i}(T,p,\{x\})=\mu_{i}(T,p) +RT \ln \left( \frac{f_{i}(T,p,\{x\})}{f_{i}(T,p)} \right)$$
Then if we substitute the LR rule we obtain the definition of an ideal mixture:
$$\mu_{i}^{IM}(T,p,\{x\})=\mu_{i}(T,p) +RT \ln x_{i}$$
In summary, the two requirements below seem to ensure that a mixture is ideal:

$\bar{V}_{i}(T,p,\{x\}) = \bar{V}_{i}(T,p)$
$\lim_\limits{p \to 0}\left( \mu_{i}(T,p,\{x\})-\mu_{i}^{IGM}(T,p,\{x\}) \right)=0$

I have two follow up questions:

Is this actually true? Does the volume condition always ensure the mixture is ideal? Or are there ideal mixtures where the volume of mixing is nonzero (i.e. $\bar{V}_{i}(T,p,\{x\}) \ne \bar{V}_{i}(T,p)$)?
Is the logic of this argument backwards? I don’t understand the
mathematical reason why we are free to define fugacity with its
second condition. If we didn’t assume that then the volume condition
would not be enough to derive the the chemical potential definition
of an ideal mixture. Is the fugacity limit requirement actually some
kind of consistency condition due to the definition of an ideal
mixture?


text:
Is the Glaser coupling reaction possible with metal alkyne complexes?
As the Glaser coupling reaction is usually done with terminal alkynes with the hydrogen being straight, I wonder if the same reaction can occur in terminal alkynes connected to the backbone of a metal alkyne complexe.
As an example, in the trimerization reaction of acetylene, which happens with the formation of a metal alkyne complete; the hydrogens are bent in a v-shape to the carbon; not straight. Alkyne Trimerisation.
Would it be possible? If it weren’t the case, what would impede this reaction?

I don't think the fact that the hydrogens are "straight" is relevant. I guess in a sense it is, because it gives information about the bonding. The fact that the hydorgen is not "straight" in a metal alkyne complex means that it is not really an alkyne, the bondig situation is completely different. Now, during the Glaser coupling, you will form a bond between the copper and the carbon of the alkyne - I am not sure you can do the same with an alkyne metal complex, because it's not really an alkyne. Maybe pre-dissociation would help?

The hydrogen being straight corresponds to its sigma bond with carbon having more carbon $2s$ character and less carbon $2p$ character, which makes the covalent overlap somewhat less and the bond more polar. Thus the hydrogen is  susceptible to deprotonation by the strong base that initiates the reaction.
If the bond angle becomes bent due to the electronic contributions from the metal, we have lost some $2s$ character in the carbon-hydrogen bond in favor of more $2p$, and the required deprotonation could be inhibited.
Note that in the alkyne trimerization shown in the question, the first two acetylene molecules are effectively reduced before the third one displaces the metal. Such a reduction is not seen in the Glaser coupling.


text:
How to rationalize independence of half-life time from the initial concentration for the first order reaction?
Using the rate expression for the first order kinetics and expressing the half-life time, it can be proven the half-life time $t_\frac 1 2$ of the first order reaction is independent of its initial concentration:
$$t_\frac 1 2 = \frac{\ln 2}{k}$$
Can we intuitively explain that the first order half-life time is independent of its starting concentration and remains constant without the aid of the derived expression for the half-life time said above?
For the zeroth order reaction, it can intuitively be explained, but for the first order reaction it seems a little hard.

It is ok   following this argument,
I can say  when the particle count goes on decreasing  half life too can go on decreasing. but how comes that the half life is "exactly" equal to initial concentration.

That is not about kinetics of the 1st order and looks rather confused. Half life cannot be equal to concentration even for the simple reason of different dimension.

I m sorry for that typo and wanted to say successive half lives are exactly equal and independent of initial concentrations.

Sure. If 10e23 units decay to 1/2 and another 10e23 units decay to 1/2, than in summary 20e23 units decay to 1/2. They do not decay faster nor slower just because there is more or less of them.

Thanks.. got your point, thinking over your answers. Treating the first order kinetics in terms of "reacting units" as you have mentioned earlier and  the fraction of units that undergoes reaction (proportionality constant) is intuitive for me to rationalize that the half life period is constant. It makes me think that molecules of first order kinetics comes in half life packets.

My professor said that for first order kinetics,between 2 successive half life times  amount of reactant  to be consumed is halved. Similarly, the rate from one period to the next one is  also halved due to first order kinetics . Therefore, half life time doesn't change. Though I understood his point, I wasn't convinced & looked for better explanation

You may want to mark Porphyrin's answer as accepted. It is nicer and he has more cool ID.

The basic assumption is that the probability of reaction of a given molecule (or disintegration of a radioactive atom) in a given time interval is independent of past history, and depends only on the time interval. If it is assumed that the time interval we choose is sufficiently small the chance of reaction is proportional to this time interval and the exponential decay law can be derived from this. The constant of proportionality is called the rate constant $k$. These assumptions mean that we cannot predict exactly when a given molecule will react but because of the proportionality (the rate constant) we can after many measurements know the form of the decay. Of course we assume that all the molecules of a given type are identical and behave in the same way and independently of one another.
We can measure the average time our type of molecule takes to decay by making making many repeated measurements and this average is the reciprocal of the rate constant and is related to the half life as $\ln(2)/k$.

This one essentially means that the infinestimal time interval dt is proportional to reaction probability dN/ N: going by this,   same time intervals mean same  reaction probability. By extended argument, for same half life times we have same extent of reaction that is to say  half life times  are constant for first order kinetics.

I didn't get the point " the average time"

@Sudhagar It is probabilistic behaviour, that has clean, exponential outcome for very large set of molecules.

I only used average time  as an example as this is mathematically straightforward to calculate and can be measured experimentally of course.

A fuller answer is here https://chemistry.stackexchange.com/questions/159111/fluorescence-what-mechanism-regulates-the-rate-of-discharge-of-energy/159171#159171

A particle has some probability to decay (Radioactive decays are ideal illustrative examples.) or react during some time interval. With more particles, the count of particles is directly proportional to particle count that will do so during that interval.
Therefore, the relative portion of particles reacting during that period does not depend on the amount nor concentration (assuming very high particle numbers converging to matter continuity). When the time interval leads to a portion being the half, it is the half life time.


text:
how to keep kappa-carrageenan lasting at room temperature for weeks or months? (art project)
I'm helping an artist make an outdoor gelly structure (not for eating) and we want to keep it from molding. The temperatures are around 10-30 degrees celsius and it's in Copenhagen, Denmark in September, and hopefully, it can last for some weeks or more.
Right now we are experimenting with kappa carrageenan with a mix of CaCl2 and we are going to add locust bean gum, pectin, and sugar to see how it behaves.
But what would we add for conservation? Sodium benzoate? and is there a limit on how much to add? (we want it as transparent as possible)
We don't wanna use some too harmful chemical since we gonna have to get rid of it and work with it.
Thanks, any help is most appreciated. Have a nice day.
Resources I found so far:
In this paper, they say they can make it last 7 months at room temperature.
The paper mentions Potassium makes it last longer:
https://idosi.org/wjdfs/wjdfs6(2)/18.pdf
food exchange:
https://cooking.stackexchange.com/questions/57752/how-much-citric-acid-to-use-to-preserve-vegetarian-jelly
general booklet on caragenan:
http://www.bisi.cz/cmsres.axd/get/cms$7CVwRhc3USVqgzxkKF96gI$2BChNrXcTq$2BOUdiEtz5TfYA$2Fg1ADRHMfXfdEjUsYQagqUs9N6byPOkok$3D

Not an answer but related: Is drying up and shrinking of gelly an issue? Is water evaporation suppressed enough?

@Poutnik thanks for bringing awreness to the issue of drying:) I'm not sure if it's an issue. I guess it would be if it stands out for long enough but I don't know how long it needs to stand out yet. I will make some tests and see if it becomes an issue. But do you know about conservation? Thanks for the comment :)

I am not particularly knowledgeable about conservants, just guessing there may be need of a strong one. This may bring another potential issue about security, if random accidental/incidental eating (small children) could use any harm.

@Poutnik good point

update: We will try to add a layer of liquid beeswax (for wood protection ) to stop it the evaporation, and hopefully it will stop the molding process as well.


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When does the angle between the planes in tetrahedral molecule with central atom deviate from 90°?
Consider an atom X bonded to four other atoms A, B, C, D in a tetrahedral fashion with sp³ hybridisation. If A, B, C, D are the same, every bond angle is 109.47°, and the angle between the planes formed by joining nuclei of X, A, B and X, C, D is exactly 90° (treat nuclei as point sized since they are very small with respect to the atoms) if pairs A, B and C, D are identical elements (symmetrical molecules like methane or dichloromethane):

If A, B, C, D are different, then the bond angle will be different to minimise repulsion and maximize stability as per VSEPR theory. But what happens to the angle between the planes? My intuition says that it will always be 90° in any molecule and the reason is that repulsions are minimised in that case.
However, this reasoning is only a qualitative one, but I wish that this could be mathematically proved or in some other way using some very good logic. In the following figure you can see that due to decrease in angle between bonding planes, B and D come close to each other and also A and C come close to each other. So they will face steric repulsion from each other leading to instability:

In case the angle between the planes is not always 90°, what are these exceptions? In organic chemistry, many kind of exceptions frequently occur, so I was trying to think of some case where steric hindrance or torsional strain or angle strain or ring strain or some other reason may cause this angle to change.
The boat configuration of cyclohexane came into mind. I knew that there is flagpole interaction between two hydrogen atoms which get very close to each other. Maybe this repulsion may cause hydrogen atoms to deviate a little to the sides, causing a little change in angle between bonding planes in this example. This could be the first exception, but I am not very sure of it.
Also, such repulsion can happen in two ways: clockwise and counterclockwise rotation of the planes. For example, in the above figure a mirror image can be formed if B comes closer to C and D comes closer to A. Both of these compounds will be mirror image of each other and therefore may create optical activity. This optical activity if it happens can prove to be a practical confirmatory test to be sure weather this thinking is correct or not.
If I am correct, are there better examples where the angle between bonding planes is different from 90°?

If I understand correctly, for the angle OP is interested in, you need to have four different substituents on the carbon, for example CHBrClF. The three H-C-X bond angles are all different magnitudes, so angle of any two planes is not 90 I believe

Of course it's gonna decrease, but bigger groups "want more space" than smaller.

Let us continue this discussion in chat.


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Can Polarizing Power and the Inductive Effect both explain the highly covalent bonds of Mn2O7?
The Mn-O bonds of Mn2O7 have more covalent character than those of MnO2. This makes sense when imagining Mn2O7 to consist of Mn7+ ions and O2- ions; the high charge of the cations give them a high polarizing power, leading to the Mn-O bonds having high covalent character.
However, in nature, bonds are not "ionic with covalent character" or "covalent with ionic character," they are simply bonds. Therefore, I am curious if it is possible to explain the difference in the covalent character of Mn2O7 and MnO2 when viewing the bonds as covalent to start. In this case, is the explanation for the covalent bonds that the Mn atoms gain a high positive charge due to the highly electronegative oxygens, allowing the Mn to pull electrons back in? In other words, the oxygens inductively pull on each other's electrons by making the manganese more positively charged, leading to all of the electrons being more evenly shared between the manganese and oxygens?
The inductive effect and polarizing power explanations both seem to be describing the same phenomenon: the manganese atoms in Mn2O7 have a high positive charge (when viewing Mn2O7 as ionic, that charge is the +7 oxidation state; when viewing Mn2O7 as covalent, that charge is the partial positive charge due to the oxygens), causing them to pull electrons better, leading to highly covalent bonds with the oxygens. In general, can we say that highly positively charged atoms in a molecule pull electrons better (and therefore have a higher electronegativity), whether we are measuring that charge by the atom's oxidation state or by looking at the charge due to polar covalent bonds? It would make sense if this were true, because oxidation states represent the charges if all bonds are ionic, which is the extreme of polar bonds. Even if an atom's charge isn't as high as its oxidation state, I imagine that a higher oxidation state still corresponds to a higher actual charge.
On a side note, why do we only talk about polarizing power in the context of ionic bonding (or at least why have I only seen it used in the context of ionic bonding, such as in Fajan's Rules)? Can we talk about polarizing power of partial charges created due to electronegativity differences? For example, in the ketone RCOR', could we say that the carbonyl carbon has some polarizing power due to its partial positive charge, causing it to withdraw electrons from the R and R' groups? I think this is simply a way of describing the inductive effect. Maybe the reason we never talk about polarizing power in this kind of context because it is simpler to sum up the effect as "the carbonyl oxygen withdraws electrons from nearby atoms through the inductive effect?"

Your understanding of the inductive effect and polarising power/polarisability is a bit inaccurate in some places.
The inductive effect is a result of the polarisation of sigma bonds (often as a result of differences in electronegativity.) This is first and foremost a covalent effect. For example, acyl chlorides are more reactive than aldehydes because the electronegative chlorine atoms pull electron density from the electrophilic carbonyl carbon, resulting in a stronger electrophile. The implicit fact that is not stated here is that this electron-withdrawing effect is through a sigma bond.
On the other hand, Fajan's Rule is a result of a cation with sufficiently high polarising power (high charge density) meeting an anion with sufficiently high polarisability (huge polarisable electron cloud often expressed as an anion of low charge density). Notice instead of dealing with the donation of electron density through a sigma bond where usually the discussion of electronegativity comes into play, Fajan's rule has little or no relation to electronegativity but rather the actual charge of the various species.
So, coming back to the question of Manganese (VII) Oxide and Manganese (IV) Oxide. From the view of inductive effects, the strength of the "covalent" bonds in these two species should inherently be the same due to the same difference in electronegativity. However, by applying Fajan's rule, it is obvious that because of the significantly higher polarising power of the Manganese (VII) species, Manganese (VII) Oxide has more covalent character.


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What are the other examples of compounds undergoing decarboxylation in the pericylic manner as shown by β-ketocarboxylic acids?
I know that β-ketocarboxylic acids undergo decarboxylation on light heating through an enol intermediate undergoing a pericyclic reaction. However, what other groups may be placed at the β position that may give the same decarboxylation reaction?
I haven't been able to find any reputable sources. However, my teacher gave us a question in which an imine at β position underwent decarboxylation following the same process as a β-ketocarboxylic acid would. Does this reaction happen and are there other groups that follow the exact decarboxylation mechanism when at the β position?


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How does the stoichiometric coefficient affect the time until a given percentage of the reactant is consumed (if at all)?
Problem from Serway's College Physics test bank:

The reaction $2NO_2\rightarrow2NO+O_2$ obeys the rate law:
$\large \frac{\Delta[O_2]}{\Delta t} = (1.40\times10^{-2})[NO_2]^2 \space at \space 500^\circ \space K $
If the initial concentration of $NO_2$ is 1.00 M, how long will it take for the [$NO_2$] to decrease to 35.8% of its initial value?
A) 45.9 s
B) 73 s
C) 128 s
D) 1.40 x 10-2 s
E) cannot be determined from this data
ANSWER: C

My work
We are given the reaction rate for $O_2$. Since for a given reaction $aA + bB \rightarrow cC + dD$, the rate of the reaction $r = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}$, I conclude that $\frac{\Delta[O_2]}{\Delta t} = -\frac{1}{2}\frac{\Delta[NO_2]}{\Delta t} \Rightarrow -\frac{\Delta[NO_2]}{\Delta t} = 2\frac{\Delta[O_2]}{\Delta t}$. Therefore:
$-\frac{\Delta[NO_2]}{\Delta t} = 2 \times (1.40\times10^{-2})[NO_2]^2 = (2.8\times10^{-2})[NO_2]^2 \Rightarrow$
$\textbf k = 2.8\times10^{-2}$
Knowning that the reaction is second-order in $NO_2$, I plug in $[NO_2] = 0.358[NO_2]_0$ into the integrated rate law and solve for t as so:
$\frac{1}{0.358[NO_2]_0} = kt + \frac{1}{[NO_2]_0}$
$\frac{1}{0.358\times1M} - \frac{1}{1M} = 2.8\times10^{-2}\times t$
$t \approx 64 s$

Answer given in the test bank
However, the textbook gives 128s as the answer, indicating that the reaction rate was NOT multiplied by 2 (from $\frac{\Delta[O_2]}{\Delta t}$ to $-\frac{\Delta[NO_2]}{\Delta t}$).
Is there a gap in my understanding of conversion between reaction rates according to the stoichiometric coefficient of the reactant/product being handled or did the test bank simply give an incorrect answer?

@Poutnik. No. If $t = 0$, the coefficient of $\ce{[NO2]_0}$  cannot be $0.358$. $t$ is not a variable in this equation.

@Poutnik. This is an equation with one unknown, $t$, which is dependent on the numerical value $0.358$. $t$ cannot be chosen at will.

@Maurice  Variables are independent and dependent, with the former possibly masked by literal value substitution. But I guess we have cleared the things and I am going to purge the related comments.

Procedural advice: Always try to stay at symbolic algebraic expressions as long as possible and apply literal values only in the last minute when you already have to.

You should not multiply the rate constant by an stoichiometric coefficient.
The rate of this chemical reactionis given by
$$ v = -\frac{1}{2}\frac{d|NO2_2|}{{dt}} = +\frac{1}{2}\frac{d|NO|}{{dt}} = +\frac{1}{1}\frac{d|O2_2|}{{dt}} = k |NO_2|^2 $$
where k is the reaction rate constant. In your exercise, $k = 1,4\times10^{-2} \;mol^{-1}\,dm^3\,s^{-1}$.
Notice that the stoichimetric coefficients appear in the reaction rate definition, thus the reaction rate does not depend on which reactant or product used to obtain it.

Check, for example, Atkins, P., De Paula, J., & Keeler, J. (2017). Atkins’ physical chemistry (11th ed.). Oxford University Press. You hava definitions and examples in section 22.2a The rates of reactions. The definition of rate


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Does the boiling point of ammonia hydroxide change with the ratio of water to ammonia?
My kids and I are designing a science experiment and we could use help with the ratio of water to ammonia in an ammonia hydroxide mixture.
A little bit of background:
Initial goal: To convert the ammonia hydroxide into gas through solar / ambient heat with enough pressure to inflate a balloon.
The how: We have a metal tank that is painted black. The balloon is wrapped around the outlet of the tank at the top of the tank. We put the ammonia hydroxide in the tank when cool outside, then wait for the day to warm (we live in the high desert) and ammonia to turn into vapor.
Initial tests will fill the balloon but does not generate enough pressure to expand the rubber/plastic to properly inflate it.
Our idea, and the genesis of this question, is that if we use a higher concentration of ammonia we will a) boil faster at lower temperatures and b) might generate more vapor and create pressure to inflate the balloon.
Apologize in advance if this question is incomplete. Happy to provide more information. And thanks for your help!

By ammonium hydroxide (which is just a commercial brand name), you must mean water solution of ammonia NH3(g), which in small extent captures H+(aq) ions, forming NH4+(aq) and as consequence of water autoionization, also OH-(aq). // By other words, there is no such thing like ammonium hydroxide boiling point. There is temperature dependent proportionality (roughly)  between gaseous ammonia pressure and concentration of ammonia solution.

So more concentrated solution and/or higher temperature leads to higher pressure of the ammonia gas ( not vapour, as ammonia boiling point is near -33 deg C.)

You may want to work backward if you haven't already to figure out how much pressure you need to generate and what the corresponding amount of ammonia would be.

Your idea of increasing the ammonia concentration to increase the total vapor pressure over the solution is physically sound. For two-component mixtures, the total pressure will be a function of both solution composition and temperature. In the case of ammonia and water, since ammonia is the more volatile component (it is a gas at standard conditions) increasing its concentration will lead to a higher vapor pressure for the solution.
Although the data are over 100 years old, a paper by Perman (J. Chem. Soc., 83, 1168 (1903)) gives some relevant information. Below is a table with the vapor pressures he measured for ammonia solutions at various temperatures and ammonia concentrations. You should be able to calculate the pressure ratios you're interested in from these data.


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Does Rubbing Alcohol + Bleach really produce Chloroform?
A lot of sources on the internet claim that mixing rubbing alcohol with chlorinated bleach produces chloroform.
Rubbing alcohol is Isopropanol, and bleach is Sodium Hypochlorite. Neither of these are the reagents for the Haloform reaction. So I ask, is chloroform really produced, and if so, how?

IIRC, dropping ethanol on calcium hypochlorite/chloride reportedly forms chloroform, not sure about isopropanol. From water treatment analysis, if water to be chlorinated contains organic substances, it frequently leads to chloroform concentration in dozens micrograms per litre, not being allowed to be distributed.

"Rubbing alcohol" is whatever blend the manufacturer uses: some percentage water (usually between 30 and 50%), denatured ethanol, isopropanol, scent, etc. Look at the label to see if it's ethanol-base or isopropanol based. https://en.wikipedia.org/wiki/Rubbing_alcohol . Note that WebMD and other sources may state only isopropanol incorrectly.

Yes, rubbing alcohol and bleach could produce chloroform when mixed, if the alcohol is isopropyl-type rubbing alcohol and the bleach is hypochlorite-type bleach. Isopropyl alcohol is indeed a substrate for the haloform reaction, according to Fuson and Bull, Chem. Rev., 15, 275 (1934) and references therein.
I can't recall the reference off the top of my head, but I believe in "dilute" solutions approaching "bleach and alcohol" the chlorination tends to continue and produces carbon tetrachloride as the main product.

The mechanism suggests that the reaction does not chlorinate chloroform. [unless there is a parallel free radical reaction happening involving Cl2]

Oh, I though it required a ketone, not an alcohol. I'm guessing the isopropanol is first oxidized to ketone and then it reacts.


text:
What does it mean that a state belongin to a given irrep transforms like $Rx$, $Ry$ or $Rz$
The present question is related to this other question I did few days ago.
Given a point group and the list of the irreps (see for example here) the meaning of an irrep which transforms like $x$ or $x^2$ is clear to me.
Instead I do not understand what is meant when it is written that an irrep transforms like $Rx$.
Let me explain a bit better. Say that the group is the symmetry group o a crystal structure. It can be represented as a set of $3\times3$ matricies $M_i$. These matrices can be used to transform the vector $v=(x,y,z)$. So I can clearly see how something which transforms like x would be affected: $v'=M_iv$ and $v'=(x',y',z')$.
I can also assume that any function of $f(x,y,z)$, like $f(x,y,z)=x^2$ would be transformed like $f(x,y,z) \rightarrow f(x',y',z')$ (is this point true?). However I do not understand how $Rx$ would transform. It is not even a defined function, but a group of operations by itself.

Assign a direction to the rotation, then perform the transformation and see if the direction of rotation is the same (symm) or reversed (antisymm).

$R_x$ etc. are rotations about the indicated axis. See for example this article in chem.libretexts or this pdf

Thanks for replying. Just it is super short and general. I extended a bit the question, hoping that this makes it easier to address more precisely the question.

If you want a function for Rx, try the x-component of angular momentum: $$L_x = y p_z − z p_y.$$ If you want a matrix, use the rotation matrix that rotates 90 deg about the x-axis, $$\begin{bmatrix}1&0&0\0&0&1\0&-1&0\end{bmatrix}$$


text:
Boiling and Evaporation
A closed tank is filled with water completely and heated from outside. What will occur boiling or evaporation or both?

Welcome to the chemistry site of the SE network.
The site expects that you write explicit compact summary of your prior effort to answer the question, based on your knowledge and on searching for existing related  info or answers. It would prevent others to tell you what you already know or what you could easily find yourself.
Effort not shown can be considered as effort not done and such a question may be  closed.  How do I ask a good question?.

https://chemistry.stackexchange.com/questions/76438/why-boiling-cant-take-place-in-closed-containers

If you fill a tank completely with liquid water (no gas bubble) and start heating it, the liquid water will try to expand faster than the vessel since the coefficient of thermal expansion of water is larger than the corresponding value for typical vessel materials (e.g. steel). Thus the pressure will increase sharply so that no evaporation can occur.
Because of the strong pressure increase, the vessel will probably fail before reaching 100 °C. When the vessel fails, the pressure will quickly drop to ambient pressure. If the temperature is still below 100 °C, no boiling can occur; there will be just slow evaporation from the surface of the warm water.
If the temperature is already above 100 °C when the vessel fails, some flash evaporation will occur when the pressure quickly drops to ambient pressure. A part of the hot liquid water flashes to steam (which can be calculated using the enthalpy balance). The temperature drops to the boiling point (100 °C) at the new pressure. After that, there will be slow evaporation from the surface of the warm water.


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Use of boron(III) oxide for drying acetone
Acetone undergoes aldol self-condensation in the presence of acids or bases:

which makes drying agents such as 3A/4A molecular sieves, barium oxide, potassium hydroxide and carbonate (basic), copper(II) sulfate or aluminium(III) oxide (acidic) less effective and promotes formation of mesityl oxide to a significant degree.
Boron(III) oxide is considered as one of the most appropriate desiccants for acetone [1, pp. 3967–3968]:

As with $\ce{Me2SO},$ the root of the difficulty is the acidic α protons, which in this
case compounds the drying problem not only by inflating apparent water content by exchange process but also by providing a pathway to self-condensation through enol intermediates. This facet of acetone chemistry makes the choice of a successful desiccant a delicate process. As Table IV shows, mild siccatives such as calcium sulfate are inept; more potent desiccants such as molecular sieves exhibit a short initial drying action but thereafter actually cause disastrous increases in water content by displacement of the condensation equilibrium. This interpretation was confirmed for molecular sieves and other basic desiccants such as barium oxide by gas chromatographic analysis which demonstrated the presence of mesityl oxide in the dried solvent (see Table IV).
In summary, while both cupric sulfate and 3A molecular sieves are clearly at least useful preliminary desiccants, the agent par excellence for acetone is powdered boric anhydride.

However, $\ce{B2O3}$ is a Lewis acid itself and reacts with water to produce boric acid, which should build up upon drying the solvent.
Why doesn't $\ce{B2O3}$ appear to be promoting aldol condensation, at least in the case of acetone?
Reference

Burfield, D. R.; Smithers, R. H. Desiccant Efficiency in Solvent Drying. 3. Dipolar Aprotic Solvents. J. Org. Chem. 1978, 43 (20), 3966–3968. DOI: 10.1021/jo00414a038.


text:
Does solid sodium hydroxide react with carbon dioxide and air moisture?
I heard a claim in the context of soapmaking that a stray grain of sodium hydroxide will decay into soda ash, i.e. sodium carbonate, on contact with air. This reaction happens in an aqueous solution, but I'm wondering how fast it would happen with only the air moisture absorbed by the hydroxide grain.
More to the point, some of the sodium hydroxide grains I have for soapmaking seems to have  stuck together into a solid lump by absorbing air moisture inside the closed container. Should I assume a nontrivial fraction of that has turned into sodium carbonate by now?

Rather reacts with aerial CO2, as compounds not containing carbon cannot decompose to compounds containing it.

a) nobody can guess how tight your closed containers are. b) what does carbonate do when you add a few drops of vinegar?

Related: https://chemistry.stackexchange.com/questions/27221/why-do-carbon-dioxide-and-sodium-hydroxide-not-form-sodium-oxide-and-carbonic-ac

Consider that, even with vastly increased CO2 in the atmosphere since the Industrial Revolution, it's << one part in a thousand. Depending on the rate of infiltration into the container, the biggest contaminant is just water, fusing the pellets together -- NaOH is very hygroscopic.


text:
Could perchlorate salts allow for underwater combustion?
So apparently perchlorate salts can act as oxidizers for combustion (basically they can replace oxygen in the fire triangle, at least I think), and the reason why you can't have a fire underwater is because there's not enough oxygen gas in the water. So if one were to mix/dissolve perchlorate salts with a liquid fuel and made it hot, could the perchlorates allow it to combust?

Sodium metal doesn't need all that. Just pop it into the water -- outdoors, and surrounded by a good firebreak -- and watch what happens!

@OscarLanzi But then it is not underwater scenario.

You need fuel, oxidant and high enough temperature. Nobody says oxygen must be in free form. E.g. asymmetric dimethylhydrazine + NO2 as binary rocket propeller. It need not to be oxygen either. And fuel need not even to be  considered combustible. E.g. sand would burn if immersed in ClF3.

By other words, the mixture would burn even in vacuum. With water, there is condition for such a reaction to be so intense that water would not manage to cool it down and stop it.

Thermite is commonly used underwater, and makes a better choice, since neither aluminium nor ferric oxide is soluble in water, as are most chlorates. https://www.youtube.com/watch?v=CO5OKAo8aIw


text:
Do you always need to balance a chemical equation to find the percentage yield and limiting reagent?

I'm doing some problems through MIT open courseware and I've got one I'm absolutely stuck on. It details a three-step synthesis for hyperzine A, gives molecular weights and diagrams for individual reactants and the formula for the final result (huperzine A).
Then it asks you to find the limiting reagent for the first reaction. Every single source I've been able to find on this says that you need to balance the equation in order to do this, but the reactants seem so complicated I would have no idea how to begin doing this. It seems like we've got C10H16O(s) reacting with C8H5O4S2F6N(aq), but the result seems to be C11H15O3S1F3 (sorry, I have no idea how to format). They don't mention any other products, so my most obvious problem is that I have no idea what happens to the nitrogen, and even getting past that I have no idea how you would balance such a complicated reaction to work out mole ratios.
Question a asks about the limiting reagent, and you are only told about how much product is produced in question b, so I'm assuming to answer question a you don't need that information.
Furthermore, question b asks about percentage yield, and again, every source I've been able to find has stated you need to know the chemical equation (which I'm interpreting to mean 'balanced') to get this.
Since the only discussion of this problem I've seen online declares it 'easy' without any further explanation, I must be missing something big, but I've been bashing my head at this for many hours and have absolutely no idea where I'm going wrong. I've even accidentally spoiled myself on the answers at this point, but they haven't helped me to get any further with it.

The chemical equation does not show all products (nitrogen disappears) so you can't balance it. Either assume a 1:1 stoichiometry of the two reactants, or look up whether Bis(trifluoromethanesulfonyl)aniline can donate two triflyl groups or just one.

Consider Formatting of chem expressions by using mhchem MathJax package, available in CH SE by default, involving MJ syntax \ce{} and \pu{}
E.g. write $\ce{H2SO4}$ or $\ce{a A <=> p P}$ or $\pu{6.022E23 mol-1}$  to get $\ce{H2SO4}$ or $\ce{a A <=> p P}$ or $\pu{6.022E23 mol-1}$ (all eventually with double dollars in the display mode like $$\ce{H2SO4}$$.

@KarstenTheis Thank you! I'll try the problem again that assumption, and if it doesn't work I'll turn to google for the stoichiometry.

You're stuck in general and inorganic chemistry. This is organics, don't balance complete sum formulas, that's pointless. You need to think in "equivalents".

@Karl Can bis(trifluoromethanesulfonyl)aniline donate two triflyl groups or just one? I tried to look it up without going to the library and did not find a definite answer. You might know.

@KarstenTheis No idea. First guess would be it transfers only one.


text:
How the optical fibre pH microsensors work?
Just browsed through the chemical sensor products of the company, called PreSens (https://www.presens.de/products/ph/sensors) in order to look for a precise method to measure pH change in a physiological solution (Dulbecco's Phosphate Buffer Saline 1M).
I just would like to know how the "Sensor Spot" and "Needle type" optical sensors work exactly. According to the brochure on the aforementioned page, the Sensor Spots seem to based on fluorescence quenching: basically a quencher (I believe the hydrogen ions or the hydronium ions? I'm not sure) reduce the detected fluorescence intensity of the fluorophore (some pH dependent dye incorporated into a membrane "spot").
I would like to know more about this mechanism, I leafed through numerous reviews, but I have not be able to find any detailed information about pH detection based on fluorescence quenching. What is the quencher here? Could you tell me what is the mechanism in this particular case?
Thank you for your time.

Could you make a bit clearer what your actual question is? The flourescence changes (is diminished, "quenched") in the presence of H+, that's it. E.g. green light is coupled into the fiber, some red fluorescence comes back out, depending on pH. Are you asking for the chemistry of the dye?

No, I won't need such detailed chemistry, just how the pH change detected in this case by the sensor (the matrix intercalated with the dye)? The detection method I would like to know. Do they detect the back-propogated fluorescence and its change in wavelenght or what?

The link you posted clearly mentions the name of the principle involved in such pH measurements. There is no electrochemistry. It is called dual lifetime referencing (DLR) method. https://bmcbiotechnol.biomedcentral.com/articles/10.1186/1472-6750-12-11

Oh, I missed that description on the bottom of the page. Thanks for letting me know.


text:
How to synthesize sodium metabisulfite?
I am looking for ways to produce sodium bisulfite from $\ce{SO2}$ and $\ce{NaOH}$, but sources talking about $\ce{Na2S2O5}$ is quite scarce.
Do you have any recommendations of articles, books or any other type of source?

Most sources talk about buying it. The alternative is heating up NaHSO3.

Wikipedia is a good start.
When you treat a solution of suitable base, say sodium hydroxide or sodium bicarbonate with sulfur dioxide, you get sodium bisulfite.
$$\ce{SO2 + NaOH → NaHSO3}$$
$$\ce{SO2 + NaHCO3 → NaHSO3 + CO2}$$
Now the same reaction can also lead to sodium metabisulfite. When the reaction is conducted in warm water, $\ce{Na2SO3}$ initially precipitates as a yellow solid but when you apply more $\ce{SO2}$, the solid dissolves and on cooling, it crystallizes to sodium metabisulfite.
$$\ce{SO2 + 2 NaOH → Na2SO3 + H2O}$$
$$\ce{SO2 + Na2SO3 → Na2S2O5}$$
Alternatively, if you have heaps of sodium bisulfite, you can simply heat it to get sodium metabisulfite.
You may refer to this book: Johnstone, H. F. (1946). "Sulfites and Pyrosulfites of the Alkali Metals". Inorganic Syntheses pt .2 pp. 162–167. doi:10.1002/9780470132333.ch49


text:
Purpose of "1H" in naming of substituted heterocyclic compouns
I had a curious thought about the following molecule's name:

For such a simple molecule it seemed awfully complicated.  From what I gathered, the "[c][1,2]oxaborinin-1-ol" stems from the following fragment:

whereby [c] indicates the bond of fusion between the trifluoromethylbenzene and 1,2-oxaboran-2-ol.
The "7-(trifluoromethyl)-3,4-dihydro" is pretty evident from the compound's IUPAC numbering, where the heteroatoms are given the lowest numbers. However, from what I understand, the "1H" designation should to the location of unsaturation (similar to the saturation at carbons 3 & 4), but this boron has no hydrogen.

I'm clearly missing some fundamental nomenclature insight.  Could anyone please help shed some light on this?

See, the parent structure used in the name is unsaturated borinine not saturated borinan - it doesn't have the 1H hydrogen that's substituted with -OH in your structure.

and https://chemistry.stackexchange.com/questions/160333/iupac-naming-of-purines-and-pyrimidines?noredirect=1&lq=1

@Mithoron Ah, I see.  So when naming these fused ring systems, is the parent structure always the aromatic compound?


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Searching for evidence-based published study that compares disinfectant residues after evaporation
I've been plowing through databases to find evidence-based research reporting on the residue remaining after evaporation of chemical disinfectants. I want to know if there is a study which states the amount of residue that disinfectants (which are commonly used for healthcare surfaces or cleanrooms) leave after evaporation in air-dry way (not via Residue on Evaporation RoE Test which includes heating the substance). These disinfectants are mostly common and include alcohol, chlorine and chlorine compounds, formaldehyde, glutaraldehyde (GTA), hydrogen peroxide (H2O2), iodophors, ortho-phthalaldehyde (OPA), peracetic acid, peracetic acid and hydrogen peroxide, phenolics, and quaternary ammonium compounds (quats).
I've found an article which states the amount of residue on some materials in healthcare equipment, but it doesn't say if the disinfectant has evaporated or not1 . There are some reports from ALS lab too, but the source is not clear enough(see image below) 2 .

C. Lerones, A. Mariscal, M. Carnero, A. García-Rodríguez, J. Fernández-Crehuet, Assessing the residual antibacterial activity of clinical materials disinfected with glutaraldehyde, o-phthalaldehyde, hydrogen peroxide or 2-bromo-2-nitro-1,3-propanediol by means of a bacterial toxicity assay,.Clinical Microbiology and Infection 2004, 10, 984–989. DOI: 10.1080/00397918908050700.

https://www.contecinc.com/articles/disinfectant-residues-mitigation-and-management/

Thanks community!

You are asking for references to reputable studies on disinfectant residues. The link you provided cites the following (which is the end of that road, it provides a citation at the end, which you consider "grey"): https://www.cleanroomtechnology.com/news/article_page/Challenging_disinfectant_residues_as_per_GMP_Annex_1/160751 In other words, it seems you want to compare the amount of residue left by different disinfectants. Is that correct?

Yes @BuckThorn ! The reason I said the citation is grey is because it's not a published study in acceptable journals. The image I used in my post is not from the link you have put in your comment. It's from ALS Lab (anecdotally) which you cannot find its source. Please also let me know what details should I provide for my question to get it reopened.

@BuckThorn
I want to see if there is actually an evidence-based published study that compares disinfectants residues.

I would edit the title then to mention "evidence-based published study that compares disinfectants residues" and put the "reference-request" tag under the question. Also, I would remove the discussion of biofilms, which imo doesn't make sense. The residues are relevant in the context of the operation of clean rooms which have to maintain very high standards with regard to residues on surfaces or the air. That's what the company you linked to is concerned with.

@BuckThorn I changed the question to what you have desired. Hope it's now admissible for the community.

I took the liberty of modifying the title and editing parts of the grammar. Btw in hindsight maybe the biofilms question makes sense, but it is unnecessary to mention, so good to remove it.


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What is the hybridization of oxygen atoms in boric acid?
I was making the $\ce{H3BO3}$ structure and tried to think of hybridization of oxygen atoms and got confused between $\mathrm{sp^2}$ and $\mathrm{sp^3}$.
Boron has empty orbitals, so the lone pairs of oxygen can do backbonding with boron, resulting in $\mathrm{sp^2}$ hybridization. But at a moment only one out of three oxygen atoms would do so. What about other two oxygens?

Whenever the lone pair electrons goes in resonance, hybridisation decreases

All three oxygen atoms backdonate a π-electron pair and thus they are all $\mathrm{sp^2}$. When you construct the molecular orbitals, you find that the π orbitals are as follows:

One orbital bonds the boron atoms to all three oxygen atoms in a double Y shape — one Y on each side of the molecular plane.

Two orbitals are nonbonding because they have electron density on the oxygen atoms only; the boron is a node. (This corresponds to two oxygen atoms having a second lone pair whenever you draw any one of the contributing valence-bond structures.)

Finally, an antibonding orbital, which has the same threefold rotational symmetry as the bonding one but the oxygen atoms contribute out of phase from the boron atoms.

Since there are six π electrons (two from each oxygen atom but none from the electron-deficient boron), only the bonding and nonbonding orbitals are occupied and thus the molecule is bonded with the planar, $\mathrm{sp^2}$-hybridized structure.

Thank you very much , this cleared my doubt .


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Can proton NMR be used to study transition metal complexes?
I prepared metal complexes with a Schiff base ligand L using cobalt(II) chloride hexahydrate, iron(III) chloride hexahydrate, and nickel(II) nitrate hexahydrate. I used 1H NMR to study the difference between them and maybe get some information about the structure of the formed complexes.
The results showed a slight upfield shift in almost every peak and the appearance of some new peaks (attached figure), which is a bit ambiguous for me, and I don't know how to interpret them:

Please be more precise ! Which Schiff base do you use ? Please describe the positions of the peaks and their intensity. Without any numerical values, we cannot help you.

Dear @Maurice, thanks for the recommendation. I added a figure, I hope its helpful.


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What will be the major product in the nitration reaction of 2-methyl-5-nitrophenol?

I know if the $\ce{-NO2}$ group be added at the ortho position with respect to the $\ce{-OH}$, hydrogen bonding between the hydroxyl and nitro groups would lead to stability.
But would that product be major considering steric factors?

Nitro groups really "don't like" to be next to each other - that leaves one position.

A Chem. Absts. search claims nitration para to OH. The patent was TL:DR. No structures. For what it is worth.

Nitration of phenol with concentrated nitric acid gives 2,4,6-trinitrophenol . So, it is quite possible here that the corresponding substituted trinitrophenol may form in the reaction. But you have also added concentrated sulphuric acid so it may cause some other product to form.
Generally electrophilic aromatic substitution of phenol gives para product in slightly higher amount than ortho product due to inductive effect of oxygen. However in case of nitration, ortho product becomes higher than para product due to intramolecular hydrogen bond formation. So I think that it should be major product in your reaction also

Can you cite a source for the nitration of phenol by concentrated nitric acid which gives benzoquinone as major product. Because I couldn't able to find it in the internet.

conc. $\ce{H_2SO_4}$ and conc. $\ce{HNO_3}$ form $\ce{NO^+}$, thus hydrogen bonding may not be the deciding factor for the kinetic pathway.

Why was this even upvoted? Nitro group is in meta  - not considering this makes any answer a waste of time. Not to mention steric effects are way more important than tenuous h-bond.

@ananta how is $\ce{NO^+}$ formed? When I woke up this morning I was sure it was $\ce{NO2^+}$.

must have been a typo, you are right $\text{NO}_{\text{2}}^+$ is formed, we are substituting the $\ce{H}$ with $\ce{NO2}$ groups


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CFSE for octahedral complexes with more than one type of ligands
How would be crystal field stabilization energy for the complex [Co(NH3)5Cl]²-? I guess that due to the presence of negative ligand Cl-  3 sets of d orbitals can be present with a dz²or dx²-y²orbital of a slightly little different energy.

Do $\ce{Cl}$ electrons interact with $\ce{e_g}$ or $\ce{t_{2g}}$ set of orbitals?

I do want to know too

When more than one type of ligands are present with some being SFL and others being WFL then, the final effect of those ligands dominate which are greater in number. So, in the given compound, ammonia is more than chlorine so effect of SFL dominates and you get inner orbital complex

I wonder .. how this answer is connected to the question.

Then what do you want to know, please clarify

How the d orbitals are split - their energy level diagram under the influence of both the ligands!


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How can I produce Red Iron oxide with electrolysis
I am producing Iron Oxide with the intention of using it for pigmentation. My attempts so far have been to use electrolysis. My first attempt I used Borax as an electrolyte and a 12 volt adapter with 3-4ish amps. The result was a typical light orange rust color, however the process was rather slow, 24 hours resulted in about a tablespoon of rust. My next attempt I used table salt instead. This was far faster, it produced nearly a cup of black sludge in 10 hours. As per the tutorials I watched on youtube, I then filtered the matter through a coffee filter and a funnel, then heated the result in a pan. The result however was a rust color covered in a layer of black crust, as if I had burnt a sheet of cookies. Powdering the result gave me a powder of very dark brown color.The Residue left over on the coffee filter and in the container I used for electrolysis are the typical rust orange color I would expect. It is my understanding that heating the black matter that results from the electrolisis usually produces a more red color. So my question(s) are:

how I can change my approach to produce red pigmentation using this process, and why may I have ended up with black residue instead of orange or red?
What are the variables that change the pigmentation of rust?

Is it the speed at which it drys? the temperature in which it is heated? the electrolyte used in the process of electrolysis?

Because I like to do things myself. And I like to learn how these things work. I Also have access to a lot of old scrap iron. Honestly this is a silly question. I'm an artist and I like to make things from scratch, why do anything I do? why build myself a kitchen island from wood I logged and milled from the forest when I could just buy one from lowes? Does it really matter why I want to do it this way?

I'll try ball milling it and see how that changes the pigmentation. Thank you for the suggestion, I will let you know how it goes

Red iron oxide is Fe2 O3 , it forms in air at roughly 700F. In a water solution you will make "rust", brown/red. Rust is a complex mixture of hydroxides and hydrated oxides such as Fe3 O4 -H2O. At temperatures higher than 700 F  you start getting black Fe3 O4 ( aka  mill scale). At very high temperature you get black Fe O. Fretting corrosion will produce Fe2 O3 at room temperature but is not a practical source.

What IS a practical source?

On the net I find "red ochre" is mostly Fe2 O3 made by roasting mixed iron oxides but no specific information.


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Question on Thermodynamics from a recently conducted Examination
This is a question from the recently conducted JEE Main exam (in India). I had solved it myself, but found the answer to be incorrect. Here's the question.
Here is how I solved it:
$\Delta H = \pu{41.1 kJ mol-1}$
$R = \pu{8.314 J mol-1 K-1}$
$w=\pu{36 g}$
$$ \Delta n_g=2$$
$$ \ce{\Delta H = \Delta U + \Delta n_gRT}$$
$$ \ce{ \Delta U = (44.1 - (2×8.31×373)/1000) \pu{J K-1}}$$
$$\ce{\Delta U = \pu{(41.1 - 6.199) J K-1}}$$
$$\ce{\Delta U = \pu{34.9 J K-1}}$$
$$Ans. 35$$
However, the answer key states that the answer will be 38, instead of 35.
I feel that the answer can be 38 only if there is 1 mole of water instead of 2.
Kindly help, and point out my mistake, if any.

Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of at least essential parts.
Suitable formatting can be done according to formatting math/chem expressions/equations.

Please transcribe the text from the images into type (see previous comment). Also, please provide a more informative title.

Your mistake is that you took $\Delta n_g=2$ when it should be $1$. Remember it is asking for the internal energy change for the reaction (this can be seen by the fact that kJ/mol is being asked, not kJ-remember dividing by moles makes this intensive so there is no dependency on number of moles taken). Thus, the moles of water taken is irrelevant and $\Delta n_g$ only depends upon the stoichiometric coefficients. Since the reaction is $\ce{H2O(l) -> H2O(g)}$, so $\Delta n_g$=1.The mass of water given is just the examiner trying to trick(or rather, test) you.
That being said, I might add that the question is poorly framed and also somewhat ambiguous. This is because of 2 reasons:-

It should be clearly mentioned that internal energy of vapourisation of the reaction is being asked and the only hint of this is kJ/mol which is easy to miss.
Additionally, it should be made explicit that internal energy of vapourisation of the reaction $\ce{H2O(l) -> H2O(g)}$ is being asked because it can vary depending on if it is $\ce{H2O(l) -> H2O(g)}$ or $\ce{2H2O(l) -> 2H2O(g)}$ or $\ce{3H2O(l) -> 3H2O(g)}$ etc.(see addendum for more)

However, some texts define enthalpy of vapourisation and internal energy of vapourisation as the $ \Delta U$ and $\Delta H$ of the reaction when 1 mol of the compound is undergoing phase transition, and then the above 2 points don't hold because if moles are taken, then stoichiometric coefficient magnitudes are irrelevant and only their ratios are important.
Addendum:
Just in case you want some depth in this topic, remember that in any calculation of $ \Delta _rH$ and $ \Delta _r U$ we
assume the number of moles of the reactants/products to be same as that of the stoichiometric coefficients. This is because by definition, $ \Delta _rH$ and $ \Delta _r U$ depend only on the reaction equation and not on particular reactions in which we can vary the number of moles being taken.If you calculate $ \Delta H$ or $ \Delta U$ of a particular reaction, then that cannot be defined as $ \Delta _rH$ or $ \Delta _r U$ unless the number of moles of every reactant and product matches the stoichiometric coefficients. Finally, note that stoichiometric coefficients need not be the simplest ratios i.e. $ \Delta _rH$ will be different for each of   $\ce{H2O(l) -> H2O(g)}$, $\ce{2H2O(l) -> 2H2O(g)}$ and $\ce{3H2O(l) -> 3H2O(g)}$ because the stoichiometric coefficients are different.


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Why does column chromatography not work like it's supposed to?
I'm a few months into working in an organic synthesis lab and I consider myself to have a solid intuition for, well, how chemistry works. I have ran at least 50 columns at this point but no matter what experiment I run, there's always some concepts that seem off.
To start, I don't think my technique is the problem. I know for a fact I am packing the column well. I dry pack but run plenty of solvent cycles through it before loading. I try to load with as little solvent as possible, to the point where the top of the sand is still dry sometimes. I load with the same solvent I use for elution, so it is not too polar in comparison. I run the column slow (like abysmally slow, 2 drips a second) to not "flush" things out. I try to perfect technique and minimize other factors.
But I still run into problems. For example, everyone uses TLC plates to test different solvent systems for maximum separation. If my desired product dot is smack in the middle with Rf of 0.5 on the plate, I'm expecting it to come out within two cycle volumes on the column (1/Rf). But regardless, it seems like it always comes out right after the first cycle and it catches me by surprise. The compounds I'm isolating have very close Rfs, and it would be great if I could at least predict when it will come out so I can switch to smaller collection tubes. I realize that not using the recommended 0.2 Rf means I will need to be very precise with collection but I don't mind. What I DO mind is that the product doesn't even match the tested Rf.
Also, it seems like if I use any solvent system more polar than the recommended 0.2 Rf TLC, the product and all the impurities will wash out together with almost no separation at all. It's like all basic concepts of chromatography have gone out the window. The separation was incredible on the TLC plate, like I'm talking the product at an Rf of 0.5 and the impurities had Rf of 0.2. Then I try a column with 50 mL cycle volume and everything started coming out within 10mL of each other?
Also, collection tube TLC doesn't make much sense either. I take a TLC of a collection tube with the product, with the product showing Rf of, say, 0.5, but the tube also contains an impurity with an Rf of 0.05. How is that even possible? According to my cycle volume calculation (CV * 1/Rf), an impurity with that Rf should come out many many collection tubes afterwards.
I just want to understand the actual reason why this is happening. I don't mind changing my procedures, but it is rather frustrating to not understand the science.

The sorbent for columns and TLC plates may not be exactly the same (plate binders, lots)  nor  at exactly the same conditions (activation, dry/wet).

On what scale are you doing this? Prep TLC may offer a better result.


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Why does equilibrium of ice and water only exist at 273K at normal pressure?
I am a high school student and I am a little confused in the concept of spontaneity of a reaction and how equilibrium works for a reaction, I got some confusions:

Let's take example of freezing of water: In my textbook its written that freezing point of water at 1 atm pressure is 0 degrees and at this temperature water and ice remains in "dynamic equilibrium". So it means this phase transition from solid to liquid is a "reversible process" and we know at equilibrium the "change in Gibbs free energy is 0". Now if I heat it "∆G would become negative" and what happens is more ice melts but the temperature doesn't changes and stopped heating so a new equilibrium got established, so does it mean here equilibrium constant is changing(as more water is forming and more ice is melting) without increase in temperature? But I used to think it only depends on temperature. So what's going on. please explain it in simple language so that a high school student can understand?

Suppose I have ice at room temperature in a closed beaker. What we observe in our daily life is whole ice melts into water but shouldn't the "equilibrium" still exist because its a reversible reaction ,so just like evaporation in a closed beaker the water reaches an equilibrium state with its vapors at all temperatures, ice should also do the same with water but it doesn't. Why?

The amount of ice or water is not important for equilibrium constant. What is important is water activity in both phases is equal. In a very informal layman way, activity is "how much water molecules announces they are there, to do some fun actions". This activity is for pure water or ice at constant T independents on amount.

Room temperature is at 293 K,fyi.

@BruhMoments This is not entirely correct.

@andselisk average,I should say.

To write down the equilibrium constant expression, you use concentrations, not volumes or mass. To illustrate this with an example you can check in your kitchen, a saturated salt solution does not get saltier by adding more salt (you have pure undissolved solid salt in either case). It does not get less salty if you remove part of the undissolved salt.
The equilibrium between liquid and solid water is a bit special because both species are pure. Adding more ice or adding more water will not change the "concentrations" or the freezing point of water.

[...] more ice melts but the temperature doesn't changes and stopped heating so a new equilibrium got established [...]

No, you still have pure water and pure ice, so the equilibrium constant is still the same, and the concentrations are still the same - it is not a new equilibrium. On the other hand, if you pour salt into the water, the liquid water concentration drops (no longer pure) and the solid water concentration remains constant, so then you are out of equilibrium, and ice will melt and the temperature will decrease. With pure water, when the temperature is higher or lower than the freezing point, ice will melt or water will freeze until we are at the freezing temperature (or we run out of ice or water).

[... from OP's comments] how the temperature will decrease??? suppose if we have ice and water at 100 degrees and I add ideal solute to the water ,the concentration of water would decrease and it will only decrease the no. of water molecules going back to the solid state but the no of solid molecules going to liquid state remains the same so equilibrium will shift forward but here also "the equilibrium constant" should remain the same if we think this way.... but it doesn't if we think of its formula why?

In ice, water has 4 hydrogen bonds. In liquid water, it has less. So transferring a water molecule from ice to water is an endothermic process, and it will cool the system down (thermal energy turns into potential energy).

[...] so just like evaporation in a closed beaker the water reaches an equilibrium state with its vapors at all temperatures, ice should also do the same with water but it doesn't.

The difference between pure liquid water and a gas mixture of water and air ("humid" air) is that the water in humid air is not pure. You can have air with higher or lower humidity. The water in a half-full closed bottle will evaporate until the humidity reaches its equilibrium value. On the other hand, liquid and solid water are both pure, and are at equilibrium (at normal pressure) only if the temperature is equal to the normal freezing point.

What we observe in our daily life is whole ice melts into water but shouldn't the "equilibrium" still exist because its a reversible reaction

Because the ambient temperature in our lives is usually higher than the freezing point of water, and because our containers are not perfectly insulated, we keep transferring heat to the container. By the way, even when ice melts, at the molecular level, some liquid water molecules still attach to the solid. The process of water molecules going from solid to liquid just happens to be faster.

[from OP's comment] does that mean the equilibrium between ice and water exists at all temperatures?? so why do textbooks says and also the phase diagrams says that it exists "only" at freezing point

If water and ice come in contact, there will be a forward and a reverse process (freezing and melting) at the particular level. When the two phases have the same temperature and these two process have the same rate, it is called equilibrium. Because the rates are the same at equilibrium, there is no net change, i.e. the amount of ice will neither increase nor decrease. The temperature where this happens is called the freezing point or melting point, even though at this point, there is not bulk freezing or melting.

[...from OP's comments] can u please explain"dynamically" what's going on when u add "ideal solute" to the water ?? I understand that the concentration of water would decrease and more ice will melt....but if we will think how the activity of the molecules actually affected on adding solute things get difficult? like if i add an ideal solute to the liquid it will not change the bonding because its ideal so what I think is the no. of water molecules going back to solid state should still remain the same??

An ideal solute will not interact different with water than a water molecule, but it can't become part of the ice lattice (it does not fit). So instead of - say - 100 water molecules bumping into the ice surface, now 99 water molecules and one other molecule bump into it, decreasing the rate of attachment of water molecules to the ice.

can u please explain"dynamically" what's going on when u add "ideal solute" to the water ?? I understand that the concentration of water would decrease and more ice will melt....but if we will think how the activity of the molecules actually affected on adding solute things get difficult? like if i add an ideal solute to the liquid it will not change the bonding because its ideal so what I think is the no. of water molecules going back to sloid state should still remain the same??

"Because the ambient temperature in our lives is usually higher than the freezing point of water, and because our containers are not perfectly insulated, we keep transferring heat to the container." does that mean the equilibrium between ice and water exists at all temperatures?? so why do textbooks says and also the phase diagrams says that it exists "only" at freezing point

"so then you are out of equilibrium, and ice will melt and the temperature will decrease. " how the temperature will decrease??? suppose if we have ice and water at 100 degrees and I add ideal solute to the water ,the concetration of water would decrease and it will only decrease the no. of water molecules going back to the solid state but the no of solid molecules going to liquid state remains the same so equilibrium will shift forward but here also "the equilibrium constant" should remain the same if we think this way.... but it doesn't if we think of its formula why?

For the first question, consider a simple reaction for example take Haber's process in which nitrogen and hydrogen combine to form ammonia. Now, at a particular temperature (T), there will be some equilibrium constant and correspondingly, a particular value of $ ∆{{G}_{0}} $. What it simply means is that if you have a mixture of hydrogen, nitrogen and ammonia at a particular temperature, then equilibrium will only be achieved at this specific composition. Some other composition is also possible at the same temperature but it will simply not be at the same temperature but it will not be at equilibrium and will have to tendency to go towards equilibrium and this rate of movement towards equilibrium will depend upon other reaction conditions. So basically in your example,when you provide heat to the mixture, the temperature of mixture suddenly increases leading to change in value of $ ∆{{G}_{0}} $. As a result, melting of ice occurs which decreases temperature. This melting of ice happens till the temperature of mixture reaches the freezing point of water. Talking about equilibrium constant, it is not simple to describe since in high school,liquids and solids have no role in reaction quotient. But gibbs free energy change is the important thing. In real life, the freezing of water or melting of ice is so fast that it almost seems as if the externally supplied heat almost immediately was converted into latent heat
For your second question, as I said, equilibrium constant is a bit difficult to describe but it will definitely be something. So basically, some amount of ice must be left at room temperature but the equilibrium constant is extremely shifted to the right because of which it is in an extremely small amount not percievable practically. (It might be so less that it is less than even a few molecules, but since matter is quantised, it would not be possible to have fractional number of molecules.)
Even when you were talking about water- water vapour equilibrium, if you have taken very small amount of water in very large container, then there will be negligible water left as liquid at equilibrium.

First some key points that are important to keep in mind when thinking about thermodynamic equilibrium: (1) thermodynamic equilibrium states describe systems that are unchanging, static; (2) an equilibrium constant is relevant only when a system can exist in either of multiple states (as you suggest); (3) the equilibrium constant for the melting of water is just that, a constant equal to the ratio (equal to 1) of the activities of water and ice at the melting point (activities are approximated by concentrations in some cases).
What you are describing (melting of ice) occurs when the system is pushed out of its original equilibrium state by adding heat at a finite rate. Melting, in other words, is not an equilibrium state, it is a transition between different states occurring when you perturb the initial equilibrium state by adding heat.
To explain how melting affects the thermodynamic equilibrium between ice and water, consider the following process: (1) you start with ice and water at equilibrium at the melting point (surroundings and the system are at that point); (2) you isolate the system (by wrapping it with insulation) and transfer some energy into the system as heat; (3) the system returns to equilibrium by redistributing the input energy and forming a new state; (4) you remove the insulation and place the system into contact again with surroundings (at the melting point). Importantly, during steps 2 and 3 the system is not in an equilibrium state. In steps 2/3, if not too much heat is added (so as to melt all of the ice), then a new equilibrium can arise between water and ice, where both are at the melting point. If too much heat is added, then the transition of ice to a liquid will be complete. With sufficient heat the temperature of the system will rise above the melting temperature, with ice no longer present (but then at step 4 the system will be cooled again).
To help wrap your head around the idea that the free energy change after melting some of the ice remains constant ($\Delta G = 0$, $K=1$), consider what would happen to the proportions of ice and water if you should mix together two different samples of ice and water, or simply add water, or ice, to a sample of ice and water, all at the melting point. Nothing! The proportions would remain constant. The point is that at equilibrium (at the melting point) there is no force driving formation of more ice or water.
Superheated ice1 (ice at a temperature above the melting point) is not an equilibrium state relative to water at the same T, because above the freezing point liquid water has a lower free energy than ice. You can in fact create superheated ice, but it will spontaneously melt to water if you remove the constraints that allowed you to create this state.
In the case of a beaker (a diathermal container, allowing heat transfer), there is no equilibrium because heat is constantly being transferred into the container, assuming the surroundings are above the melting point (although if heat is added slowly enough then the system is approximately at equilibrium at any given point). The way of thinking around this is to consider the Gibbs free energy at the initial and final equilibrium states of the system, before and after changes due to addition of heat.
References

Schubert, G. and Lingenfelter, R.E., Superheated Ice Formed by the Freezing of Superheated Water. Science 1970, 168 (3930) pp. 469-470, DOI: 10.1126/science.168.3930.4.