text:
not understanding buffering region vs equivalence point
"During the titration of a weak polyprotic acid with a strong base, the first stoichiometric equivalence point occurs when the moles of added base are equal to the moles of the initial acid species.  Prior to the equivalence point, a buffering region forms in which both the weak acid and its conjugate base are present and the pH is determined by the pKa and the ratio of the acid and base species concentrations.  When the concentrations of the acid and base species are equal, pH = pKa."
The bolded statements seem like they're saying the same thing. What is the difference between the buffering region and the equivalence point?
Isn't the buffering region when [HA] = [A-], so that in the HH equation, pH = pka? The first bolded statement makes it sound like the first equivalence point is also when [HA] = [A-].

You're absolutely correct. The word base was used in two different ways which is confusing.

This is a matter of definitions which you must learn. Let's say that the titrant is NaOH and that we have a monoprotic acid.
Equivalence point - The equivalence point occurs when the moles of added base are equal to the moles of the initial acid species.
Notice that this doesn't infer anything about the pH of the solution.
Now for a polyprotic acid there are multiple equivalence points. The first equivalence point occurs when the moles of added base are equal to the moles of the initial acid species. The second occurs when twice as many moles of added base are added as compared to the initial acid species. The third three times as much base as acid and so on.
Again notice that this doesn't infer anything about the pH of the solution at the multiple equivalence points.
When the concentrations of the acid and base species are equal, pH = pKa."
This is a poor statement. Let's let the acid be $\ce{HA}$ so the conjugate base is $\ce{A^-}$. The statement would have been much better expressed as:
When the concentrations of the acid and its conjugate base are equal, pH = pKa."
Now to wrap this up, imagine doing this experiment. We start off with $\ce{HA}$ in solution. As we add more base we get to the pH where $\ce{[HA] = [A^-]}$ and where (1) there is maximum buffering capacity, and (2) the pH = pKa of the acid. We now add more base and we finally get to the equivalence point.
PS - For titrations you should also know the endpoint which is where the chemist gets some sort of signal to stop the titration.

I started reading about chemistry 2 months ago, so take this with a grain of salt. I also learned this in norwegian, so terms migt be different..
As far as i know, the equivalence point is when there is no longer any of the reactant, because it has all reacted with the added base/acid. Whereas the buffering/halfway titration point is when, as they say, there are equal amounts of H- as HA, making the solution a buffer.
Did i understand your question right?

That is not the correct definition for the equivalence point.

I know the definition is what you stated above, but is it not also true that at the equivalence point, all the initial acid has reacted with added base(which naturally have to be the same amount)?

Not necessarily. At the equivalence point we'd assume that a strong acid and a strong base would completely react. However that wouldn't be true for weak acid/strong base, strong acid/weak base, or weak acid/weak base.

I see.. I’ll have to talk to my teacher about this, then. She kind of gave the impression that the weak base would completely react because the strong acid would continue to remove the OH- as a product, thereby making the reaction continue.. I’m sure I’ll get it soon. Thanks!

Well what would the pH be for a 0.1 molar solution of sodium acetate? It isn't going to have a pH of 7.


text:
Bromination of but-3-en-1-ol in basic medium and Baldwin's rules

After formation of the bromocarbocation the alkoxide anion attacks the bromocarbocation which leads to ring formation and can proceed through two pathways
5 endo tet cyclization is disfavored but 4 exo tet cyclization is favored according to Baldwin's rules; however, the answer given in my text book is the 5 membered ring.

Your reasoning is totally correct according to Baldwin's Rule - the 4-exo-product would be more stable than the 5-endo-product ( see : This for more details).
But your method may be wrong in the given question the product A was first formed and then a ring closure step was performed using $\ce{KOH}$ as shown below :

here the 5-exo-trig ring closure is favoured according to Baldwin's rule .
Let me also address something that may be a doubt ( It was mine ) i.e. why doesn't the oxygen atom attack the bromonium Ion in the first step ? this is because even if it will attack you will also form a acid $\ce{HBr}$ which will cause the cleavage in the cyclic ester we just formed returning us to the original reagent hence it doesn't attack.

Upon closer reading, I do want to point out some inaccuracies. (1) Baldwin's rules do not say anything about the stability of products, but only whether the route for their formation is kinetically possible. The 4-exo-tet pathway is "favoured", and the 5-endo-tet is "disfavoured". Of course, it does so happen that the five-membered ring is more stable than the four-membered ring. That, however, is not a consequence of Baldwin's rules. (2) Also, there are no trig ring closures; SN2 takes place at a tetrahedral carbon, hence tet.

(3) Finally, and maybe more critically: in the SN2 step, both possibilities - 4-exo-tet, as well as 5-exo-tet - are "favoured", according to Baldwin's rules. Therefore, invoking Baldwin's rules is not sufficient to explain which pathway is preferred, as the rules do not distinguish between which is "more favoured". The preference for 5-membered ring formation over 4-membered ring formation is explained by the balance between enthalpy and entropy of activation, see https://chemistry.stackexchange.com/a/107070/16683 as well as pp 805–807 in Clayden 2nd ed

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text:
How can you use ICE tables to solve multiple coupled equilibria?
If I have a problem involving multiple coupled equilibrium reactions, such as

Calcium fluoride, $\ce{CaF2}$, has a molar solubility of $\pu{2.1e−4 mol L−1}$ at pH = 7.00. By what factor does its molar solubility increase in a solution with pH = 3.00? The p$K_{\mathrm{a}}$ of $\ce{HF}$ is 3.17.

The relevant reactions are:
$$\ce{CaF2(s) <=> Ca^2+(aq) + 2 F-(aq)}$$ and
$$\ce{HF(aq) <=> H+(aq) + F-(aq)}$$
They are coupled because fluoride occurs in both of them.
Is there a way to use ICE tables to organize the information (stoichiometry, initial concentrations, mass balance) as a way to solve the problem?
For example, I could try to set up one ICE table for each reaction (the column for $\ce{H+}$ is strange because in the problem, the pH is set to a value through an unspecified mechanism):
$$
\begin{array}{|c|c|c|}
\hline
 &[\ce{Ca^2+}]  & [\ce{F-}]  \\
\hline
I &  \pu{2.1e−4} & \pu{4.2e−4} \\
\hline
C &  +x  & +2x  \\
\hline
E &  \pu{2.1e−4}+x & \pu{4.2e−4}+2x \\
\hline
\end{array}
$$
and
$$
\begin{array}{|c|c|c|}
\hline
 &[\ce{HF}]  & [\ce{H+}]  & [\ce{F-}]  \\
\hline
I &  0 &  \text{N/A} & \pu{4.2e−4} \\
\hline
C &  +x &\text{N/A} & -x \\
\hline
E &  +x & 10^{-3.00} &  \pu{4.2e−4} - x\\
\hline
\end{array}
$$
However, how do the fluoride concentrations "talk to each other" in the two tables? Is the $x$ in one table the same as the $x$ in the other table?

Your "initial state" seems to be pH 7 so you do know the [H+] at the start.

Yes, but the pH is adjusted externally, so we can't take the change in [H+] to deduce how much [HF] and [F-] are changing. For that reason, I left that box as N/A. I could have left the entire column out instead.

https://chemistry.stackexchange.com/questions/110607/q-36-2018-molar-solubility-of-caf2-at-ph3-given-molar-solubility-at-ph7-and-pka

The way to use ICE tables in this case would be to combine the ICE tables, and use "$x$" for the changes due to one reaction and "$y$" for the changes due to the other. For each reaction, you have one unknown (how far it reacted, i.e. $x$ and $y$), so you need two pieces of information to solve it, in this case the two equilibrium constants.
Here is the combined ICE table:
$$
\begin{array}{|c|c|c|c|c|}
\hline
 &[\ce{Ca^2+}]  & [\ce{F-}]  & [\ce{H+}]&[\ce{HF}]   \\
\hline
I &  \pu{2.1e−4} & \pu{4.2e−4} & \text{N/A} & 0 \\
\hline
C &  +x  & +2x-y & \text{N/A} & +y  \\
\hline
E &  \pu{2.1e−4}+x & \pu{4.2e−4}+2x-y & 10^{-3.00} & +y \\
\hline
\end{array}
$$
Now, you can use the equilibrium constants for the two reactions to solve for $x$ and $y$, giving you the equilibrium concentrations.
What the combined table shows is that the fluoride concentration depends on both reactions, so you can't first deal with one equilibrium and then with the other but have to solve a system of two equations with two unknowns.
Once you combine the tables, you could also have initial conditions where nothing is dissolved yet, to get easier expressions (I'm using $p$ and $q$ because they will be different from $x$ and $y$):
$$
\begin{array}{|c|c|c|c|c|}
\hline
 &[\ce{Ca^2+}]  & [\ce{F-}]  & [\ce{H+}]&[\ce{HF}]   \\
\hline
I &  0 & 0 & \text{N/A} & 0 \\
\hline
C &  +p  & +2p-q & \text{N/A} & +q  \\
\hline
E &  +p & +2p-q & 10^{-3.00} & +q \\
\hline
\end{array}
$$

Alternatively, given the cross promotion of a related question, we can consider doing away with ICE tables all together.
$$K_{\mathrm{sp}} = \ce{[Ca^{2+}][F-]^{2}}=\pu{3.7e−11}$$
$$K_{a} = \frac{\ce{[H+][F-]}}{\ce{[HF]}}=10^{-3.17}$$
At this point, the unknowns are $\ce{[Ca^{2+}]}$, $\ce{[F-]}$, $\ce{[HF]}$. $\ce{[H+]}$ is known from the pH and it's $10^{-3}$.
We have two equations and 3 unknowns. The final piece that's missing is that because the source of all fluorine and calcium in this system is from calcium fluoride, we may impose the stoichiometry of the solid on the species present:
$$2\ce{[Ca^{2+}]} = \ce{[F-]} + \ce{[HF]}$$
Three equations and three unknowns. I cheated and used WolframAlpha:
$$\ce{[Ca^{2+}]} \approx 3.8\times 10^{-4}$$ This represents an approximately 1.8 fold increase in solubility.
(I sanity checked the calcium ion concentration for a neutral solution using this method and got back the molar solubility.)

Solving it is not too bad. The ratio of fluoride to hydrofluoric acid is $10^{-0.17}$ (from pKa and pH), so you know which fraction of total soluble fluorine is fluoride, so you know the proportion of calcium ions and fluoride ions. You can plug this back into the solubility product to get the calcium concentration.

@KarstenTheis Honestly, I was too lazy to do it by hand. I just noticed that WolframAlpha gave approximate values... I'll look again.

Yeah, you're right. The approximations are because I plugged in decimals into the system.

@ZheHow to derive the  following equation,explain ?$$\ce{2[Ca^{2+}]=[F−] +[HF]}$$

@AdnanAL-Amleh The question asks about solubility, so that means all calcium and fluorine derive from dissolution of calcium fluoride. Given the composition of that compound, there must be twice as much fluorine as calcium at any point in time.

@ZheCan we take into consideration the hydroylsis of $\ce{F-}$ and the following charge balance :$$\ce{[H+] + 2[Ca^{2+}]=[F-] +[OH-]}$$ and the mass balance :$$[\ce{F-}]_\mathrm{I}= 4\cdot{2}\times{10^{-4}}=[\ce{F-}]_e +[\ce{HF}]_e$$ correct and advice.

But you don't know what the starting concentration is. If you want to do it by looking at changes, use the other answer.

@KarstenTheisKindly to express and change the following sentence to  equation and numbers  "so you know which fraction of total soluble fluorine is fluoride, so you know the proportion of calcium ions and fluoride ions."

just to remediate the ambiguous.thanks.

@AdnanAL-Amleh It's just division, and frankly, it's not ambiguous. At this point, it's probably more helpful if you try to work through these difficulties yourself than have everything spelled out.

@ZheI am not live in western civilization, sometimes I face difficulties in understanding the intention of the writer,  so when he expresses the  idea  by equations and numbers it seems to be clearer

Unfortunately, we have words because equations cannot always express what we mean.

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text:
Feasibility and Entropy
Suppose you've got some reaction that has a positive $\Delta H$, and also has positive $\Delta S_\mathrm{sys}$. Due to the equation
$$\Delta S_\mathrm{tot} = \Delta S_\mathrm{sys} - \frac{\Delta H}{T}$$
a low enough temperature will cause $\Delta S_\mathrm{tot}$ to be negative. At this point the reaction won't happen anymore because second law and whatnot.
What exactly is happening at the molecular level that prevents molecules from reacting with each other? The orientation of the molecules is more or less random, and the Maxwell–Boltzmann distribution means there ought to be some particles with sufficient kinetic energy, so what's happening? If the reactants are left long enough, 2 molecules with acceptable orientations & kinetic energies should whack into each other.
A while ago I found the "micro-states per macro-state" explanation of entropy. I get that entropy isn't a property of a single particle, or even of a specific arrangement (microstate) of particles (please correct if wrong). Honestly this just makes me even more clueless as to my above question.
I hope I haven't made this too unclear.

If it takes a trillion years for such a process to happen, do you still think of it as "happening"?

Pretty much. It's random, so it could happen within 10 seconds, it just probably won't. Shouldn't 2nd law mean it would never even be possible for it to happen, under any circumstance?

Your equation assumes that this is occuring at constant P. Your formula for $\Delta S_{tot}$ is not a general one, but a specific one. Entropy is usually defined in terms of heat. Heat is only equivalent to enthalpy when enthalpy is a valid. In textbooks we always end up using enthalpy when we have constant T and P, and we use internal energy U, when we are given constant T and V. This is because we are also given at the back of the textbook tables so we can look up U and H for given T and V or P respectively. You will have better luck understanding  thermo if you use fully general equations.

If you think that, then that would be a misunderstanding. Positive entropy change doesn't mean impossible. It just mean very unlikely.

To start, it's important to differentiate kinetics (which describe the rate at which reactions happen) and thermodynamics (which describe what net reaction can happen, even if it doesn't happen on the time scale of the universe's existence).
Your equations are all thermodynamic equations, so let's start there. I'm going to assume that you meant standard enthalpy and entropy. Then if we multiply through by $-T$, we end up with the definition of standard Gibbs free energy:
$\Delta G^\circ = -T\Delta S_{tot}=\Delta H^\circ_{sys} - T\Delta S^\circ_{sys}$.
If $\Delta G^\circ<0$, we say the reaction is spontaneous, but only in the thermodynamic sense that it can happen given infinite time. An important point is that this statement refers to the net reaction involved. The reverse reaction (which necessarily has a positive $\Delta G^\circ$) can happen, but always at a rate less than that of the forward reaction, such that the net reaction observed is the forward reaction. Because we're talking about thermodynamics and not kinetics, we can't make any conclusion about what those rates will be in absolute terms, only that the forward will be relatively faster than the reverse.
A different way of saying that is that if we set up a situation in which all reactants and products are at standard states (generally defined as 1 M solutes, 1 bar gases, pure solids, pure liquids), the concentration changes over time will be increases in products and decreases in reactants until the reaction reaches equilibrium. For a reaction that reaches equilibrium quickly, both the forward and reverse reaction happen quickly, but the forward slightly more quickly. At equilibrium, both reactions are still occurring, but now at exactly the same rate.
For a concrete example, consider the dissolution of a soluble compound like NaCl. If we have a solution of 1 M NaCl in water and we add excess solid NaCl, more NaCl will dissolve, but even when the solution becomes saturated (equilibrium), ions are constantly leaving the solid and going into solution while other ions are depositing from solution to the solid at the same rate.
In your hypothetical case where a change in T results in $\Delta G^\circ$ becoming positive, the reactions again are still happening. The difference is just that the reverse is faster than the forward when things are at standard states. It turns out that dissolution of NaCl is one of those cases you described, where $\Delta H^\circ >0$ and $\Delta S^\circ>0$. That means that if you made a 1 M solution and then cooled it, at a certain point solid would precipitate out, meaning that you've gotten to the point where at the standard state, the reverse reaction is faster than the forward reaction. At this temp, $\Delta G^\circ$ for dissolution is positive. If you stay at that temp, you can wait infinite time, and the solid will never fully redissolve.
All of that said, it sounds like your question might really be about kinetics, that is, how much can you slow down a reaction by cooling it? That question has little to do with entropy and instead relates to activation energies.

So for the cold NaCl solution, entropy change of dissolution would be negative. But in a really, really small time frame, an ion might come off the lattice. In the next moment, sure, 2 ions may rejoin, but in the first moment the overall entropy change is negative. Doesn't this violate 2nd law?

@Hal Gee you are just looking at those two ions neglecting what happens to those already in solution. If they were alone, their going in solution doesn't violate anything.

Ok, but if you pick a moment during which the only thing that is happening to the lattice is an ion coming off, isn't this a negative entropy change? The rest of the ions in solution during that moment aren't doing anything, so they shouldn't make a difference.

The second law is a statement of the statistical behavior of a large ensemble over time. It does not apply to a momentary fluctuation of a single atom.

Assume the entropy change of the system associated with a reaction is independent of temperature and positive, as you stated. Assume also that the reaction is endothermic, with constant enthalpy change, as you stated. Finally say that you lower T so much that
$$S_\mathrm{tot} = \Delta S_\mathrm{sys} - \frac{\Delta H}{T}=\Delta S_\mathrm{sys}+\Delta S_\mathrm{surr}<0$$
The reason a reaction stops occurring (is no longer spontaneous) is that you need to transfer heat from the surroundings to the system in order to have enough energy to make the required change to the system (this is the meaning of "endothermic"). However, the loss of entropy of the surroundings associated with this heat transfer would exceed the gain in entropy of the system during the course of the transformation. Now, you can simply say that "this violates the second law of thermodynamics": you cannot spontaneously take heat away from (reduce the entropy of) the surroundings without a correspondingly equal or greater increase in the entropy of the system. If this sounds tautological, it is because it is a fundamental law. The proof is in the fact that the opposite has never been observed.


text:
Spraying aluminium hydroxide nanoparticles onto PLA as fire retardant?
Several scientific papers study FR-PLA (fire resistant PLA) in which Aluminium Hydroxide nanoparticles are melt blended into the PLA (example source: Advances in Flame Retardant Poly(Lactic Acid). Would spraying the nanoparticles (dispersed in a liquid -source - maybe non flammable silicon oil) onto a 3D printed PLA help with fire resistance or not make a difference?

You mean you first print your thing, and then you cover it with spray of nanoparticles containing oil?

@Gimelist yes correct, we've printed the thing already so too late for us to reprint with the nanoparticle inside. Trying to find a fire/heat retardant as surface treatment for PLA.

I suspect this is too late. The whole point of embedding the nanoparticles inside PLA is that it remains inside until needed (when temperatures rise above ~300 °).
By applying it as a coat, it can be easily scratched or rubbed off because PLA is so soft or flexible. You are also running into the problem of shedding of Al(OH)3 NP everywhere which may be a problem.
A coat of sprayed Al(OH)3 NP is only going to be a few micrometres thick, or you will need to use some glue to make them stick, which complicates thing and changes the appearance of your product. On the other hand, embedded Al(OH)3 NP are present throughout the entire thickness of the product. During fire, your sprayed NP might dehydrate just from the heat of radiation without even coming into contact with fire.
This is a bad solution.


text:
Calculating Equilibrium Partial Pressures Given Kp and Mass

For the reaction
  $$\ce{AsCl5(g) <=> AsCl3(g) + Cl2(g)}$$
  at $550$ K, the equilibrium constant ($K_p$) is $9.81$. Suppose that $3.150 \ g$ $\ce{AsCl5}$ is placed in an evacuated $600$ ml bulb, which is then heated to $550$K.
What is the partial pressure of $\ce{AsCl5}$ at equilibrium?

So at the moment, I understand that
$$\ce{K_p=$$\frac{(P_{AsCl3})(P_{Cl_2})}{(P_{AsCl_5})}$$}$$
And that using a rearranged Ideal Gas Law I can get
$$\ce{P=$$\frac{n(0.0125mol AsCl5)*R*T(550K)}{V (0.6L)}$$}$$
Giving me $$\ce{0.94 atm}$$
but I'm not sure where to go from here. Can somebody help by pointing me in the right direction?

Say you start with $A$ moles of $\ce{AsCl5}$, and then you reach a temperature $T$, for which the equilibrium constant is $K_p$.
In order to achieve equilibrium, some amount of forward reaction ($\ce{AsCl5 -> AcCl3 + Cl2}$) happens to consume $x$ moles of $\ce{AsCl5}$. Based on the stoichiometry of the equation, you know that exactly $x$ new moles of the products are formed.
$$K_p=\frac{P_{\ce{AsCl3}}P_{\ce{Cl2}}}{P_{\ce{AsCl5}}}$$
Since for $V$, $R$, $T$ are constant, we can replace the above equation by:
$$K_p=\frac{RT}{V}\frac{n_{\ce{AsCl3}}n_{\ce{Cl2}}}{n_{\ce{AsCl5}}}=\frac{RT}{V}\frac{x^2}{A-x}$$
Solve this for $x$, and substitute in the ideal gas equation to get the answer.

$\pu{0.0125mol}~\ce{AsCl5}$ is the initial amount of $ \ce{AsCl5}$before the system reaches the equilibrium .The actual amount of $\ce{AsCl5}$ at equlibrium equal $\pu{0.001mol}$.

$$K_\mathrm{p}=  K_\mathrm{C}\cdot{(\mathrm{R}}\cdot{\mathrm{T}})^{\Delta{n}}$$


text:
Is honey really a supersaturated solution? Does heating to un-crystalize redissolve it or melt it?
In the SciShow video Honey: Bacteria's Worst Enemy after about 00:30 the narrator says:

Honey is only about 17% water. Most, but not all of what remains is sugar. The two main types of sugar in honey are glucose and fructose. Like all sugars, glucose and fructose are sticky — they attract water.
Honey is technically a supersaturated solution, meaning it contains more sugar than would normally dissolve at that temperature. That’s why it eventually gets all crystally in the pantry — over time, sugar comes out of the solution.

In my experience when old honey "gets all crystally in the pantry" I've placed the jar in a hot water bath, and eventually the honey's viscous liquidy consistency is restored.
When I do this, am I redissolving the sugar, or melting it?
I'm asking because even hot, with only 17% water it's hard to imagine it can become an unsaturated solution.

Honey is indeed a complex mixture containing more than hundred compounds.
As for Wikipedia and depending on the point of view it is

a supersatured liquid solution
a viscous supercooled liquid (in the sense that it can get so viscous as to appear solid, without affecting its status of being a supersatured solution, and undergoes glass transition).

https://en.wikipedia.org/wiki/Honey#Physical_and_chemical_properties
The facts that you describe in the question are less surprising if we consider that:

with respect to sugars crystallization, of which at least  there are two different ones, glucose and fructose, the rest of the non-sugar components must be considered to be impurities
impurities, even in traces, often hamper the crystallisation of a compound, even in simple mixture of a single compound and the above traces. This is common after organic synthesis, in which often a viscous "oil" is attained that might crystallize only upon prolonged storage or a careful removal of the disturbing trace compound(s)
finally and most important, the solubility of sugars in water is very high, and very sensitive to temperature. For instance, at room temperature glucose is already soluble in the reason of 90 g per 100 ml of water, that means a saturated solution already contains about fifty percent weight per weight of sugar

A table is here (I didn't cross check the values):
http://www.mpcfaculty.net/mark_bishop/supersaturated.htm
All this, viscosity included, makes the attainment of a supersatured solution particularly easy, as in the kitchen in the case of sucrose:
https://sciencing.com/make-supersaturated-solution-sugar-6199355.html
As such, heating crystallized honey does indeed dissolve the sugar, and a supersaturated solution is attained upon subsequent cooling.

In the circumstances the other answer is more than valid. What is peculiar is the easy to maintain the super-saturation regime (ie the resistance to crystallisation). @uhoh

You might add the high viscosity to the list. Just about everything is slow in honey.

@Karl I thought the •supercooled entry did the job. Shall I put it more in connection with slow crystallisation?

I think so. Supercooled water is still very much like water, and the viscosity of sugar solutions does not rise differently above the solubility limit.

@Karl yes but water isn't that viscous and does not show glass transition. Anyway I have emphasised the viscosity aspect further. Thx.

Melting and dissolving are all the same when you look at mixtures close to saturation.
You can say water lowers the melting point of the sugar, or that the solubility of sugar increases with temperature. Different description, same fact.
What makes this seem different from e.g. a salt water solution is that the molten (i.e. non-crystalline) sugar is fully miscible with water, which is only possible because the m.p. of sugar is not so far above the Bp. of water. If you go to high pressures (like in earth's mantle), the situation between (supercritical) water and rock is probably very similar.

Is this a general principle in chemistry (for supersaturated solutions), or does it mostly apply to saturated solutions of sugar?

I would call it a non-principle. ;-) If you think about a saltwater solution, you would clearly use the word "solubility". There's a large miscibility gap. Molten sugar and water however are completely miscible, the Bp of one and Mp of the other are closeby.


text:
Why can't a weaker acid protonate the conjugate base of a stronger acid?
So I've heard that, supposedly, a weaker acid cannot protonate the conjugate base of a stronger acid to give that stronger acid. For instance,  adding ethanoic acid to sodium sulphate will not result in sulphuric acid and sodium ethanoate. Is this true? If so, why?

This is not true. Acid strength depends on medium (the same acid can be weak in one solvent and strong in another), and of course you can chemically convert, say, $\ce{H2S}$ to $\ce{H2SO4}$.

I interpret the question to mean that given two hypothetical acids $\ce{HWeak}$ and $\ce{HStrong}$, why does the following reaction not occur:
$$
\ce{HWeak} \ + \ \ce{Strong-} \overset{?}{\rightarrow} \ce{Weak-} \ + \ \ce{HStrong}
$$
?
The answer lies in the definition of "strong and weak" acids. To recap: A Brønsted–Lowry acid is a substance capable of dissociating an $\ce{H+}$ ion, commonly called a proton. A strong acid is thus an acid that does so easily, or equivalently, whose conjugate anion has a very low affinity for protons. A weak acid has a conjugate anion that more readily accepts a proton.
The wording above already shows that acids are in principle always about equilibria:
$$
\ce{HAcid} \rightleftharpoons \ce{H+} \ + \ \ce{Acid-}
$$
meaning that the reaction can occur in either direction. For an acid, the equilibrium lies on the right side of the equation and more strongly so for a strong acid.
This means that the equilibrium
$$
\ce{HWeak} \ + \ \ce{Strong-} \rightleftharpoons \ce{Weak-} \ + \ \ce{HStrong}
$$
tends to the left side, but is of course subject to all factors affecting such an equilibrium. If you would continously remove the strong acid by some means or if you start with a strong acid concentration of zero, some reaction to the right will occur. However, if there is another base (i.e. acceptor of proton such as water), $\ce{HStrong}$ would probably pass the proton to it.
To summarize, the answer is: you can make some strong acid with a weak one, but not in large yields. Typically, the reaction will occur the other way around due to the way strong and weak acids are defined.

Say you have a sodium chloride solution and you bubble in some carbon dioxide.  The latter is only a weak acid, but it does generate some solvated hydrogen ions in solution and there are also chloride ions from the salt.  These are all you need to claim that there is now hydrochloric acid in your solution!  But there's a catch: you can claim only as much hydrochloric acid as the solvated hydrogen ions you generated, which is not a lot.  You're overlooking the true major constituents if you emphasize this point of view.
Here's the moral of the story:  You can make a strong acid from a weak one.  But, the amount of strong acid you can capture at equilibrium will be limited.  The amounts of the various acids you can identify must all be consistent with generating the same concentration of protons, and perforce that corresponds to less strong acid than weak acid.


text:
Why is a very small peak with larger m/z not considered to be the molecular ion?
With regards to the following spectrum:

When asked for the M/Z for the molecular ion, the peak at 58 was taken, not 59.
What causes the peak at 59, and why isn't it taken as the molecular ion?

The gist here is that nobody can just "know" what the various peaks are. Rather you look at the spectrum and start to look for clues to put together a consistent explanation. So you have a peak at 58 and 43, the difference is 15 where you also have a large peak. 15 is the mass of a methyl group so... and on the analysis goes.

Without knowing more details, it is hard to guess, but at this $m/z$, it seems likely that the peak is the result of one $^{12}\ce{C}$ being substituted by one $^{13}\ce{C}$. It is more useful to assume a uniform mass of 12 for carbon when analyzing such a spectrum.
Note that with more carbons in a larger molecule, it becomes more likely that at least one of them is a $^{13}\ce{C}$. Thus, the spectra become more complicated that way.
This whole analysis goes out the window if this is not an organic molecule, though similar patterns may arise for other molecules - it depends on the isotopes and their distribution.

It looks like the mass spectrum of butane, so I think it's quite a safe assumption that it's organic ;)

The peak at $m/z = 59$ with lower intensity in respect to the one at $m / z =58$ (the molecular ion) is not overseen.  Mass spectroscopy is capable to deliver information about the isotopic composition of your sample, too.  Since -- accounting over all carbon atoms of your sample -- about 1.1% are of the non-radioactive isotope of $\ce{^{13}C}$, you may use this information to determine the carbon atoms in total present.
The story indeed contiues further, as to establish by this "isotope fingerprint" if you have one or two chlorines per molecule present ($\ce{^{35}Cl}$ and $\ce{^{37}Cl}$), bromines, presence / absence of nitrogen, etc.:
(source)


text:
Why is fluorine a oxidising agent?
An oxidizing agent pulls the electron cloud of the substance being oxidized towards itself, for example:
$$\ce{F2 + 2 X- -> 2F- + X2}\qquad (\ce{X} = \ce{Cl}, \ce{Br}, \ce{I})$$
We also know that fluorine has very high charge/mass ratio as it is very small in size and its electron gain enthalpy is know to be very high, i.e gain of electron is favourable thermodynamically.
Also, many sources state that:

Fluorine has low bond dissociation enthalpy and high hydration enthalpy due to which fluorine is a good oxidizing agent.
Fluorine is achieving octet, so electron gain is favorable.
I found another source stating that bond dissociation enthalpy is the major driving factor.

My questions(along with reasons why fluorine shouldn't be oxidising agent)

What has hydration enthalpy and bond dissociation enthalpy got to do with oxidizing power?
(As we know that lithium is the best reducing agent, and it also has high hydration enthalpy but we can’t say it is an oxidising agent.)
What am I missing here? What is the reason for such high standard electrode potential for fluorine?
Why wouldn't the electron cloud of fluorine atom repel the electron cloud of substance being oxidized? It seems paradoxical that the fluorine pulls electron cloud towards itself so easily.

Even in the post Why does chlorine have a higher electron affinity than fluorine? it is mentioned that electron cloud is repelled.
Is there any better explanation other than "octet is being achieved", such that it resolves the repulsion issue, to explain why fluorine is a good oxidising agent?

look up Born-Haber cycle, hydration energy effects will make sense.

Maybe write it this way to gain some insight:
$$\ce{F-F(g) + 2Cl-(aq) -> 2 F-(aq) + Cl-Cl(g)}$$

If we consider
$\ce{F2(g) -> 2 F(g)}$ with $\Delta H_\mathrm{F, bond}$
$\ce{2 F(g) + 2 e- -> 2 F-(g)}$ with $\Delta H_\mathrm{F, aff}$
$\ce{2 F-(g)  ->[H2O] 2 F-(aq)}$ with $\Delta H_\mathrm{F, hydr}$
and respectively
$\ce{Cl2(g) -> 2 Cl(g)}$ with $\Delta H_\mathrm{Cl, bond}$
$\ce{2 Cl(g) + 2 e- -> 2 Cl-(g)}$ with $\Delta H_\mathrm{Cl, aff}$
$\ce{2 Cl-(g)  ->[H2O] Cl-(aq)}$ with $\Delta H_\mathrm{Cl, hydr}$
Then overall reaction enthalpy for $\ce{ F2(g) + 2 Cl-(aq) -> Cl2(g) + 2 F-(aq)}$ is
$$\Delta H_\mathrm{r} = \left({\Delta H_\mathrm{F, bond} - \Delta H_\mathrm{Cl, bond}}\right) + \left({\Delta H_\mathrm{F, aff} - \Delta H_\mathrm{Cl, aff}}\right) + \left({\Delta H_\mathrm{F, hydr} - \Delta H_\mathrm{Cl, hydr}}\right)$$
The differences between fluorine and chlorine enthalpies of atomization(bond breaking ), electron affinity and ion hydration lead to the overall reaction enthalpy, that is pushing reaction toward chloride oxidation.
The truth is, that the equilibrium is driven by Gibbs energy, not enthalpy,
$$\Delta G^{\circ}_\mathrm{r}=\Delta H^{\circ}_\mathrm{r} - T \cdot \Delta S^{\circ}_\mathrm{r}=-RT\ln{K}$$
but for the rough reasoning you get the picture. Futhermore, the entropy change will not be major factor, due analogical structures.

Electron Affinity
Using Slater's rules we can calculate the effective Charge that an electron added to $\ce{F*}$ would encounter. $$Z_{eff} = 9 - (0.35 \cdot 7 + 0.85 \cdot 2) = 2.4$$
This is higher than the effective charge of any other element before fluorine, which makes it more favorable to have an electron added to fluorine, rather than leaving it where it was.
For any element after fluorine, the lowest-energy orbital available is still a lot higher in energy than the 2p-orbital in flourine, because of the higher quantum number and the greater distance to the core. Which again makes it more favorable to add an electron to fluorine rather than leaving it in another atom. (Disregarding noble gases here)
Size
But wait, there is the effect of flourines small size. And it is true that electron-clouds do repel each other through Coulomb-forces and, when very close, through Pauli-repulsion. And this effect does lead to the situation of fluorine having a slightly less (but similar and still impressive) negative electron affinity than chlorine.
So we can conclude that if we had a single atom each of fluorine and chlorine and a spare electron, it would rather be added to the chlorine atom.
Dissociation energy
But if we talk about oxidising agents, we do not talk about single atoms. Fluorine is naturally a molecule and as you mentioned, its dissociation energy is extraordinarily low for a single covalent bond. This anomaly is another result of its small size and the resulting repulsion of the electron clouds of the two fluorine atoms. (Actually, it has to do with the overlap of fluorines orbitals, which is more complicated but still a result of size)
For an oxidation to take place, the molecular fluorine has to undergo homolytic dissociation.
$$\ce{F_2 -> 2F*}$$
This means that the dissociation energy needs to be provided for the reaction to take place. With a lower dissociation energy, the kinetic barrier is lower as well as the process being thermodynamically more favorable.
Conclusion
If there were no anomalistic effects resulting from its small size, fluorine would be the best elemental oxidising agent, resulting from its position in the periodic table. However there are effects. But luckily for fluorine the effect of repulsion counters the effect of the lower dissociation energy, which leads to it still being the best elemental oxidising agent.
The hydration enthalpy is relevant in liquids only, which is not necessarily the case for oxidations, so it should not be regarded when considering fluorines oxidising potential.

Welcome to Chemistry.se! This does unfortunately not answer the main question of the post. It also reads more like speculation and conjecture. It would be nice, if you back up your claims with references.


text:
Comparing acidity in 3,4-dihydroxyfuran-2(5H)-one and 2,3-dihydroxycyclopent-2-en-1-one
I have two compounds with two hydrogens namely (a),(b),(c) and (d). I have to compare the acidity of the following hydrogens.

My approach:
I removed the hydrogen atoms one by one and compared the stabilities, as there are most resonating structures for (a) it will be most stable (most acidic too). Then followed by (b), (c) and (d).
But the answer is given different without any further explanation. Does anybody have the right approach to tackle these type of questions?

b and d should be many orders of magnitude weaker then a and c BTW both compounds are probably as acidic as typical carboxylic acid.

https://en.wikipedia.org/wiki/Chemistry_of_ascorbic_acid#Acidity

Order is different from what you're saying.

c must be more acidic than a, for sure. does that go well with your textbook's answers?

And you should get better textbook.

@Mithoron Can you provide a complete answer? I'm not expecting commenting and taunting.

I think you really should, even from this photo alone it doesn't look good. As far as "full answer" goes I'd need to find actual, experimental data. You'd be surprised how big difference can make stereoelectronic effects, which can be hardly seen from simple picture.

This is a question which was asked in a test to compare the acidity of hydrogens (order not values) among these compounds the information is restricted to this and no further data was given.


text:
Naming this organic compound
What’s the name of the following compound?
$$\ce{CH3C(p-ClC6H4)2CH(Br)CH3}$$
My teacher said that she strongly thinks( with another teacher) that it should be 3-bromo-2,2-bis-(4-chlorophenyl)butane, though many other books and other students (similar to me) stated that it should be 2-bromo-3,3-bis-(4-chlorophenyl)butane.
She asked us try to find it out. Any organic chemistry experts out there?

I think your teacher is correct, numbering of carbons is done acording to the lowest locant rule.

The red numbering gives you : 2, 3, 3
The green numbering gives you : 2, 2, 3
We see that the first point of difference between the two numberings occurs at second number, so the green numbering is preferred.
Consequently, the correct name should be : 3-Bromo-2,2-bis(4-chlorophenyl)butane


text:
Does the reaction become more or less favorable as temperature increases?
Suppose a reaction has $\Delta S > 0$ and $\Delta H < 0$.
The equation $\Delta G = \Delta H - T\Delta S$ predicts that as the temperature increases, the reaction becomes more favorable because the $T\Delta S$ term becomes larger and $\Delta G$ becomes more negative.
However, the equation $$\ln(\frac{K_2}{K_1}) = -\frac{\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})$$ as well as Le Châtelier's principle, predict that as the temperature increases, K will decrease (and the reaction shift to the left, thereby becoming less favorable and $\Delta G$ becomes more positive).
These two claims seem to be inconsistent, but I can't identify the error in my reasoning.

https://en.wikipedia.org/wiki/Van_%27t_Hoff_equation
may be helpful.

You should use the van't Hoff equation together with the standard enthalpy of reaction $\Delta H^\circ$ for the process to predict the response to a temperature change. This should provide you with a prediction consistent with Le Châtelier's principle that an endothermic reaction is pushed in the direction of products as $T$ increases.
The general equation for the Gibbs' free energy $$\Delta G = \Delta H - T \Delta S$$ should not be relied upon to determine the response of the reaction equilibrium constant to $T$. The expression is valid at constant $T$, but this is not the problem with using this equation, rather it is its generality: it is not limited to describing reactions at equilibrium. Note that $\Delta G = 0$ for a reaction at equilibrium.
On the other hand, you could consider using $$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$$to guide you, and the result should be consistent with analysis based on the van't Hoff equation. It is simpler however to work with the van't Hoff equation directly.  In fact, assuming $\Delta H^\circ$ and $\Delta S^\circ$ are independent of $T$ you can use fairly straightforward mathematical manipulations to arrive at the usual integrated form of the van't Hoff equation.

Why shouldn't the equation for the Gibb's free energy be used to determine the response to T?

@DrPepper Well, I give one reason in the answer: you can use $\Delta G^\circ$, but for a reaction at equilibrium  $\Delta G = 0$ which is not very useful.

If you were given delta H and delta S, and wanted to calculate delta G at a specific temperature, then you would use delta G = delta H - T * delta S. Doesn't this determine the equation's response to temperature? Why is it incorrect?

@DrPepper Well, $\frac{\partial \Delta G}{\partial T} = - \Delta S$, so no....

I don't understand. Can you please clarify?

$dG = dH - TdS -SdT$. The equation $dG = dH - TdS$ holds when T is constant.

The equation $\Delta G = \Delta H - T\Delta S$ predicts that as the temperature increases, the reaction becomes more favorable because the $T\Delta S$ term becomes larger and $\Delta G$ becomes more negative.

It does predict that $\Delta G$ becomes more negative (for example for the standard state). This also means that the maximal work available from the reaction becomes larger. If you equate favorability with $\Delta G$ or maximum work, you could also say the reaction becomes more favorable. It does not mean, however, that the equilibrium constant increases. This is because the equilibrium constant depends on both $\Delta G$ and the temperature. Both terms change because we are changing the temperature and $\Delta G$ is temperature-dependent:
$$ \mathrm{ln}(K) = - \frac{\Delta G^\circ}{RT}$$
It is fine to equate more negative $\Delta G^\circ$ with larger equilibrium constant when comparing two processes at the same temperature, though.

However, the equation $$\ln(\frac{K_2}{K_1}) = -\frac{\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})$$ as well as Le Châtelier's principle, predict that as the temperature increases, K will decrease (and the reaction shift to the left, thereby becoming less favorable and $\Delta G$ becomes more positive).

This is all true except for the last bit about $\Delta G$ becoming more positive. $\Delta G$ at equilibrium concentrations will be zero at both temperatures, and $\Delta G^\circ$ will become more negative, as you correctly stated in your question (assuming we can neglect the temperature dependence of enthalpy and entropy of reaction). Again, the misconception is linking changes in $\Delta G$ to changes of K at two different temperatures.


text:
What will be the sign of work done?
Consider the following question:

The enthalpy of vapour is $10\ \mathrm{kcal}/\mathrm{mol}$ higher
than liquid at $350\ \mathrm{K}$ and $2\ \mathrm{atm}$. Under given
conditions, the difference in internal energy of vapour and liquid
will be:
(a) $10700\ \mathrm{cal}$
(b) $-9300\ \mathrm{cal}$
(c) $9300\ \mathrm{cal}$
(d) $-10700\ \mathrm{cal}$

My solution is as follows:
Let the internal energy of the liquid be $U_\mathrm L$ and that of vapor be $U_\mathrm G$. We are required to calculate $U_\mathrm G - U_\mathrm L$. As per the question,
$\because H_\mathrm G = U_\mathrm G + W$, where $H_\mathrm G$ is enthalpy of vapor.
$10000 + U_\mathrm L = U_\mathrm G + W$
Where $W$ is the work done due to the expansion of vapor as the liquid changes state. The magnitude of work done $|W| = \Delta n_\mathrm gRT = 1 \times 2 \times 350 = 700\ \mathrm{cal}$.
Now the real question starts, what is the sign of $W$? As per the IUPAC convention, work of expansion is negative, hence $W = -700\ \mathrm{cal}$. So our answer will come out to be,
$10000 + U_\mathrm L = U_\mathrm G - 700$
$\implies 10000 + 700 = U_\mathrm G - U_\mathrm L$
$\therefore U_\mathrm G - U_\mathrm L = 10700\ \mathrm{cal}$
But the terse solution given to this question is:

As per the solution $W = +700\ \mathrm{cal}$, i.e. they have considered the work of expansion to be positive.
Where are we wrong? Or are we right?

Analyzing the problem by considering the work done is not the easiest place to start solving this problem. It's easier to consider the definition of enthalpy in terms of internal energy and $pV$:
$$\Delta H \equiv  \Delta U  + \Delta(pV)$$
It follows that
$$ \Delta U = \Delta H - \Delta(pV)$$
Taking $p$ as constant and equal to $\pu{2 atm}$
$$ \Delta U = \Delta H - p(V_g-V_l)$$
Assuming $V_l$ to be much smaller than $V_g$,
$$ \Delta U = \Delta H - pV_g$$
You can equate $-pV_g$ with the work $W$ required to expand the gas. This is consistent with the usual convention in chemistry: work of expansion is negative (as you write). The correct result follows.

Although the IUPAC convention is for work done by a system to be negative for expansion, many physicists and engineers use the opposite convention. What's important is that you use the same convention throughout your calculation. You wrote that H = U + W, which is true only if work done by the system is regarded as positive. To be consistent with the IUPAC convention that you used later in the calculation, this should be instead H=U - W.
On a side note, since work represents a change in energy, not a state function, this equation should be written in terms of changes: $\Delta H = \Delta U - w$. Also note that this equation is only valid for the condition of constant pressure.


text:
Absorbance maxima of polyenes

Can anyone tell me how to approach this question from USNCO 2013?

The question boils down to: which molecule has the longest conjugated $\pi$-system? The answer is (A). For such a system, the particle-in-a-box model is a decent approximation and one can show that larger systems absorb at lower energies, i.e. longer wave-lengths.


text:
Diffusion across a thin film

Given this concentration profile, I am to (1) determine at which side substance A will diffuse out of the membrane and (2) chose one of the following profiles below which represents the concentration of a substance A at steady-state.

(1) Diffusion occurs from an area of high concentration to an area of low concentration. Therefore, substance A would diffuse from the left to the right. However, in the question when they ask "at which side will the substance diffuse out of the membrane", should I answer at the right side or that the substance stays within the membrane?
(2) I know that the concentration profile at steady-state should be linear, however I'm wondering what the difference between (A) and (B) is. I assume that (A) is the correct answer to this question since substance A has a higher concentration than the concentrations at the membrane wall and hence the slope should start a little above the concentration at the wall. Is this correct or does steady-state affect the concentration profile of the substance?

Is the membrane homogenous? Does the membrane have preferential interactions with the solute compared to the solute:solvent interactions? If the answer is no to both, how about having the solute diffuse out on both sides until a steady state has been reached?

@A.K. - Looked at your answer which is much better so I deleted mine.

@KarstenTheis - I didn't remember that diffusion according to Fick's law was considered to be a steady state. Since the concentration on both sides of the membrane will be changing until equilibrium, as well as  the rate of diffusion across the membrane, I was thinking of that as more of a quasi-steady state. // My imagination got the better of me and I was wondering about something like an ion-pumping biological membrane to get the weird hyperbolic profile.

If it is a real system, it probably is quasi-steady state. I don't have a good example of a real steady state system one could actually accomplish, but you can come close by making a membrane with a small cross-section and well-stirred very large reservoirs on both sides.

(1) determine at which side substance A will diffuse out of the membrane

should I answer at the right side or that the substance stays within the membrane?

Neither, because the concentration is higher in the membrane than either side, the substance will flow out of both sides initially at least. After a while the material will flow from left to right (high concentration to low concentration) and converge towards steady state.

(2) chose one of the following profiles below which represents the concentration of a substance A at steady-state.

First thing - When doing diffusion problems immediately refer to Fick's laws of diffusion.
Fick's first law assumes a steady state system and is given as:
$$J = -c D\frac{dx}{dz}\quad \text{or more simply} \quad J = -D\frac{\Delta C(x)}{\Delta z} \quad \text{or more generally}\quad J = -cD\ \nabla x$$
Where $J$ is flux (mols through an cross-sectional area per time), $c$ is total concentration, $C(x)$ is the concentration at point $x$, $D$ is the diffusion constant, $x$ is the mole fraction and $z$ is position.
For steady state to occur, $J$ must by definition be constant through the entire membrane and thus $-D\frac{\Delta C(x)}{\Delta z}$ must be constant through the entire interface. Since $D$ is already a fixed constant, we can assume $\frac{\Delta C(x)}{\Delta z}$ must also be constant. Thus the plot with a constant slope is the steady state profile which is B.
Q.E.D.

I agree with your analysis and choosing (B), however, the question suggests that diffusion only occurs from one side of the membrane so I think that the question is erroneous as none of the figures will fit this. Also no prohibition mentioned for material diffusing into the membrane.

We aren't told whether the thin film is the same phase (e.g. a porous membrane) or a different phase than the bulk phases L and R.  Without that information, it is hard to know if (a) or (b) is the correct answer, isn't it?  E.g. if the film was a different phase with a different solubility / activity coefficient for the solute, wouldn't (a) be possible?

@CurtF. You are correct that the underlying assumption of no difference in chemical potential is not trivial. I actually considered this before tendering an answer and it should be remembered that diffusion is ultimately driven by chemical potential not concentration. Part of why I chose to omit it is that the question itself seems a bit simplistic and thus chemical potential/activities would be too far. In answering a question of MaxW's made me realize that what we call diffusion across a membrane is actually effusion a physical process.


text:
relaxed dihedral PES scan: symmetry of the system not reflected
Using G16 and performing a relaxed PES scan on a dihedral angle, I performed 72 steps where one phenyl group of a symmetric terphenyl is rotated about the C(aryl)-C(phenyl) bond by 5° in every step. I'm using the BP86 functional with the 6-31G basis set. I observe that -180° to 0° will not be the same as 0° to +180°.
In my opinion, rotation of a phenyl group (with a mirror plane which would be perpendicular to the C6 axis in benzene) should result in energies at x° being the exact same as for x°+180°. In the first graphics below, you can see one whole rotation (black dots, scan points 1 to 73) and the scan points 38 to 73 shifted by +180° (red dots). You can see that there is a shift on the abscissa, meaning that the first 180° of rotation about the aryl-phenyl bond are not the same as the second 180°. Interestingly, when rotating more than 360° (3-5 times 360°), the dots will be at the exact same position as 360° earlier - so x° and x°+360° will be the same in every case, while x° and x°+180° are not. Don't think that this is an issue related to convergence?
edit: @TAR86 suggested to have a look at the nuclear repulsion energies to check whether this is an issue related to the geometries produced. The respective image is added below.
edit2: @Martin - マーチン♦ requested an G16 input file to reproduce the problem. I've added the respective file below.

%chk=job.chk
%mem=8gb
%NProcShared=16
#p opt=(modredundant) BP86/6-31G geom=(gic)

dihedral pes scan

0 1
6        5.226229852      2.461878014      0.364189364
6        6.517648681      1.947188819      0.628146249
1        7.127277746      2.400165503      1.416350388
6        7.024836811      0.874189127     -0.119447737
6        6.252002950      0.287970075     -1.134646502
1        6.649922190     -0.550963401     -1.716497030
6        4.963870485      0.783187734     -1.399800460
6        4.455606763      1.856498426     -0.656174766
1        3.460115638      2.254235558     -0.885310716
1        8.028604788      0.493012088      0.098455632
1        4.355515157      0.336549951     -2.194552005
6        5.437910365      4.695855989      1.613350721
6        4.667674122      3.560511364      1.202847170
6        3.339228627      3.414455715      1.655493257
1        2.762287260      2.537249136      1.343693331
6        2.778852196      4.366478962      2.509604334
1        1.728615174      4.291477665      2.812827707
6        3.587531970      5.372572404      3.043834082
1        3.154198853      6.017546127      3.811431515
6        4.962035000      5.490387000      2.719069000
6        5.793408000      6.483337000      3.479780000
6        7.155794401      6.716767843      3.150348234
1        7.690505145      6.040409950      2.477397785
6        7.891233889      7.760301086      3.724625384
6        7.281499938      8.654025495      4.618188277
1        7.854319283      9.475639651      5.062586031
6        5.919284422      8.491819231      4.906055341
6        5.183645944      7.448680994      4.325851576
1        4.106339318      7.414359241      4.527417701
1        8.948095638      7.878562451      3.459649110
1        5.404779334      9.198550838      5.566858766
15       6.836613173      5.254444911      0.492426546
6        6.402468333      4.749060218     -1.232735776
1        5.331395678      4.884562656     -1.457090749
1        6.692781337      3.708789948     -1.428936030
1        6.992612351      5.408972053     -1.892849893
6        6.570730658      7.079097928      0.321764243
1        5.553588175      7.291894513     -0.050476225
1        7.307098080      7.438494888     -0.417840357
1        6.732539811      7.615443003      1.266231462
8        8.252395391      4.848887467      0.884390392

interplanar=Dihedral(22,21,20,12)
interplanar(NSteps=72, StepSize=5.00000)

Could be a guess issue or something about the geometries. To detect the latter, examine the same plot based on nuclear repulsion energies.

@TAR86: I've added a plot showing the respective nuclear repulsion energies. The geometries are of x°+180° are not the exact same - which I don't understand.

Relaxed PES scan should mean that some geometry optimization takes place before the evaluation of the final, plotted energy, right? There's all sorts of issues that can arise during that optimization and small differences in the input can be amplified.

This could very well be an issue due to the internal definition of the redundant coordinates; I'm not sure how Gaussian constructs them and if they are truly equivalent at these points. Could you add the input file, which you use to generate the energies, so that one could try to reproduce the problem.

I've added the G16 input file above.


text:
Volume of dried gas at STP

A $54.0\ \mathrm{mL}$ sample of oxygen is collected over water at $20\ \mathrm{^\circ C}$ and $770\ \mathrm{Torr}$ pressure. What is the volume of the dry gas at STP?

By using the combined gas law, I came up with the answer $\pu{51.2 mL}$.
$770 - 21.1 = 748.9\ \mathrm{Torr}$ pressure for oxygen
$21.1$ = vapor pressure of water at $23\ \mathrm{^\circ C}$
$\frac{P_1V_1}{ T_1 }= \frac{P_2V_2 }{ T_2}$
$\frac{748.9V_1}{273 }= \frac{770\times54}{296}$
$V_1 = \pu{51.2 mL}$
Is this how it's supposed to be done?

No. What are the conditions for STP?

@MaxW ohh I see now. STP would be 760 torr and 273 K so P1 should be 760 and P2 should be 748.9. Would that be right?

That is correct. However I'll point out that STP is a bit different now than 760 mm Hg and O C. It is now about 750 mm Hg at 0 C.

So $$\frac{P_{STP}V_{STP}}{T_{STP}} = \frac{PV}{T}$$

I like your comment about not closing it MaxW, but my own comment is why don't you put your answer in the answer section :P

@user75030 see this post for details on how the definition of STP has changed.

@CharlieCrown - I wasn't trying to answer the question per se but rather I was trying to lead the OP to the answer.


text:
Is there any sort of correlation between melting point and the degrees of freedom for the molecules of a substance?
Suppose I had a solid whose molecules did not have rotational freedom. But if I were able to make it so that the molecules had rotational freedom ceteris paribus, is there a way to tell how the melting point would change compared to the original solid. Here I assume that the molecules have rotational freedom in the liquid phase always.

It should increase as for entropy of the melting should diminish. Assuming nothing else change in the forces leading to the solid, which seems magic.


text:
Why do reactions happen at all?
What makes some atoms/compounds reactive and some noble? What is the fundamental truth behind a reaction? It seems that two species can undergo number of combination to give a compound. Say Sodium and Chlorine, it seems intutive to have number of combinations from this pair such as $\ce{Na2Cl}$, $\ce{NaCl2}$, $\ce{Na3Cl2}$, and even do not react at all. But why $\ce{NaCl}$?
I had been carrying the concept of octet rule, which I now find to be exception in some cases like $\ce{AlCl3}$. So, what determines the products of a reaction? Or is it that all possible combination are formed but only few are major products, in this case $\ce{Na2Cl}$, $\ce{NaCl2}$ etc. are formed but $\ce{NaCl}$ is major product.
I think it has to do something with energy but cannot figure out what.

entropy likes chaos

There is a natural tendency for a system to come to rest at the lowest-energy state.

Intuitive? Chemistry depends on observation. Specific to the general, not the other way around. Remember Galileo and the Tower of Pisa!

What makes some atoms/compounds reactive and some noble?

Unpaired electrons and how tightly these electrons are bound to their respective nuclei.

Why do reactions happen at all? What is the fundamental truth behind a reaction?

Because atoms are more stable together than they are apart, and systems will tend towards their lowest energy state.

But why NaCl?

This is really a matter of ratios, which is the basis of chemistry. Different elements share their electrons in different ways under different circumstances. Some can donate their electrons ($\ce{Na}$, $\ce{Al}$, $\ce{Au}$), some accept electrons ($\ce{Cl}$, $\ce{O}$, $\ce{N}$), and others are somewhere in between ($\ce{C}$, etc.). Sodium and chlorine are very consistent with how they share electrons. Sodium has 1 valence electron, chlorine lacks 1 valence electron, and they combine in a one-to-one ratio. The formula unit of their salt is $\ce{NaCl}$.
When considering the salt of aluminum and chlorine, ask how many valence electrons does each atom have/need. I think you might begin to see the pattern.

$[1]$ Bond energy diagram


text:
Why does liquid water form when we exhale on a mirror?
In the descriptions below, I always assume external pressure to be constant at 1 atm, the condition where daily observations are made.
1) When I exhale on a mirror, liquid water forms on the mirror. That's condensation. Obviously, the temperature of mirror must be < 100 °C, so water vapor condensing on mirror makes sense.  However, in that case why do we have water vapor in our breath when our body temperature is also < 100 °C in the first place?
2) One reason for (1) may be like 'water vapors in air is in equilibrium with liquid water', so some water vapors can exist although T < 100 °C. If this is correct, based on this reasoning, then why doesn't ice exist at T > 0 °C? [p = 1 atm only]
3) If liquid water can evaporate into gas at T < 100 °C, then why doesn't ice turn into liquid at T < 0 °C? (I never use the term 'melt' here, just like evaporate≠boil) (both have hydrogen bonds, intermolecular forces should be the same?)
4) When I exhale on a wall, no water droplets form. Why do water droplets form on mirror but not on wall (just an example)? The mirror must have the same temperature as that of the wall, both of them must have achieved thermal equilibrium with their surroundings a long time ago.

Posting as a comment because I have no references to back this up, perhaps someone else knows more. I have understood the difference in gas:liquid vs. liquid:solid balances to be caused by the molecules being more separated in the gas phase. If a molecule of water has enough kinetic energy to escape from ice crystal but remains bound as liquid, it is so close to the ice that it would settle back quickly. But if it has enough energy to go flying as gas molecule, it takes time until it settles back somewhere, and the equilibrium between escaping and settling back determines the concentrations.

"If liquid water can evaporate into gas at T < 100 °C, then why doesn't ice turn into liquid at T < 0 °C?" is probably worthy as a question of its own (if done well).

If your look at the pressure melting point Ice can turn into liquid at T < 0 °C by raising the pressure.

@jpa your answer to part (3) makes sense to me.
It would be good if someone knows more can actually confirm it.

@A.K.  If I refer to the phase diagram, it shows that at 100kPa and 0°C<T< 100°C water should exist as pure liquid, yet commonly said that pure liquid has vapor pressure i.e be in equilibrium with its vapor in a closed container.Some water exists as vapor. This makes me wonder what does the word 'liquid' actually means in the phase diagram. Label it 'liquid' despite there being both 'liquid' and 'vapor'.

When considering pure water (no air present), "has vapor pressure" doesn't mean that liquid and vapor are present in equilibrium. Liquid will only transform into vapor when the vapor pressure exceeds ambient pressure: e.g. at 100C, vapor pressure reaches 1 atm, and water boils. I find it helpful to think of vapor pressure as describing how much the water "wants to" be in vapor form, but ambient pressure describes how strong the forces are preventing that from happening.

@GeoffreyBrent Huh? I thought they will still be some water vapor in a closed container at <100C normal pressure without air?

@The99sLearner No. There is vapor pressure but all the water is in liquid phase.

@GeoffreyBrent so what cause the pressure? air?

@The99sLearner don't think of "vapor pressure" as "pressure created by vapor". Think of it as "the strength of the water that is attempting to become vapor". If that strength is powerful enough to overcome ambient pressure, then water will become vapor. If it isn't, then the water remains liquid, but the vapor pressure still exists, and it's useful for quantifying how close the water is to boiling. (Again, I'm talking specifically about an airless scenario here.)

Why do we have water vapors when our body temperature is also <100°C in the first place?

At normal pressure, water boils at 100°C, meaning that bubbles of pure steam form under water. At lower temperatures, water molecules reversibly move from the liquid to the gas phase and back. The higher the temperature, the higher the vapor pressure, and the higher the equilibrium concentration (partial pressure) of water in air. On the geological scale, there is no equilibrium, and we experience different temperatures, different humidity (related to partial pressure of water in air), and different pressures depending on location.

Why does ice not exist at T>0°C

Ice, as a pure solid, and water, as a pure liquid, have defined concentration at given pressure and temperature (the equilibrium constant expression for melting includes neither liquid nor solid water, it is simply 1). Above the melting temperature, all the ice melts, there is no equilibrium. This is different from the liquid:gas equilibrium, which exists at temperatures below the boiling point, with lower and lower concentrations of the gas as the temperature drops (here, the equilibrium constant expression included the concentration or partial pressure of water vapor).

If liquid water can evaporate into gas at T<100°C, then why not ice turns into liquid at T<0°C?

Again, this has to do with pure liquids and pure solids having a constant (or nearly constant) concentration. If you add salt to the liquid, however, ice will turn into liquid below the freezing point (effectively lowering the concentration of water in the liquid). Also, the surface of the ice melts at lower temperature than the bulk, so even for pure water, there can be liquid at temperatures below the bulk freezing point.

When I exhale on a wall, no water droplets form.

I would do an experiment and check. I think the water droplets are easier to see on a mirror. Try a grand piano (i.e a smooth surface painted with shiny black paint), you might see the water droplets there as well. Or take a long hot shower and check whether water droplets form on surfaces other than a mirror.

It should be noted that ice does turn into a gas at T<0°C readily enough.

w.r.t the last paragraph: how about a tiled wall? With normal fairly shiny tiles in a cool room the fogging should be clear, especially if you arrange a reflection of a light

@ChrisH I tried tiled wall before, it feels slippery after exhalation, but at first I actually tried concrete wall which I saw and felt no presence of water.

@The99sLearner I suspect concrete soaks up the water more than many other surfaces. On a porous surface, and you'll be unlikely to feel the moisture easily.

As @Roland says, plus the macroscopic surface roughness of concrete makes both setting and front the condensation much harder than on smooth tiles. I've just tried on a painted wall (matte paint over plaster) and I can feel a difference quite easily where breath moisture has condensed. Seeing it is also possible at some angles.

@The99sLearner Try a sideways raking light on the surface. This may make it easier to see the slight change in texture caused by the condensation. Though on coarse concrete this still may be hard to detect.

Water and vapour is not usually in equilibrium with liquid water under normal conditions. Condensation happens because the local situation is far from equilibrium: a cold mirror causes a local condensation because the temperature near the mirror is below the temperature of the surrounding air and the water vapour concentration in breath is higher than the concentration in the surrounding air. As the mirror warms or the surrounding air absorbs the local excess the condensation evaporates.

@matt_black: I edited accordingly to say it is a reversible reaction rather than liquid water and water vapor are at equilibrium. Condensation occurs because the partial pressure of water in our breath is higher than the vapor pressure of water at the temperature of the mirror. Once the air near the mirror mixes with the dry air in the room, the water will evaporate again. If the air is humid like in a bathroom after someone took a hot shower, the water will not evaporate.

@Luaan How? Because I never observe my ice cubes in refrigerator suddenly turn into gas. Are you saying solid has vapor pressure?

@KarstenTheis 'so even for pure water, there can be liquid at temperatures below the bulk freezing point',so actually some ice does turn into liquid at T<0°C?

@The99sLearner Yes, solids do have vapour pressure, but you're not going to see ice sublimating in your freezer. Water ice only sublimates at low or high pressures (though I'm not entirely sure if water ice close to freezing point can also be considered as sublimating - it's usually considered to go through the liquid phase IIRC). Dry ice readily sublimates at standard pressure, though. Water ice sublimates quite easily under sunlight too.

@Luaan Ah I see, because I am considering p=1atm only

The major difference is that in vapor phase, the water can leave the container (so you can tell it's in a different form), and which prevents it from immediately rejoining the liquid.
In both solid/liquid phases, the water remains in the rest of the bulk.  A few molecules of liquid in a solid or a few molecules of a solid in a liquid won't be detectable, and any created will almost immediately return to the thermodynamically favored state.

why do we have water vapors when our body temperature is also <100°C in the first place?

Water exists in different forms in equilibrium.  At body temperature, the equilibrium is for vapor until the vapor pressure exceeds about 7kPa.  So it's expected for there to be a certain quantity of vapor present.

why not ice exists at T>0°C? [p=1atm only]

Let's imagine that there is an equilibrium between liquid-vapor and between liquid-solid.
When a bit of vapor forms, the vapor can move away from the bulk liquid.  There it can be detected on its own (or mixed with other atmospheric components).
If a few molecules were to form bonds similar to that of ice within the water, the same doesn't happen.  They remain within the bulk and wouldn't be detectable.  Ice is a bulk phenomenon and doesn't really behave like familiar ice until you've got a lot of molecules bonded together in one place, and that's not favored at room temperature.

If liquid water can evaporate into gas at T<100°C, then why not ice turns into liquid at T<0°C

It does.  There is an equilibrium between the liquid and the solid forms.  Especially near the freezing point, there will be some molecules that leave the bonding and are not part of the ice crystal.  But because the equilibrium points to the solid form, they will be only a tiny part of the mass and will not be seen.
In fact, there will still be some vapor as well.  Even below 0°C, the vapor pressure is still positive.

Why water droplets form on mirror but not on wall(just an example)?

The finish of the glass makes a collection of even tiny drops visible.  The appearance of the mirror/glass normally is dominated by light from specular reflection.  Condensation droplets scatter the light, creating more diffuse reflections.  A normal wall will already show diffuse reflection, so adding a little bit of condensation doesn't change that.  You have to actually see the drops to notice them, and that only happens when the drops get a bit bigger.

If I look at phase diagram of water, under that conditions we actually have 'liquid', while I understand there should be vapor pressure in equilibrium, details from phase diagram contradicts with the fact. What is the meaning of 'liquid' in the phase diagram then?

In that region of the diagram, it is thermodynamically favorable for the liquid phase to exist in bulk.  Outside of that region, even with sufficient amounts of water, any bulk liquid will tend to transition to vapor or solid.

can my ice actually disappear completely at T<0°C, just like liquid water evaporates completely leaving empty bucket?

Absolutely.  Leave an ice cube in your freezer alone and it will sublimate away.  The vapor pressure is still positive, so some amount of it turns to vapor.  That vapor is replaced by dry air from the condenser coils and the ice cube shrinks over time.

*Numbers refer to parts of question in this post. My further questions are based on your answer.

'At body temperature, the equilibrium is for vapor until the vapor pressure exceeds about 7kPa.' If I look at phase diagram of water, under that conditions we actually have 'liquid', while I understand there should be vapor pressure in equilibrium, details from phase diagram contradicts with the fact. What is the meaning of 'liquid' in the phase diagram then?
Answer by BowlOfRed

'Especially near the freezing point, there will be some molecules that leave the bonding and are not part of the ice crystal.' Assuming this is true, if I somehow manage to remove that part of water which is 'not part of the ice crystal',can my ice actually disappear completely at T<0°C, just like liquid water evaporates completely leaving empty bucket?
Answer by Geoffrey Brent and jeffB

@The99sLearner You're missing one important thing - phase diagrams show what happens to a material in an ideal closed container. That is, if you have a full can of water at 32 °C at standard pressure, it will be liquid, and if it's at 110 °C, it will be a vapor. But the same is not true if you have an open can - water at 32 °C will then tend towards an equilibrium given roughly by the vapor pressure of the liquid and the partial pressure of the gas. The same is true of ice, but there the equilibrium tends to be skewed towards liquid/gas - it's hard for the solid phase to form spontaneously.

@The99sLearner All liquids always evaporate (or solidify) - even liquids like mercury at standard conditions (though that takes thousands of hours per gram!). Strictly speaking this is true even in the closed container, however, you need to account for the fact that the inside is not going to stay "standard" - any liquid that evaporates greatly increases the pressure (1L of water produces about 1700L of steam at the same pressure!), which makes it liquify again almost instantly. So there's still an equilibrium between the liquid and gas, but I doubt the gas is even measurable.

Liquid water has a vapor pressure that increases with temperature. 100°C is special only because that's the temperature where water's vapor pressure equals atmospheric pressure. When that happens, water can boil.
Even while water isn't boiling, though, it can evaporate, because its vapor pressure is still non-zero. In fact, this is true for ice as well.
Meteorologists use the concept of dewpoint. It's a characteristic of an air mass containing water vapor, and it represents the temperature at which that air mass becomes saturated with water vapor. (Technically, at that point, the partial pressure of water vapor in the air equals the vapor pressure of water at the same temperature.)
If you expose a surface to air, and the surface temperature is below the air's dewpoint, condensation will form on the surface.
As it turns out, air that you exhale is quite moist; its dewpoint is high. Breathe onto a surface below that dewpoint temperature, and you'll see condensation.
So why don't you see condensation on a wall? As another answer already stated, wall material doesn't conduct heat as well as glass, and it has lower heat capacity. As a result, your breath quickly warms the surface until it's above the dewpoint, and condensation stops.
Edit to add: The point @Luaan makes below is probably even more important for many wall materials! A surface that absorbs water won't show condensation as quickly as one that is impermeable.

It should also be noted that when you spray water on a wall, it gets wet - while on a mirror, it gives you droplets. If you breathe on a surface that doesn't wet (laquered wood, some plastics, metal...), you'll get condensation, though not always very visible. A surface that does get wet will absorb the water instead (by the time it's wet enough to show actual droplets, you're probably deep in moldy trouble :)).

When I exhale on a wall, no water droplets form. Why do water droplets form on mirror but not on wall (just an example)? The mirror must have the same temperature as that of the wall, both of them must have achieved thermal equilibrium with their surroundings a long time ago.

Yes, the mirror and the wall will be at the same temperature, but that's only part of the story.
Put one hand on a mirror, and the other on the wall next to that mirror. Which feels colder?
Or, leave a wooden block and a glass cup in water at about 70C/160F, long enough for them to reach equilibrium. Then pick them out quickly with tongs and hold one in each hand. Which feels hotter?
The hot glass is at the same temperature as the wooden block. But the glass is a much better conductor of heat, which means that when you pick it up, it's much better at transmitting that heat into your fingers - so hot glass feels hotter than hot wood at the same temperature, and cold glass feels colder than cold plasterboard at the same temperature.
The same thing is going on here: although the mirror is at the same temperature as the wall, the mirror is much better at cooling the air that passes over it, so you'll see far more condensation.
(Heat capacity is also a consideration here, but the difference in heat capacity between wood/plasterboard and glass is much less than the difference in conductivity.)

It has been almost two years, yet reading this question again will still confuse me. The answers and comments given by the users are really helpful and excellent, but different answers/comments explain different parts of the questions to different extents. I decided to write a compilation based on the answers/comments which gave the most convincing explanation, IMO.

jeffB's answer

Dewpoint is the temperature characteristic of an air sample, such that, when the air is at this temperature, it is saturated with water vapour, and if cooled further, condensation of water will happen, and liquid water formed. The air sample from our body has higher water content than air sample from surrounding and so higher dewpoint. When this air sample is exposed to mirror which is at slightly lower temperature, that lower temperature is already below the dewpoint of the air sample, co condensation happens.
Actually this question is nothing special; the answer by Karsten Theis explained this well too. Our body temperature higher, so vapour pressure higher; mirror lower the temperature, so vapour pressure will decrease, so water vapour condenses, that's all.
$\space$$\space$$\space$2 and 3. Karsten Theis's answer
The theoretical view is that, for liquid-vapour case, you are on the line (means that liquid and vapour at equilibrium) in phase diagram of water, and when the vapour pressure changed due to temperature changed, you will still be on the line (equilibrium preserved). For the solid-liquid case, your pressure is fixed (the pressure is just pressure experienced by the condensed phases), when you change T from 0 $^\mathrm{o}$C the equilibrium between solid and liquid will just be broken, and you are no longer on the line, you are on the solid region with 2 degree of freedom. The discrepancy in behaviour is because the pressure of vapour can change when temperature change and so preserving equilibrium, and there is no similar analogous behaviour for solid-liquid case.
The physical reason by BowlOfRed and the comment by jpa (on the main question) do make sense. If a molecule of water has enough kinetic energy to escape from ice crystal but remains bound as liquid, it is so close to the ice that it would settle back quickly.

Geoffrey Brent's answer

Argument based on heat conductance and heat capacity: in essence, mirror can 'spread' heat well. Exhaling on a wall, the local temperature at the point of contact will increase fast, reducing the efficiency of condensing water vapour further.
PS: IMHO all four answers help at different parts of the question, so I am not sure which to accept. But I do consider this question solved. I will accept the most upvoted answer.


text:
Imine as substituent nomenclature
I am trying to name the following compound:

I thought the name would be: (E)-3-[(4-methylphenyl)methylideneamino]benzoic acid
I stayed away from benzylidene since P-57.1.2 indicates no substitution on this preferred prefix. However, Chemdraw shows the name as: (E)-3-((4-methylbenzylidene)amino)benzoic acid
Is my Chemdraw outdated? If so, is my name correct or is there a better way to name this compound? I am also a bit confused as to when to use parenthesis, brackets, or braces with substituent names, am I using them correctly? If not, why? Also should it be "amino" or "aza", I am confused as to when to use one or the other (similar to "oxa" vs "oxo", I think; unrelated but I also don't get the difference between those).

Your assessment is correct. The rules concerning the prefixes ‘benzyl’ and ‘benzylidene’ have been changed. In the 1993 IUPAC Recommandations, these prefixes could be substituted, but only on the ring. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), these prefixes cannot be substituted for preferred IUPAC names, although for general nomenclature restricted substitution is permitted (see P-29.6.2).
Except for the use of enclosing marks to set off parts of the name dealing with specific structural features, the suggested name is correct. The preferred IUPAC name is 3-{(E)-[(4-methylphenyl)methylidene]amino}benzoic acid.

Note that enclosing marks are used around compound (see Rule P-29.1.2) and complex (see Rule P-29.1.3) prefixes. When multiple types of enclosing marks are required, the nesting order is {[({[( )]})]}, etc.
Concerning the use of ‘amino’ in this case instead of ‘aza’, Rule P-62.2.3 explains

(…)
Preferred IUPAC names for prefixes corresponding to $\ce{-NHR}$, $\ce{-NRR'}$, or $\ce{-NR2}$ are formed by prefixing the names of the groups $\ce{R}$ and $\ce{R'}$ to the prefix ‘amino’, for example ‘methylamino’ for $\ce{-NH-CH3}$. Prefixes such as azanyl and azanylidene may be used in general nomenclature.


text:
Is there a difference between equilibrium and steady state?
The term equilibrium is used in the context of reversible reactions that reach a point where concentrations no longer change. The term steady-state is used in enzyme kinetics when the concentration of the enzyme-substrate complex no longer changes (or hardly changes, in case of a quasi steady state). It is also used to describe multi-step biochemical pathways. Is there a difference between the two, given that both concern a situation where concentrations don't change over time?

From a (simplified) biochemistry perspective when an organism reaches equilibrium that means it died. Organisms strive for steady state consumption of energy to stay alive.

Yes, equilibrium and steady-state are distinct concepts.
A reaction is at equilibrium if reactants and products are both present, the forward and reverse rates are equal and the concentrations don't change over time. If this is the only reaction in a closed, isolated system, the entropy in the system is constant.
Steady-state implies a system that is not at equilibrium (entropy increases). A species is said to be at steady state when the rate of reactions (or more general, processes) that form the species is equal to the rate of reactions (or processes) that remove the species.
In both cases, there are rates ($\mathrm{rate}_1$ and $\mathrm{rate}_2$) that are equal. For an equilibrium, the forward and reverse rate of the same reaction are equal to each other. For a steady state, the rates of processes leading to increase of the concentration of a species are equal to the rates of processes leading to decrease of the concentration of the same species.
$$\ce{A <=>[rate_1][rate_2] B}\ \ \ \ \    vs \ \ \ \ \  \ce{source->[rate_1]C->[rate_2]sink}   $$
For an equilibrium, all concentrations are constant over time. For a steady-state, there is a net reaction, so some amounts change (the amount of source and sink), while at least one species - the one at steady state - has a constant concentration as long as the conditions of steady state prevail.

@KarstenTheis: In a closed system, where the source equal the sink, does the steady-state mean the same thing as equilibrium?

@TheLastCipher I think of steady-state as non-equilibrium. It is possible that some folks define steady-state such that equilibrium is a special case of steady state. Certainly, if the source equals the sink, the system will attain equilibrium eventually.

In short, equilibrium is a static process characterized by an equilibrium constant and does not have a time component.  Steady state is a kinetic process characterized by rate constants and determined by the half-life of the substance with respect to the process.  Time is an important dimension here.
An example of a steady state process is seen in the formation of an enzyme-substrate complex as described so nicely by Karsten Theis in the post above.
Another example is the administration of a drug to a person.  The drug is continuously absorbed into the systemic circulation, continuously be distributed to the target site(s), continuously be metabolized and continuously be removed by excretion out of the body.  These processes are collectively known as the 4 pharmacokinetic parameters.


text:
Why does increase in pressure cause a increase in boiling point
When we increase the atomospheric pressure pressure above the solution , the boiling point of a solution increases. Why does this happen?

Welcome to Chem SE! Do note that while posting homework questions on SE is not discouraged, you must make a mention of your attempt at the problem, and where you got stuck.

Welcome again to Chemistry.SE (SE = stack exchange)! Please take the [tour] to get familiar with this site. You have not posted a lot of context about your question, it did not even form a complete sentence. This may then appear to some of us as a homework question according to our definition. Nevertheless, questions that do not make the effort to explain where the actual problem in understanding is, are often ill-received, down-voted and closed.

Definitions and facts
Boiling point - when the vapour pressure of the solution is equal to the atompheric pressure, then it is said to be boiling.
Vapour pressure - sometimes molecules escape the solution to convert into the gas phase, also molecules in the gas phase combine to turn back into liquid phase. Hence there exists a equilibrium between gas phase and liquid phase.

The pressure exerted by this gas phase is called vapour pressure, and the vapour pressure increases with temperature.
For a more accurate definition checkout vapour pressure on Wikipedia
and Chemistry.se for an intuitive explanation of boiling points.
Conclusions and interpretations.
So if you increase atmospheric pressure, the solution needs to exert greater amount of vapour pressure to boil.
As mentioned earlier, you need to heat the liquid further so that the vapour pressure matches with the atmospheric pressure.
Hence the boiling point increases.
Sources: Wikipedia, NCERT Chemistry textbook for class 12.

I have edited your question for a better format. Please use markup sparingly. Please have a look at the licenses of the image you have used and edit your post accordingly to comply with them.

Sorry, I meant answer. All of the above applied still.

I would take a look at this thread A boiling point question.
A crude answer to this is that it takes work for a molecule to "squeeze" or "jump" from one phase into another. Liquids are tightly packed but vapor is quickly moving.
A possible analogy is that moving into the liquid is like squeezing a marble into a full jar of marbles whereas moving into the vapor phase is like trying to run through spaced out but fast moving traffic. Neither is necessarily easier than the other, and actually at equilibrium both require the same amount of effort.
As pressure increases the liquid gets even more packed, and in the vapor the traffic gets even faster moving.
What does this have to do with the boiling point?
Boiling is the process in which molecules move from the liquid into the vapor phase. When the pressure is higher it is harder to move into the vapor.  Thus, more energy is required. We get this energy by transferring energy to the liquid in the form of heat. This heat increases the temperature. Hence, to reach a boil at a higher pressure, we need a higher temperature. Temperature is really a measure of how energetically molecules are moving. As pressure increases they need more energy (temperature) to move about and jump into the vapor (traffic).

Why we refer temperature as temperature of the system but we refer to pressure as pressure of surroundings? Shouldn't both temperature and pressure be properties of the system?

Suppose there is equilibrium between molecules leaving the liquid (to become gas) and condensing from the gas (to become liquid). When a gas, the molecules occupy a far greater volume than they do when a liquid. Therefore, increasing the pressure displaces the equilibrium (c.f. Le Chatelier) towards liquid, and causes condensation. This would be true at any temperature, but at the boiling point this condensation requires increased temperature to resume boiling.


text:
Cobalt electroplating over nickel-plated steel
From my research, I understand that cobalt can be used on top of a nickel layer to improve durability of the nickel plating on a steel object.
I am using a nickel anode and nickel acetate as electrolyte.
Would I be able to use a cobalt anode and cobalt acetate as electrolyte?
If I buy $\ce{CoSO4 * 7H2O}$ but don't have a cobalt anode, can I use a stainless steel or titanium anode?

in the past cobalt was used as a brighterner for Nickel plating electrolyites, now since Cobalt is about 5 times more expensive than Nickel, is hardly used but you can add about 30 grams per litter of CoSO4 to your regular nickel plating solution and you will end up with a Nickel/Cobalt alloy that is stronger and whiter than plain Nickel, also keep in mind that in these type of baths Cobalt plates out faster than Nickel, that means that a 10% Cobalt Sulfate(90% Nickel Sulfate) plating bath will produce about a 70% Nickel/ 30% Cobalt alloy plating.


text:
Is liquid water in equilibrium with water vapor at room temperature?
This question was inspired by Why does liquid water form when we exhale on a mirror?.
This question is different from Why does water evaporate at room temperature? because it asks about whether an equilibrium is established. At room temperature under certain circumstances, water molecules leave the liquid phase and end up in the gas phase, as explained in the answers to the linked question. Whether an equilibrium is established is a related but distinct question.
Consider the reversible process of water moving from the liquid phase to the gas phase:
$$\ce{H2O(l) <=> H2O(g)}$$
At room temperature and atmospheric pressure, for which of the following scenarios does this process reach equilibrium?

A glass filled with water
A bottle of water after you poured half of the water out and recapped it.
A zip lock bag filled with water (no air).
Humid air in a zip-lock bag.

According to Wikipedia,

[...]chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time, so that there is no observable change in the properties of the system. Usually, this state results when the forward reaction proceeds at the same rate as the reverse reaction. The reaction rates of the forward and backward reactions are generally not zero, but equal.

I am pretty confident in my answers for 1. and 2., but I am having trouble with 3. and 4.
Scenario 1. is an open system, and unless the air is at 100% humidity, the water will evaporate. This means the forward rate is larger than the reverse rate, and it is not at equilibrium.
Scenario 2. is a closed system, and if the temperature is constant, the water will evaporate until the air is at 100% humidity and the equilibrium is reached.
Scenarios 3. and 4 are not clear to me. Would you say these are at equilibrium because there are no changes over time? Or would you say these are not at equilibrium because either product (3.) or reactant (4.) is absent? What about if there is a tiny air bubble in scenario 3., or the bag is wet on the inside in scenario 4., would that make a difference?

Are the "systems" in equilibrium with their surroundings? Why is the answer not obvious?

The "systems" are in thermal equilibrium with their surroundings. Depending on your level of knowledge and experience, answers will be obvious or not. Even if they are, it is not trivial to explain them to someone who is new to the concept of equilibrium.

The top answer to this question seems to do provide an answer that encompasses all of your scenarios : https://chemistry.stackexchange.com/questions/7449/why-does-water-evaporate-at-room-temperature?rq=1

Possible duplicate of Why does water evaporate at room temperature?

Related question: https://chemistry.stackexchange.com/questions/59541/is-a-liquid-in-a-container-always-in-equilibrium-with-its-vapour?rq=1

It depends on the circumstances, but mostly not
Under most normal atmospheric circumstances water in an open vessel is not in equilibrium with water vapour in the atmosphere. We know this because water usually evaporates, albeit slowly.
One way to measure whether water is in equilibrium is to measure the humidity which is a measure of the maximum concentration of water vapour under the prevailing temperature and pressure. Humidity is mostly less than 100% which means that the air could have more water vapour in it before the vapour spontaneously condenses. This isn't always true, which is why we get fog and why some countries are unbearable to live in.
At normal temperatures the equilibrium between water and water vapour is established slowly. Any vessel containing water that is open to the atmosphere will try to establish that equilibrium, but that won't be complete until the air is 100% saturated with vapour (humidity=100%). And you'd have to saturate the entire atmosphere in the room or, if the room is open, the entire planet's atmosphere to achieve equilibrium.
That this doesn't happen quickly is obvious. But it can happen locally (we call that weather). It it obviously not true across the whole planet at the same time or the planet would be called "sauna" and not earth.
The point is that the atmosphere is rarely 100% saturated with water vapour, so nothing open to the atmosphere will normally be in equilibrium with liquid water. A close bottle left long enough might establish equilibrium (with 100% humidity) inside the closed bottle, but any exchange with the surrounding air will lower the humidity and the equilibrium might take some time to re-establish.
The key to understanding this is that the kinetics of establishing the equilibrium are slow and the normal atmosphere is not saturated with water vapour. This means that the air above a glass containing open water will have a slightly higher concentration of vapour than the air surrounding the glass but with not be fully in equilibrium because the room surrounding the glass is big. And the atmosphere around the room is even bigger and clearly isn't in equilibrium (we have weather and wind to demonstrate that). The only systems where equilibrium will be established are closed ones which have been left a reasonably long time. atmospheric exchange causes the first; slow kinetics the second of these constraints.

I changed the question because it got closed in its original form. I incorporated some of your points (concept of 100% humidity mostly) into the edits. Unfortunately, now it looks like you answered the parts I'm less confused about.