Calculating concentration of dissolved gas from its mole fraction I came across the following (solved) problem in my textbook:

  The Henry's law constant for $\ce{O2}$ dissolved in water is $\pu{4.34 \times 10^4 atm}$ at $\pu{25 ^\circ C}$. If the partial pressure of oxygen in air is $\pu{0.4 atm}$, calculate the concentration (in terms of molarity, $\pu M$) of dissolved oxygen in water, in equilibrium with air at $\pu{25 ^\circ C}$.

The following is my attempt at a solution to the above problem:

From Henry's law, we have that

$$p{\ce{O2}} = K_\mathrm H \cdot x_\ce{O2}$$

where $x_\ce{O2}$ is the mole fraction of oxygen dissolved in the solution, $K_\mathrm H$ is the Henry's law constant, and $p_{\ce{O2}}$ is the partial pressure of oxygen over the liquid.

Given that the Henry's law constant for a mixture of water and some amount of dissolved oxygen is $\pu{4.34 \times 10^4 atm}$, this gives us

$$x_\ce{O2} = p_{\ce{O2}}/K_\mathrm H = 0.4/(4.34 \times 10^4) = 9.2 \times 10^{-6}$$

To calculate the concentration of oxygen in this mixture in terms of its molarity, I need to know the number of moles of oxygen present in the solution and also the number liters of water in which the oxygen is dissolved.

To calculate the number of moles of dissolved oxygen, I can use the following equation:

$$x_\ce{O2} = \frac{n_\ce{O2}}{n_\ce{O2} + n_\ce{H2O}}$$

However, no information regarding the number of moles of the solvent, water, has been provided.

Hence, I checked the solutions provided by my textbook and found that they have arbitrarily assumed that we are dealing with $\pu{1 L}$ of water.

My question is this: are we allowed to arbitrarily take the volume of water as $\pu{1 L}$ even when absolutely nothing has been mentioned about this in the problem?

The gist is that as long as you have a specified mole fraction, the concentration of oxygen does not depend on the amount of water present. This is what is called an "intensive property". Intensive properties don't depend on how much stuff there is, and therefore, you can arbitrarily specify any volume you like. In order to make the calculations simpler, this specified volume is usually some unit quantity, like $\pu{1 L}$.

It is somewhat analogous to determination of an empirical formula, in that it doesn't matter how many grams of the compound is present, you will always end up with the same empirical formula. A common tactic is to assume $\pu{100 g}$ of compound is present, such that if there is $x\%$ of carbon by mass, there is $x\ \pu{g}$ and therefore $x/\pu{12 mol}$ of carbon.

Here, if you have $\pu{2 L}$ of water, then $n_{\ce{H2O}}$ is double of that if you only have $\pu{1 L}$. But since you have twice as much water, to maintain the mole fraction (that you determined via Henry's law), you have to dissolve twice as much oxygen - otherwise the mole fraction won't be the same. So $n_{\ce{O2}}$ is double of that if you only have $\pu{1 L}$. The factor of 2 will cancel out in the equation you gave:

$$\begin{align}
x_{\ce{O2}} &= \frac{2n_{\ce{O2}}}{2n_{\ce{O2}} + 2n_{\ce{H2O}}} \\
&= \frac{n_{\ce{O2}}}{n_{\ce{O2}} + n_{\ce{H2O}}}
\end{align}$$

If you are really uncomfortable with assuming the volume, then simply assign an unknown value: $k$ litres, for example. You should find that $k$ cancels out and you are left with the exact same value for the molar concentration of $\ce{O2}$.

There is, perhaps, a more "elegant" way to prove this. Note that

$$\begin{align}
\frac{1}{x_{\ce{O2}}} &= \frac{n_{\ce{O2}} + n_{\ce{H2O}}}{n_{\ce{O2}}} \\
&= 1 + \frac{n_{\ce{H2O}}}{n_{\ce{O2}}} \\
&= 1 + \frac{V_{\ce{H2O}} \times [(\pu{1000 g L-1})/(\pu{18 g mol-1})]}{n_{\ce{O2}}} \\
&= 1 + \frac{\pu{55.56 mol L-1}}{c_{\ce{O2}}}
\end{align}$$

where $V$ is the volume and $c = n/V$ the molar concentration. You can see that the molar concentration is uniquely determined by the mass fraction $x$ and the exact value of $V$ doesn't matter at all.


Synthesis Golf III: Chloramphenicol A full FAQ post has been written on meta.chem.SE, explaining the premise of synthesis golf and the 'rules'. Please take a look at this before answering (if you haven't already).

This third round of golf concerns the synthesis of chloramphenicol 2,2-dichloro-*N*-[(1*R*,2*R*)-1,3-dihydroxy-1-(4-nitrophenyl)propan-2-yl]acetamide, an antibiotic on the World Health Organisations list of essential medicines:

InChI=1S/C11H12Cl2N2O5/c12-10(13)11(18)14-8(5-16)9(17)6-1-3-7(4-2-6)15(19)20/h1-4,8-10,16-17H,5H2,(H,14,18)/t8-,9-/m1/s1

In addition to being a useful target, chloramphenicol was chosen as
during previous rounds of synthesis-golf, several people commented that they'd like to see targets amenable to biosynthetic transformations, and scaleable process routes. This target very much fits both of those criteria (the drug is manufactured on a huge scale, and several enzymatic routes have been developed).

The challenge is to propose a route to chloramphenicol, including a method for setting the absolute and relative stereochemistry of the 1,2-hydroxyamine functionality. You may start from anything commercially available in the Sigma Aldrich catalogue, but to keep things interesting, no starting material may contain more than 8 carbon atoms.

Approach:
The approach taken in this proposed synthesis of chloramphenicol is largely based on Greene's synthesis of paclitaxel, using dihydroxylation methodology proposed by Sharpless.
Whilst not particularly elegant (I was hoping for some kind of asymmetric hydrogenation to set the stereocentres), it does get to the product in a respectable 7 steps (including setting both stereocentres, as required), and avoids having to introduce protecting/directing groups at any stage.
Forward synthesis:

Step 1: Horner Wadsworth Emmons olefination

The first step in this proposed synthesis is a Horner Wadsworth Emmons olefination to introduce the (E)-alkene required for the planned dihydroxylation. Without the nitro-group, the exact reaction has been done many times and with good selectivity, so no reason to assume that this HWE would be particularly challenging (the nitro group clearly makes the aldehyde electron poor).

Step 2/3: Sharpless dihydroxylation, epoxidation sequence

With the (E)-alkene formed, a Sharpless asymmetric dihydroxylation can take place using AD-MIX-B (taking the aryl group as 'large' and the ester as 'medium'). This step (without the nitro) is present in the Greene paclitaxel synthesis (above) and initially gives 82% e.e, though a single recrystallisation of the product brings this up to >95%.
Once recrystallised, treatment with mesyl chloride and subsequent heating (same pot) affords the necessary epoxide.

Step 4/5: Epoxide opening and reduction

Now comes the slightly dubious step....
In the paclitaxel synthesis, the epoxide is opened with sodium azide at the benzylic position (expected, the aromatic ring can stabilise the developing partial positive charge here), however in our case, we have an incredibly electron withdrawing nitro group in the 4-position, completely shutting down any stabilisation at the benzylic position.
Based on this lack of benzylic stabilisation, we could reasonable assume that it will take place with the opposite regioselectivity to the paclitaxel case. There is some evidence for this in the literature, (including where there is an ester present on the other side, but sadly not using azide as the nucleophile), and a cursory flick through an organic textbook will tell you that SN2 adjacent to carbonyls is generally fast.
Assuming it works fine, a Staudinger reduction converts the azide to the secondary amine with retention of stereochemistry.

Step 6: Ester reduction

Lithium borohydride is a mild and selective reagent, able to reduce esters in the present of various other functionality, including nitro groups. One draw-back of this is that lithium borohydride is significantly slower than its counterparts (not to mention more expensive), but, using LAH or other highly reactive hydride species would result in side reactions involving the nitro group.

Step 7: Amide bond formation

The required acyl chloride is commercially available, and should do the required reaction in place of the possibly competitive ester formations (nitrogen LP higher in energy).
Conclusion:
7 steps overall starting from the cheap, commercially available 4-nitrobenzaldehyde. The majority of steps have good precedence on similar systems (ignoring the epoxide opening, which may/may not be fatal), and all of the proposed reactions can (and have been) demonstrated to be possible on large scale (again, ignoring the epoxide opening, but TMS-azide is often substituted in the case of process syntheses involving azide).


How to prevent losing solvent into the vacuum pump in rotovap solvent collection? This is a practical question about the use of rotary evaporators ("rotovap"). So I'm asking for answers from people who actually use the things.

A rotovap typically uses a cold condenser to condense the solvent into a collection flask. But the distillation is typically done under weak vacuum as well. So the solvent sitting in the collection flask seem susceptible to re-vaporizing and getting sucked into the vacuum pump. Now I have a cold trap in front of my pump that should be cold enough (-50C) to trap most solvents under weak vacuum.

I think the answer is that if the vacuum is weak enough to vaporize the solvent under the hot water bath, but not under the cooler ambient air temperature, then the loss to the pump should be minimal. Indeed, this is what the instructions for the rotovap suggest.

  Question: In practice, what are the risks to losing solvent into the pump?

Honestly, I never bothered to actually measure the volume of the condensed solvent in order to compare it with the original volume.

The concept described in the question, that is using a cooling trap with dry ice + acetone, is good practise and typically good enough to retain the evaporated solvent and protect the membrane pump. In rare cases, I have used a second, flat Dewar filled with dry ice + acetone on a "Laborboy" to cool the condensate-collecting flask.


Which acid/base does a strong base/acid react when added to a buffer solution? I'm in high school chemistry and just learning about buffer solutions. For example, given a buffered solution of $\ce{CH3COOH}$ and $\ce{NaCH3COO}$ with $K_a = 1.8 \times 10^{-5}$, we might want to find what the pH will become if $\ce{0.010 M KOH}$ is added. Given the dissociation of acetic acid:

$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$

My question is, which acid ($\ce{CH3COOH}$ or the conjugate $\ce{H3O+}$) will the base $\ce{KOH}$ react with? My teacher said that the base will react with $\ce{H3O+}$ to shift equilibrium in the forward direction, but why wouldn't it react with $\ce{CH3COOH}$ since it also is an acid?

Is there a specific rule that an added acid/base will react only with the base/acid in the product side of the reaction?

$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$

  My question is, which acid (CH3COOH or the conjugate H3O+) will the base KOH react with?

You have to satisfy two equilibria, the one you wrote and the auto-dissociation of water.

$$\ce{H3O+ + OH- <=> 2 H2O}$$

If you add up the two reaction, you get a third one:

$$\ce{CH3COOH + OH- <=> H2O + CH3COO-}$$

You can go either way (use the second or the third reaction) to explain that $\ce{OH-}$ is part of a neutralization reaction. As a consequence, the concentration of $\ce{CH3COO-}$ increases and that of $\ce{CH3COOH}$ and hydronium decreases.

Hydroxide reacts with hydronium

In this case, you would invoke the second reaction. There is more hydroxide (you are adding sufficient KOH to raise the concentration to 0.01 M) than hydronium (at about pH = 5, the buffer will contain about $\pu{e-5}$ M). As the hydronium gets neutralized, the weak acid equilibrium (first reaction) will shift forward (not reverse - typo in the question), making more hydronium, which will get neutralized etc.

Hydroxide reacts with acetic acid

In a different explanation that results in the same answer, you could say the hydroxide turns some of the weak acid into weak base (third reaction). As a result, the weak acid / weak base concentration ratio will change, causing a net reverse shift in the weak acid equilibrium (first reaction) and lowering the hydronium concentration.

Which way is better?

It is personal preference. If we knew how fast the two reactions go, we might choose the faster one, but we don't. For figuring out the equilibrium concentrations and the pH, it does not matter.

  Is there a specific rule that an added acid/base will react only with the base/acid in the product side of the reaction?

No, and you can always combine two equilibrium reactions with overlapping species (like the hydronium ion in the first and second reaction) to arrive at a third that also has to be at equilibrium.


Why is heating increasing the solubility of sodium hydroxide? \begin{align}
\ce{NaOH &-> Na+ + OH-} & \Delta H^\circ &= \pu{-44.51 kJ/mol}
\end{align}

The dissolution of sodium hydroxide in water is an exothermic process, and so, according to Le Chatelier’s principle, cooling the container should shift the reaction to the right. Shouldn’t this mean that cooling the container will increase the solubility?

According to a solubility chart for sodium hydroxide, heating the container will increase the solubility; why is this the case?

The question to the Newton - Ask a Scientist program by Danna C Griffiths (to be found via the Internet Archive) has an answer for it.

What I understood from the explanation on that site:

Solubility is defined as the concentration of the solute in a saturated solution. So when considering the increase in solubility with temperature, you have to check the enthalpy of solution of $\ce{NaOH}$ in a saturated solution of $\ce{NaOH}$.

When you dissolve $\ce{NaOH}$ in pure water, the process is highly exothermic, but as the concentration of $\ce{NaOH}$ increases, the process becomes less exothermic and eventually near the saturation point it becomes endothermic (it becomes less favourable for more $\ce{NaOH}$ to dissolve).

So according to Le Chatelier’s Principle, although the dissolution process is overall exothermic, since solubility is determined only at the saturation point, the solubility increases with increase in temperature.


Why there is an Upper Explosive concentration Limit? If we have a mixture of combustible gas (let's say it's methane) and air, what is the physical explanation for having a upper concentration limit?
I found this document:
Lower and upper explosive limits

For example, methane has 5% and 15%.
Let's say we have a 30% concentration of methane... There is still enough oxygen, so that some of the methane could burn.
There won't be any combustion at all, or there is some combustion, but no explosion?

In simple terms, an explosion takes place when the reaction front propagates faster than the speed of sound. If it is slower, the expanding gas cloud can push away the unreacted mixture before it reaches ignition temperature.

With too much methane, the combustion reaction slows down, because the additional material takes up energy for heating, but cannot react.

A very simplistic microscopic picture would be that a hot methane molecule too often does not hit a cold oxygen molecule, with wich it could react to produce more heat, but a cold methane molecule, leaving two lukewarm methanes, than won't react any further.


Stereochemical outcomes in opening of vinyl epoxides Explain the outcomes of the following reactions. In each case the major product is shown.

Firstly, I don't understand why the chloride always attacks on the right hand side of the epoxide. Also, regarding the stereochemistry for (ii), I assume the chlorine can open the epoxide as it is antiperiplanar to the left hand side of the epoxide?

While @orthocresol was busy finding the Carreira paper, I was busy constructing my own thoughts on the stereochemical consequences of the reactions of 1 (iii) and 5 (ii). Yes, it is an issue of conformation. It is far easier to rationalize the results after the fact. To predict the most reactive conformation of epoxide 1 the Newman projection 2, viewed along the red bond of 1, places the two carbon chains anti to one another. Intermediate 3 undergoes straightforward SN2 displacement at the allylic site.
In the case of epoxide 5 (ii), the two alkyl chains maintain the anti positions in conformation 6 but now the carbon-bound chlorine acts as a neighboring group in an intramolecular SN2 displacement to form intermediate 7. This species suffers a second SN2 displacement to afford 8 with retention of stereochemistry at the reacting site. Compound (i) follows the same pathway as 1 (iii).    The role of water is to cleave the silylether.


Synthesis of potassium trioxalatocobaltate(III) trihydrate I've synthesized the compound potassium trioxalatocobaltate(III) trihydrate and for it I have two reaction equations. One of them is:

$$\ce{CoCO3 + 3K2C2O4.2H2O -> K2[Co(C2O4)3] + 2CO2}$$

and the other:

$$\ce{K2[Co(C2O4)3] + PbO2 -> K3[Co(C2O4)3].3H2O}$$

I see that the lead dioxide oxidizes $\ce{Co^2+}$ to $\ce{Co^3+}$ in the second step but I can't figure out what happens in the first. Does cobalt(II) make a ligand complex with the oxalate group?
One of the beginning steps was a also to mix oxalic acid and potassium oxalate. I assume (but I am not sure) that it was done to dissolve the potassium oxalate as other acids might have led to an unwanted reaction that would hinder the reaction following reaction with $\ce{CoCO3}$. I wanted to know that this was in fact the case.

It appears that Sørensen was the first to publish the synthesis of trioxalatocobaltate(III) trihydrate [1].
A summary written in English (and that should answer both of your questions) including chemical reactions can be found in the very first volume of Inorganic Syntheses1 [2, p. 37]:

C. POTASSIUM TRIOXALATOCOBALTIATE
$$\ce{K3[Co(C2O4)3]·3H2O}$$
$$\ce{H2C2O4 + CoCO3 → CoC2O4 + H2O + CO2↑}$$
$$\ce{2 CoC2O4 + 4 K2C2O4 + PbO2 + 4 HC2H3O2 → 2 K3[Co(C2O4)3] + 2 KC2H3O2 + Pb(C2H3O2)2 + 2 H2O}$$
Twenty-three and eight-tenths grams (0.2 mol) of cobalt carbonate is dissolved in a solution of 25.2 g. of oxalic acid $(\ce{H2C2O4·2H2O})$ and 73.7 g. of potassium oxalate $(\ce{K2CrO4·H2O})$ in 500 cc. of hot water.
When the solution has cooled to 40°C., while it is vigorously stirred, 23.9 g. of lead dioxide (see synthesis 16) is added slowly, followed by 25 ml. of glacial acetic acid added a drop at a time.
The stirring is continued for an hour, during which time the color changes from red to deep green.
After the unused lead dioxide is filtered out, the trioxalatocobaltiate is precipitated by the addition of 500 ml. of alcohol.
The material appears as emerald-green needles, which are sensitive to both light and heat; The yield is 70 g. (71 per cent).

Notes

Please disregard obsolete typography and notations used in this textbook such as full stops after unit symbols, as well as the lack of white space between the numerical value and a degree symbol (as seen in “40°C.”) and use of notations such as “cc.” (cubic centimeter).

References

Sörensen, S. P. L. Kritische Präparatenstudien. III. Darstellung von Einigen Kobaltidverbindungen. Z. Anorg. Chem. 1896, 11 (1), 1–5. https://doi.org/10/fh47nq. (in German)
Inorganic Syntheses; Booth, H. S., Ed.; John Wiley & Sons, Inc.: Hoboken, NJ, USA, 1939; Vol. 1. ISBN:9780470131602, https://doi.org/10.1002/9780470132326.


Formation of Bonding and Anti-bonding Molecular Orbitals If two orbitals combine in-phase, a bonding molecular orbital is formed. When they combine out-of-phase, an anti-bonding molecular orbital is formed. For a single molecule, how are both orbitals present at the same time? They are formed differently.

First, remember that orbitals are a more or less theoretical construct, based on mathematical models. With that in mind, remember that an orbital in the strictest quantum-mechanical sense is a wave function of one and exactly one electron. (I am separating α and β spins here.) Since an orbital somehow nicely corresponds to one possible particle state, it may seem logical to postulate a ‘rule of conservation of orbitals’ — and exactly that happens in quantum chemistry.

So if we take one 1s orbital of one hydrogen atom and another 1s orbital of another hydrogen atom, we need to create two new orbitals by linear combination if we choose to do so. The two orbitals we create correspond to σ and σ*, respectively. (In a spintastic sense, we would have taken two orbitals of each hydrogen, one corresponding to α the other to β. We would receive four orbitals: $\unicode[Times]{x3c3}_\unicode[Times]{x3b1}, \unicode[Times]{x3c3}_\unicode[Times]{x3b2}, \unicode[Times]{x3c3}^*_\unicode[Times]{x3b1}$ and $\unicode[Times]{x3c3}^*_\unicode[Times]{x3b2}$. The electrons now occupy the lowest energy orbitals which are $\unicode[Times]{x3c3}_\unicode[Times]{x3b1}$ and $\unicode[Times]{x3c3}_\unicode[Times]{x3b2}$.)


Stereochemistry of meso compounds I got tripped up by this question; I thought these were all enantiomers when in fact they were all meso compounds.

How can one prevent this from happening?

How does one rotate the Fischer projection, for example, in the first problem; I understand that one has to hold one group in place, but then what?

To rotate on a single bond in a Fischer Projection, all 3 groups move in the same direction. $\frac {1} {2\ \ \ 3} \rightarrow \frac {2} {3\ \ \ 1}$. See this page for more information.


Formation Constants; Lewis Acidity I still distinctly remember reading about how Lewis acidity/basicity was characterized using kinetics. The reason given was that because formation constants are so large, these reactions are not equilibrium reactions and completion reactions. So we just characterize strength using the rate of reaction/formation. This at least makes partial sense; formation constants are huge:

http://bilbo.chm.uri.edu/CHM112/tables/Kftable.htm

EDIT: I have found what I read:

  Many reactions of nucleophiles are not reversible. A bond forms, a bond breaks, and that’s the end of the reaction. The problem with this from a measurement standpoint is that we often can’t determine an equilibrium constant for a reaction. And if we can’t do that, then we can’t develop a reactivity scale based on equilibria.

  If we can’t measure equilibria, then what do we do? Well, we use the next best measurement available: to measure reaction rates.

1) Does the quoted text make sense?

2) Am I interpreting it correctly or out of context?

3) Reference to previously answered question: Lewis base strength and the hydrogen proton

I think I got this figured out.

Acidity and basicity are thermodynamic properties.

Nucleophilicity/electrophilicity are kinetic properties.

Lewis acids are electrophiles.

Lewis bases are nucleophiles.

We can use electrophile/nucleophile in describing Lewis acids/bases. But care should be taken to ensure we don't conflate kinetics with thermodynamics.

I think I finally got this issue resolved.


LeChatelier's Principle - Phosphofructokinase in Glycolysis In glycolysis, PFK is allosterically regulated by ATP, F6P, etc.

When the concentration of F6P increases, the concentration of ATP increases too. I understand that high levels of ATP shift the eq towards the T-state, and this decreases the affinity of PFK for F6P.  But is there a principle in Le Chatelier's which explains how the increase in conc of one reactant affects the conc of another reactant?

The reaction regulated by PFK is: $F6P$ + $ATP$ <-> $F1,6BP$ + $ADP$

Thanks!

PFK plays a central role in the control of glycolysis because it
  catalyses one of the pathway’s rate-determining reactions. The control
  of Phosphofructokinase (PFK) is exquisitely complex.

To analyse this complicated enzymatic reaction pathway, we have to revisit important aspects of metabolic control mechanism:

Thus the flux (rate of flow) of intermediates through a metabolic pathway is constant; that is, the rates of synthesis and breakdown of each pathway intermediate maintain it at a constant concentration.
Regulation of the steady state (homeostasis) must be maintained in the face of changes in flux through the pathway in response to changes in demand. In an open system, such as metabolism, the steady state is the state of maximum thermodynamic efficiency.

This explanation provides a somewhat agreement of Le Chatlier’s principle. (i.e. where a dynamic equilibrium is affected by changing conditions, the position of equilibrium moves to counteract the change)

Consider this simple hypothesis:

  J = vf – vr

  Since a metabolic pathway is a series of enzyme-catalysed reactions,
  it is easiest to describe the flux of metabolites through the pathway
  by considering its reaction steps individually. The flux of
  metabolites, J, through each reaction step is the rate of the forward
  reaction, vf, less that of the reverse reaction, vr:

  At equilibrium, by definition, there is no flux (J=0), although vf and
  vr may be quite large. At the other extreme, in reactions that are far
  from equilibrium, vf>> vr, so that the flux is essentially equal to
  the rate of the forward reaction, J~vf.

The flux throughout a steady-state pathway is constant and is set (generated) by the pathway’s rate determining step (or steps).

Consequently, control of flux through a metabolic pathway requires:

that the flux through this flux-generating step vary in response to the organism’s metabolic requirements and
that this change in flux be communicated throughout the pathway to maintain a steady state

Phosphofructokinase is a non-equilibrium enzyme of glycolysis

One important thing to take note of is that the enzyme phosphofructokinase, a highly controlled enzyme functioning far from equilibrium (but ironically itself operates via an equilibrium of two conformational states), evidently is the major target for regulating glycolysis.

PFK is a tetrameric enzyme with two conformational states, R and T, that are in equilibrium. The substrate site binds ATP equally well in either conformation, but the inhibitor site binds ATP almost exclusively in the T state. The other substrate of PFK, F6P, preferentially binds to the R state. Consequently, at high concentrations, ATP acts as a heterotrophic allosteric inhibitor of PFK by binding to the T state, thereby shifting the T <=> R equilibrium in favour of the T state and thus decreasing PFK’s affinity for F6P.

Increase in concentration of PFK does not increase the rate of glycolysis.

In this scenario I believe the Le Chatlier’s principle applies little here. The reason is because the control is very complicated Although PFK has a major role in regulating the flux through glycolysis, it is controlled, in vivo, by factors outside the pathway.

An increase in the in vivo concentration of PFK will therefore not increase the flux through the pathway because these controlling factors adjust the catalytic activity of PFK only to meet the needs of the cell.

Let’s consider another illustration showing enzyme activity versus substrate concentration:

The inhibition of PFK by ATP is relieved by AMP. This results from AMP’s preferential binding to the R state of PFK. If a PFK solution containing 1 mMATP and 0.5 mM F6P is brought to 0.1 mM in AMP, the activity of PFK rises from 10 to 50% of its maximal activity, a 5-fold increase. when [F6P] =0.5 mM (the dashed line), the enzyme is nearly maximally active, but in the presence of 1 mM ATP, the activity drops to 15% of its original level (a nearly 7-fold decrease).

However, measurements of [ATP] in vivo at various levels of metabolic activity indicate that [ATP] varies <10% between rest and vigorous exertion. Yet there is no known allosteric mechanism that can account for a 100-fold change in flux of a nonequilibrium reaction with only 10% change in effector concentration.

Thus, some other mechanism, or mechanisms, must be responsible for controlling glycolytic flux. Let’s discuss the action of adenylate cyclase; adenylate cyclase catalyses the following reaction:

2ADP <=> ATP + AMP (K=0.44)

as a result of the adenylate kinase reaction, a 10% decrease in [ATP] will cause over a 4-fold increase in [AMP]. Consequently, a metabolic signal consisting of a decrease in [ATP] too small to relieve PFK inhibition is amplified significantly by the adenylate kinase reaction, which increases [AMP] by an amount sufficient to produce a much larger increase in PFK activity.

In summary the control of the PFK (non-equilibrium enzyme) has limited Le Chatlier’s principle application as other crucial regulation mechanism exist through the reaction pathway. Even if you were to consider temperature changes, these control enzymes operate far from equilibrium (these have very negative delta G°) and the actual free energy changes in vivo differ from experimental ones making it difficult to simulate actual effects in equilibrium states. It is more likely based on supply-demand basis. One evidence of this is seen through the variation of metabolic flux  throughout glycolysis (may vary by 100-fold or more), depending on the metabolic demand for ATP.

However, for pathways which operate close to equilibrium it may be applied, for example, the delta G° for aldolase is endergonic, whereas under physiological conditions in heart muscle it is close to zero, indicating that the in vivo activity of aldolase is sufficient to equilibrate its substrates and products. Their forward and reverse rates are much faster than the actual flux through the pathway (although their forward rates must be at least slightly greater than their reverse rates). Consequently, these near-equilibrium reactions are very sensitive to changes in the concentration of pathway intermediates.

References

Biochemistry Voet and Voet


Why do we use ethanol in SN2 reaction? Solomons and Fryhle has an example in which they have showed a very simple substitution reaction. The question was to synthesize methyl iodide. What they have done is taken methyl chloride and produced methyl iodide in the presence of $\ce{NaI}$ and ethanol through an $\mathrm{S_N2}$ mechanism.

Why did they use ethanol? I have read that $\mathrm{S_N2}$ reactions are favored under polar aprotic solvents and clearly ethanol is not that. Also, since it's a methyl and all substitution reactions on methyl go through $\mathrm{S_N2}$; so how would it affect my reaction if I would't use ethanol?

Usually, nucleophilic substitution reactions of anionic nucleophiles occur more rapidly in polar aprotic solvents. In such solvents (e.g. DMSO) the typical relative reactivity order is
$$ \ce{SCN-} < \ce{I-} < \ce{Br-} < \ce{Cl-} < \ce{F-}$$

In protic hydrogen-bonding solvents (e.g. ethanol), anions are solvated. Therefore, the reactivity of nucleophiles is decreased. Since hard nucleophiles are more strongly solvated than soft nucleophiles, the relative reactivity of soft anions is increased in such solvents. The typical relative reactivity order is
$$ \ce{F-} < \ce{Cl-} < \ce{Br-} < \ce{I-} < \ce{SCN-}$$

In the given example, $\ce{-Cl}$ shall be replaced by $\ce{-I}$; and in alcohols, the nucleophilicity of $\ce{I-}$ is higher than the nucleophilicity of $\ce{Cl-}$.


Boiling point of liquids If we can have liquid - vapour equilibrium at any temperature what is the 'point' behind boiling point.Boiling point is just a special case of liquid - vapour equilibrium where partial pressure equals atmospheric pressure.Why is it told in my textbook that boiling point is where the liquid and vapour coexist. At any temperature we can have liquid and vapour in equilibrium.

During boiling process when vapourisation take place from the bulk of the liquid does that increase the vapour composition which alters the equilibrium constant without the change in temperature?..

If we look at the phase diagram boiling point lies on phase boundary between liquid and gas ...so does that mean boiling point is when liquid and gas coexist..

Liquid and vapour can coexist at many combinations of temperature and pressure. But for most of them they are not in thermodynamic quilibrium.

For a given temperuture, there is only one (vapour partial ) pressure value with both phases in equilibrium.

For a given vapour partial  pressure, there is only one temperature value with both phases in equilibrium.

This temperature is also the liquid boiling point at the given external ( total ) pressure.

Standard boiling point is then temperature, when vapour partial pressure is equal external pressure 1 atm.

Note that boiling is not state of thermodynamic equilibrium, as there is net change of phase mass and thermal energy.

If partial pressure is higher than the equilibrium pressure for given temperature, the vapour condensates until the pressure reaches the liquid-vapour equilibrium curve of the phase diagram.

If partial pressure is lower than the equilibrium pressure for given temperature, the liquid evaporates until the pressure reaches the liquid-vapour equilibrium curve of the phase diagram.

If total pressure is lower than the vapour pressure for given temperature, the liquid boils until the total pressure reaches the saturated  vapour pressure.


Not sure why, but I can't modify/balance this equation I'm working on modeling a few reactions for my research. For example:

$$\ce{CO2 + H2S + CH4 <=> CH3SH + H2O}$$

was one I grabbed from the literature. However, this reaction is taking place in the aqueous phase around neutral pH, so most of the $\ce{CO2}$ will actually be $\ce{HCO3-}$. So in an attempt to model modified versions, I swap out the $\ce{CO2}$:

$$\ce{HCO3- + H2S + CH4 <=> CH3SH}$$

I've tried a redox-equation balancing approach but this hasn't really gotten me anywhere- sulfur doesn't change oxidation states.

I think my brain might just be fried, but can anyone lend any insight on this?

For starters, You have a problem because of a lack of charge balancing. You should also consider that at neutral pH $\ce{H2S}$ is present as $\ce{HS-}$ and $\ce{CH3SH}$ is present as $\ce{CH3S-}$.  Modifying the reaction to account for charge you get:
$$\ce{HCO3- + HS- + CH4 <=> CH3S-}$$

This will give you a start.  The other piece that will help is what the $\ce{HCO3-}$ molecule becomes. For the sake of being able to provide an answer I will assume formate ($\ce{HCO2-})$ is formed. Thus the reaction is now:
$$\ce{HCO3- + HS- + CH4 <=> CH3S- + HCO2-}$$

We know that the molecule of $\ce{CH4}$ becomes $\ce{CH3S-}$ and the molecule of $\ce{HCO3-}$ becomes $\ce{HCO2-}$ thus to balance the lost oxygen of $\ce{HCO3-}$ and the lost two hydrogens of $\ce{CH4}$ we add water to the products.
$$\ce{HCO3- + HS- + CH4 <=> CH3S- + HCO2- + H2O}$$

This equation serendipitously requires no coefficients to balance.


Why aren't there 2 products formed? The question is asking for the major organic product. This seems like an $S_N1$ reaction to me as there is a weak base in $\ce{CH3OH}$ and heat. If it is, shouldn't there be 2 products. One with the $\ce{OCH3}$ in the equatorial position (shown in the answer key here) and another with the $\ce{OCH3}$ in the axial position. I think there are 2 products because the C attached to Cl is an asymmetric center, thus there will be an $\ce{OCH3}$ that is wedge and an $\ce{OCH3}$ that is dash.

A lone pair on the ring oxygen forces Chlorine leave. The anomeric carbon in the resulting oxocarbenium ion is sp 2 hybridized, so that part of the molecule is planar. (An oxocarbenium ion has a positive charge that is shared by a carbon
and an oxygen.)

When the alcohol approaches from the top of the plane, equitorial isomer is forrmed. Similarly if alcohol apporoches from bottom of sp2 hybridised oxo carbenium ion , then ,axial isomer is formed .So the given chloro compound should give two isomers (as shown in figure below 2& 3)..

Which is the major product ?

Substituents on a cyclohexane ring prefer the equatorial position to relieve steric strain. cyclohexane ring in chair form with substituent in equitorial position ring-flipps to chair in which the substituent is axial.

When an endocyclic heteroatom ( oxygen, nitrogen, or sulfur) is  at a position adjacent to the methoxyoxy group,the hetroatom substituent now prefers the axial orientation (despite of the associated steric strain).The equilibrium preference of most such compounds for the alpha-isomer is referred to as the anomeric effect.

Such behavior is explained as

In α-methyl glucoside, the non bonding HOMO with a pair of electrons on the ring oxygen is antiperiplanar to the antibonding LUMO of C-O bond in methoxy group. This allows hyperconjugation between them and thus by stabilizing the α-form.
In axial position,such interactions are possible.Whereas, in equitorial the methoxy group is at equatorial position and cannot involve in hyperconjugation since it is not antiperiplanar to the lone pair on ring oxygen. Therefore  equitorial isomer is less stable than the axial isomer.

Other Explanations:

The  dipole interactions of the two ether functions are  in the equatorial conformer are nearly parallel;while, in the axial conformer these dipoles are oriented in opposite directions.

image source :https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/carbhyd2.htm#anomr

Electrostatic Repulsion: Decreased electrostatic interactions are favored

All the above combination of factors result in equitorial isomer being less stable than the axial isomer.

Referance

https://www.scripps.edu/baran/images/grpmtgpdf/Krawczuk_Nov_05.pdf

Organic Chemistry ,SEVENTH EDITION , Paula Yurkanis Bruice


What is a food-safe level of exposure of plastic containers to heat? There are many articles about how heat can trigger the release of potentially harmful carcinogens from plastic containers into the food or liquid they contain.  Specifics they fail to mention, though, are 1) the safe level of exposure and 2) the rate of BPA release per measurement of time at a constant level of heat.  While I will never drink from a water bottle that has been in my car exposed to summer heat over the course of days or weeks, I have been known to forget my water in my car as I head into work where I'll only end up retrieving it hours later when I go to lunch.  I also store bottled water in my garage which may reach temps of 90 degrees Fahrenheit or slightly higher in the summer.

So I am basically inquiring not only about the safe level of exposure to chemicals by quantity, but also the rate of chemical release over time based on heat level.  This would allow me to produce a practical ballpark idea as to whether or not the water is safe to drink.

You ask:

  I'd like to know if the information passed around the internet regarding the release of BPAs into the contents of plastic containers is grossly over-exaggerated or if there is a grain of truth to it that I should be concerned about to a degree.

In the United States, the Food and Drug Administration (FDA) has published  Bisphenol A (BPA): Use in Food Contact Application, in which they state:

  FDA’s current perspective, based on its most recent safety assessment, is that BPA is safe at the current levels occurring in foods. Based on FDA’s ongoing safety review of scientific evidence, the available information continues to support the safety of BPA for the currently approved uses in food containers and packaging.

From the link (above), you can go to the 2014 Updated safety assessment of Bisphenol A (BPA) for use in food contact applications, where they conclude that the "no observed adverse effect level (NOAEL)" for BPA is 5000 $\mu\mathrm{g}$ per kg of body weight per day ($\mu\mathrm{g}$/kg-bw/day) and that exposure estimates are "0.2 $\mu\mathrm{g}$/kg-bw/day (mean) and 0.5
$\mu\mathrm{g}$/kg-bw/day (90$^{\rm th}$ percentile) for the adult US population aged 2 years and older." The estimated dietary intake (EDI) is given as 1.1 $\mu\mathrm{g}$/kg-bw/day for population under 2 years old and 0.5 $\mu\mathrm{g}$/kg-bw/day for population greater than 2 years old. Thus, the exposure levels are on the order of a thousand times less than the NOAEL.

The European Food Safety Authority (EFSA) has recently (January, 2015) revised the tolerable daily intake (TDI) of bisphenol-A (BPA) from 50 $\mu\mathrm{g}$/kg-bw/day to 4 $\mu\mathrm{g}$/kg-bw/day. See this press release and this FAQ for more details.

The summary from the press release is:

  EFSA’s comprehensive re-evaluation of bisphenol A (BPA) exposure and toxicity concludes that BPA poses no health risk to consumers of any age group (including unborn children, infants and adolescents) at current exposure levels. Exposure from the diet or from a combination of sources (diet, dust, cosmetics and thermal paper) is considerably under the safe level (the “tolerable daily intake” or TDI).

The scientific opinion on the risks to public health related to the presence of bisphenol A (BPA) in foodstuffs, used by the EFSA to bolster the down-revision of the TDI for BPA, has been summarized in a very nice form suitable for the lay public, available here.

What about heating?

In one publication (Assessment of Bisphenol A Released from Reusable Plastic, Aluminium and Stainless Steel Water Bottles. J. E. Cooper, E. L. Kendig, and S. M. Belcher. Chemosphere 85(6), 943-947 (2011)), the authors state:

  At room temperature the concentration of BPA migrating from polycarbonate bottles ranged from 0.2–0.3 mg/L. Under identical conditions BPA migration from aluminium bottles lined with epoxy-based resins was variable depending on manufacturer ranging from 0.08 to 1.9 mg/L. Boiling water significantly increased migration of BPA from the epoxy lined bottles. No detectable BPA contamination was observed in water stored in bottles made from Tritan™ copolyester plastic, uncoated stainless steel, or aluminium lined with EcoCare™. The results from this study demonstrate that when used according to manufactures’ recommendations reusable water bottles constructed from “BPA-free” alternative materials are suitable for consumption of beverages free of BPA contamination.

and they conclude

  As shown previously by our results and those of many others, BPA migrates into water stored in polycarbonate plastic, especially when heated to high temperatures. Relatively high levels of BPA are found to be released into drinking water from reusable aluminium bottles with an epoxy-based liner. The amount of BPA released from the epoxy coating was greatly increased by heating. No detectable BPA was observed in water to migrate from unlined stainless steel, EcoCare™ copolyester lined aluminium, or copolyester Tritan™ plastic water bottles. To avoid exposures to BPA unlined stainless steel, copolyester lined aluminium or copolyester plastic drinking bottles should be used.

So, should you be worried? In my opinion, that's largely a function of whether or not you use plastic that contains BPA. A few years ago, it might have been difficult to find BPA-free plastics, but there is no shortage of BPA-free items available now. It's easy enough these days to remove any doubt from the equation by simply using containers that don't contain BPA.


What pulls the electrons from iron in rust formation? In the first step of formation of rust in iron, there is something that pulls electrons from the iron lattice, resulting in the formation of $\ce{Fe^{2+}}$. The reaction is:

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

After that positive iron is repelled from the lattice, it is dissolved into water and a lot of reactions take place.

What is the thing that pulls the electrons from iron? (I guess it must be something positive, right?)

The iron is being oxidized in this process. You just need something to act as an oxidizing agent. Dioxygen will do quite nicely:

$$\ce{O2 + 4e- -> 2O^{2-}}$$


Does sheer amount affect reaction rates? For example, in the iodine and starch reaction, if I use 20 mL of a set concentration of each, will it be slower to completion if I use 40 mL of each at the same concentration?

If you define "completion" by a certain percentage of conversion then no, they will behave exactly the same, provided that things like cooling and stirring are the same in both systems.


Entropy change of a rubber band When I stretch a rubber band, the fibers straighten out causing lesser randomness and more order. This should imply the entropy of the rubber band has decreased, but how does the entropy of the whole universe increase here? Does this mean a rubber band gets colder on stretching? I have seen an experiment where it gets hotter. Doesn't entropy of the whole universe increase only in spontaneous processes? Here I have to pull it hence its not spontaneous? In a nutshell, I'm really confused.

You are treating the rubber band as an isolated system when in fact you are acting upon it.  The entropy increase occurs in your body as glucose is oxidize into water and carbon dioxide to produce the energy needed for your muscles to contract to stretch the rubber band.


Equation for reaction between chlorine fluoride and potassium bromide The equation given in the answer key was

$$\ce{ClF + 2KBr → KCl + KF + Br2}$$

However won't the $\ce{KCl}$ also react with $\ce{ClF}$ to form $\ce{KF}$ and $\ce{Cl2}$, resulting in the net reaction being

$$\ce{2ClF + 2KBr → 2KF + Cl2 + Br2}~?$$

Let's subtract the two reactions from each other to simplify things a bit:

$$\ce{2ClF + 2KBr -> 2KF + Cl2 + Br2}\ \ \ \ \ \ \ \ \ \ [1]$$

$$\ce{ClF + 2KBr -> KCl + KF + Br2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [2]$$

$$\ce{ClF + KCl -> KF + Cl2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ [3] = [1]-[2]]$$

Now we can discuss [2] and [3] separately by splitting them up into half reactions.

Reaction 2

$$\ce{ClF + 2KBr -> KCl + KF + Br2}$$
Potassium is a spectator ion. The reduction half reaction is:

$$\ce{ClF + 2e- -> Cl- + F-}$$

The oxidation half reaction is:

$$\ce{2Br- -> Br2 + 2e-}$$

Reaction 3

$$\ce{ClF + KCl -> KF + Cl2}$$

Potassium is a spectator ion. Fluorine does not change oxidation state. Chlorine undergoes comproportionation (oxidation states +1 and -1 to oxidation state of 0). The reduction half reaction is:

$$\ce{ClF + e- -> 1/2 Cl2 + F-}$$

The oxidation half reaction is:

$$\ce{Cl- -> 1/2 Cl2 + e-}$$

Which reaction happens?

That depends on the reduction potentials, the kinetics of the two reactions and the concentration of species present. I'm not sure how you would figure it out without experimental (or computational) data.

Comparing the oxidation half reactions, you could make an argument about electronegativities of bromine vs chlorine, but I have no insight to compare the two reducation half reactions, which are very different.

The Wikipedia article on chlorine monofluoride states:

  Many of its properties are intermediate between its parent halogens, $\ce{Cl2}$ and $\ce{F2}$.

If you substitute $\ce{Cl2}$ for $\ce{ClF}$ in reactions 2 and 3, they both look reasonable (reaction 3 becomes a self-exchange, i.e. equilibrium constant = 1, whereas reaction 2 reduces chlorine and oxidizes bromine, products are favored).


How should I measure out a solution given in nanorods per ml to make a solution of a set number of mols? I am trying to determine the number of nanorods that I would need to measure out to get a 1 mM solution.  For the gold nanorods that we ordered, it was mentioned that the single nanorods dimensions are 40 nm length by 10 nm in diameter.

The product's concentration is $6.3\times10^{11}$ nanorods per ml.

How would I measure out the solution to obtain a 1 mM gold nanorod solution?

In order to avoid any possible misunderstanding, we should first clarify that “nanorods” (and thus also “nanorods per millilitre”) is not the name of a unit. A nanorod is a very small (so-called nanoscale) object.

The number of nanorods in one millilitre of the given solution is $N_\text{nanorods}=6.3\times10^{11}$, i.e.

$$\frac{N_\text{nanorods}}{V_\text{solution}}=\frac{6.3\times10^{11}}{1\ \mathrm{ml}}=6.3\times10^{14}\ \mathrm{l^{-1}}$$
where $V_\text{solution}$ is the volume of the solution.

Since

$$c=\frac nV$$

and

$$N_{\mathrm A}=\frac Nn=6.022\,140\,76\times10^{23}\ \mathrm{mol^{-1}}$$

where $c$ is concentration, $n$ is amount of substance, and $N_{\mathrm A}$ is the Avogadro constant, the concentration of nanorods in the given solution is

$$\begin{align}c_\text{nanorods}&=\frac{n_\text{nanorods}}{V_\text{solution}}\\[6pt]&=\frac1{N_{\mathrm A}}\cdot\frac{N_\text{nanorods}}{V_\text{solution}}\\[6pt]&=\frac1{6.022\,140\,76\times10^{23}\ \mathrm{mol^{-1}}}\cdot6.3\times10^{14}\ \mathrm{l^{-1}}\\[6pt]&=1.0\times10^{-9}\ \mathrm{mol\ l^{-1}}\\[6pt]&=1.0\times10^{-6}\ \mathrm{mmol\ l^{-1}}\end{align}$$

The given nanorods are approximately cylinders with a length of $l=40\ \mathrm{nm}=4.0\times10^{-8}\ \mathrm m$ and a diameter of $d=10\ \mathrm{nm}=1.0\times10^{-8}\ \mathrm m$, i.e. a radius of $r=5\times10^{-9}\ \mathrm m$.

Therefore, the volume of one nanorod is approximately

$$\begin{align}V_\text{nanorod}&=\pi\cdot r^2\cdot l\\[6pt]&=\pi\times\left(5\times10^{-9}\ \mathrm m\right)^2\times4.0\times10^{-8}\ \mathrm m\\[6pt]&=3.14\times10^{-24}\ \mathrm{m^3}\end{align}$$

The density of gold is $\rho=19.32\ \mathrm{g\ cm^{-3}}=1.932\times10^4\ \mathrm{kg\ m^{-3}}$. Thus, the mass of gold in one nanorod is approximately

$$\begin{align}m_{\ce{Au}}&=\rho\cdot V_\text{nanorod}\\[6pt]&=1.932\times10^4\ \mathrm{kg\ m^{-3}}\times3.14\times10^{-24}\ \mathrm{m^3}\\[6pt]&=6.07\times10^{-20}\ \mathrm{kg}\end{align}$$

The molar mass of gold is $M_{\ce{Au}}=196.97\ \mathrm{g\ mol^{-1}}=0.196\,97\ \mathrm{kg\ mol^{-1}}$. Thus, the amount of gold in one nanorod is approximately

$$\begin{align}n_{\ce{Au}}&=\frac{m_{\ce{Au}}}{M_{\ce{Au}}}\\[6pt]&=\frac{6.07\times10^{-20}\ \mathrm{kg}}{0.196\,97\ \mathrm{kg\ mol^{-1}}}\\[6pt]&=3.08\times10^{-19}\ \mathrm{mol}\end{align}$$

Note that this amount is equivalent to only about $190\,000$ gold atoms since

$$\begin{align}N_{\ce{Au}}&=N_\mathrm A\cdot n_{\ce{Au}}\\[6pt]&=6.022\,140\,76\times10^{23}\ \mathrm{mol^{-1}}\times3.08\times10^{-19}\ \mathrm{mol}\\[6pt]&=1.9\times10^5\end{align}$$

Therefore, the concentration of gold in the given solution can be estimated as follows:

$$\begin{align}c_{\ce{Au}}&=n_{\ce{Au}}\cdot\frac{N_\text{nanorods}}{V_\text{solution}}\\[6pt]&=3.08\times10^{-19}\ \mathrm{mol}\times6.3\times10^{14}\ \mathrm{l^{-1}}\\[6pt]&=1.9\times10^{-4}\ \mathrm{mol\ l^{-1}}\\[6pt]&=0.19\ \mathrm{mmol\ l^{-1}}\end{align}$$

or

$$\begin{align}c_{\ce{Au}}&=N_{\ce{Au}}\cdot c_\text{nanorods}\\[6pt]&=1.9\times10^5\times1.0\times10^{-9}\ \mathrm{mol\ l^{-1}}\\[6pt]&=1.9\times10^{-4}\ \mathrm{mol\ l^{-1}}\\[6pt]&=0.19\ \mathrm{mmol\ l^{-1}}\end{align}$$


pH probe bulb - what is happening within the glass? I am trying to understand how the glass bulb of a pH electrode of a pH meter works - the glass bulb itself. Not the reference electrode or the rest of the electrode (HCl, Ag/AgCl wire, etc...), the math, or the equilibrium yet. For this question just the glass bulb of the pH electrode.

Here is what I have learned so far. This is my current understanding. I'm not saying it is right - but it's where I am right now.

The thin-walled glass bulb has a conductive solution inside, and the outside should be kept in liquid at all times as well. This hydrates a thin layer of the glass on the outside and the inside. I assume these layers are manufactured differently, otherwise, the glass should be uniformly hydrated after a long enough time. It is important that the middle layer of the glass remains very low conductivity so that a potential difference can be maintained; there are likely other reasons as well.

The glass is amorphous and in this case, the exterior layers are somewhat porous, so there is a large volume of Si-O groups exposed to the solution. Protons will stick to these groups and establish a negative charge on the outside of the glass. The number is related to the pH or hydronium concentration of the solution on the outside of the electrode.

edit: I have just started to read this early discussion, where the idea that the glass itself may behave as a sort of buffer:

  Hughes (3) has pointed out that the hydrogen ion concentration in the glass phase may be held relatively constant by the buffer action of the glass which is a mixture of the salt of a weak acid $(\ce{Na2SiO3})$
   with the anhydride of that acid (excess $\ce{SiO2})$.

Note: The hydrated layer is also called a "gel layer", but it is not clear if this is formed naturally as the glass hydrates, or if there is a special gel-enabling material applied to each surface during manufacture.

Do these have to be specially prepared layers of more porous, hydratable glass on the inside and outside of the glass bulb? If so, roughly speaking how is this done? If not, what does limit the depth of the hydrated layer?
When inserted into an acid/base solution, is it just protons diffusing into the hydrated layer of the pH probe bulb by jumping between Si-O sites, or is it the hydronium ions in solution that is diffusing into the glass?
Why is it this often called an ion exchange process? (e.g. not in the Mettler link but in the other two links below, and several random textbooks pulled from a library shelf). Are there Li or Na ions in the glass that are moving? What is being "exchanged"?

below: From A Guide to pH Measurement – the Theory and Practice of pH Applications, Mettler, Toledo.

below: From Theory and Practice of pH Measurement.

below: From The Glass pH Electrode by Petr Vany´sek.

My reference for all information and pictures is Harris' Quantitative Chemical Analysis, 9th ed., pp 347-9. I think it'll be worth your while to consult those pages, but I'll try to summarize the important points here.

An ion-selective electrode is characterized by a thin membrane that, well, selectively binds ions. The glass electrode is an ion-selective electrode for $\ce{H+}$ made of amorphous silicate glass, which consists of connected $\ce{SiO4}$ tetrahedra.

Presumably, no special preparative techniques are required for the glass, and the depth of the hydrated gel layer is mediated by the strength and range of the intermolecular interactions between water and the glass.

Protons are the main ions that bind to the layer, leading to the selectivity of the electrode. They diffuse between the solution and the hydrated gel layer and displace the metal ions originally present on the surface of the glass, which describes an ion-exchange process. Note that they cannot, however, diffuse through the inner glass layer.

A few side remarks.

Equilibrium is reached when the favorable binding of protons to the glass surface is balanced by the unfavorable electrostatic repulsion and chemical potential gradient that result from diffusion into the hydrated gel layer. This provides an equation relating the potential difference to the pH of the solution and allows for $\ce{pH}$ measurement.
Something has to be able to move through the inner layer of the glass membrane to conduct a current and hence allow for a measurement of the potential difference. It turns out that sodium ions can move through this inner layer, but only sluggishly---the resistance of the glass membrane is about $10^8\,\Omega$.


The true shape of p orbitals I'm trying to model  $\mathrm{p}$ orbitals using a 3D program. What confuses me is that there are 2 shapes of $\mathrm{p}$ orbitals. I'm not sure which one is the accurate model. And what is the reason to have 2 types of shape?

Shape 1

Shape 2

There is a misconception here. A p orbital is a 3D-function, and these functions don't have shapes, they have values at any point in space. If you describe an electron distribution with one of these functions, you can plot contours at a chosen value, and these contours have shapes. For example, you could choose a contour level such that an electron has a 90% chance of being inside the shape defined by the contour level.

[OP...] What confuses me is that there are 2 shapes of p orbitals.

There are actually four common graphs depicting p orbitals, and a lot of conceptual drawings that don't have an exact mathematical shape. The graphs differ in whether they show the wave function or the square of it (the latter is proportional to the probability of finding the electron in that location). They also differ in whether they use a cartesian coordinate system, or plot the angular dependence in a polar plot.
The one you probably want is a contour plot of the probability density (you could also say the electron density). Here is an example from an article in the Journal of Chemical Education (center panel):

The shape of the contour lines of the wavefunction are the same, but you can have positive and negative values (and because of the squaring involved, the contour lines also have different values). Your "shape 2" is a contour plot of the wavefunction, with two contours of equal magnitude and opposite sign.
Polar plots
The other type of depiction is a polar plot, where the distance from the origin is the value of a function. This is different from a Cartesian coordinate system, where the nucleus is at the origin, and the distance from the origin corresponds to the distance from the nucleus.
If you plot the angular component of the wavefunction (you can think of this as keeping the distance from the nucleus constant, and sampling different angles), you get what looks like two spheres. A polar plot has no good way of dealing with negative values, so often one part of the curve is shown in a different color.

If you plot the square of the angular component, you get a shape closer to your "shape 1". For technical reasons, the plots are rotated by 90 degrees (the angle zero is defined along the x-axis in Wolfram Alpha, which I used to plot these).

So what does your "shape 1" represent? The left-most panel looks like it wants to show an electron density, but it is very inaccurate. There are other artists' interpretations floating around, some that look like they are actual graphs of functions, such as this one:

It is not clear which function, if any, this shows a graph of.

[OP...] The true shape of p orbitals

There is none. You have a choice of representations, and it depends on the context which one, if any, makes sense.


Why does the sulfone ring in tazobactam open when the lactam is hydrolyzed? One of the drugs I work with is a beta-lactam (4-membered ring with an amide bond) fused to a sulfone ring, tazobactam.

It's relatively stable in water; the lactam is not significantly hydrolyzed without being catalyzed.  When the lactam is hydrolyzed, however, typically by an enzyme, and the lactam ring opened, the sulfone ring also opens cleaving the bond between the sulfur and bridgehead carbon.  A paper by Kuzin gives the following reaction scheme (this is part of figure 2, click for the whole thing)

and says of it (emphasis mine):

  The Ser70-bound moiety (2) is thought to be the initial intermediate [...] Ring opening after departure of the sulfinate from C5 produces a reactive imine (3) and the more inert tautomeric enamine forms (4, 5).

But why is the sulfuryl (sulfone, sulfonyl(?)...) leaving only when the lactam is opened and not spontaneously beforehand?

Other fused bicyclic beta-lactam drugs don't open the other ring, where in lieu of the sulfone there is commonly a thioether (as in penicillin).  The sulfone group seems more electron hungry, but I don't see how the broken lactam bond affects this.

I've heard some people suggest that the lone pair on the nitrogen is able to make this happen, but why doesn't it then happen in the intact drug?

Here's a draft:
First, read this paper. Considering we're under alkaline conditions (blood pH between 7.35-7.45) we'll take their alkaline mechanism proposal.
Mechanism

This is obviously incomplete, with good reason. Without knowing what the active site of the enzyme looks like (except for Ser70), it's difficult to devise of a mechanism. I could imagine another alkoxy attacking the sp3 carbon between S and N, and shifting two electrons to the sulfur thus generating your SO2- in one step, but that would require a proper active site.
The geometry of the molecule inside the active site of the enzyme is obviously optimal for this reaction to proceed. I'd like to know more. For some reason my school doesn't give me access to this paper.


K in Gibbs free energy When we write for an equilibrium that $$\Delta G^\circ = -RT \ln(K)$$  Which type of  $K$ , ie is it $K_\text{concentration}, K_\text{pressure}$ or $K_{\chi}$ because all these have different numeric values which will give different values of $\Delta G^\circ$ which is obviously not possible.

Edit: In the question marked as being duplicate of, they still don't tell which $K$ it is. Even if we divide by the appropriate standard value, and make it unitless, $G$ will still give different values based on $K_\text{concentration}$, $K_\text{pressure}$ or $K_{\chi}$.
Also can $K_{\chi}$ also be used as it is dimensionless less too?

$K$, the standard thermodynamic equilibrium constant, is computed from $\Delta G^\circ$, using

$$\Delta G^\circ = -RT\log K \tag1$$

Generally speaking, $K$ in equation (1) is unitless. Its value depends on the specified reference standard states and $T$ (and obviously on the equilibrium activities of reactants and products).  Both $K$ and $\Delta G^\circ$ change (and are certainly allowed to) if you change the choice of standard states for any species.

You can also compute an equilibrium constant from a value of $\Delta G^\circ$. To do this reverse operation you need to know the standard state ($p^\circ=\pu{1 bar}$, $c^\circ=\pu{1 M}$, or $m^\circ=\pu{1 molal}$, $\chi^\circ=\pu{1}$) associated with each species in the equilibrium equation. At equilibrium each product/reactant is at the same temperature and in the same phase as in its associated reference standard state (if not at the same partial pressure, concentration, or mole fraction).

When you use either $K$ or $\Delta G^\circ$ in practice, you need to know the standard state of each species (this is often apparent from additional information that is provided with these parameters). If you are working with a gas-phase reaction then you can refer to $K$ as $K_p$. If you are using standard states then all reference states are $\pu{1 bar}$ $^\dagger$, and partial pressures computed directly from $K_p$ are in $\pu{bar}$ units (these can obviously be converted later).

To compute $K_c$ from say $K_p$ you have to convert units with an appropriate conversion factor. This is described in the linked post. Same applies to problems where $K_\chi$ is provided, where concentrations are described in mole fraction units. Note that if you change units, for instance to compute $K_c$, then you have (perhaps implicitly) changed the reference state (perhaps to a nonstandard state), and in that case $\Delta G^\circ$ also changes. This is also explained in an answer to the linked post.

$\dagger$ Components whose activity is constant can usually be ignored


What does "reaction coordinate" mean? What does it mean when a reaction has TWO reaction coordinates? I'm familiar with the concept of a reaction coordinate from high school chemistry as some generic conformational parameter that all of the intermediate states of a chemical reaction lie upon. Recently, though, I've seen some presentations where chemists presented plots of the free energy with respect to two reaction coordinates, and my understanding is that the plot was of a measurement (I could be wrong, though).

What exactly does the reaction coordinate mean? What does it mean to have more than one reaction coordinate for the same reaction? Is the free energy as a function of the reaction coordinate a measurable quantity?

The reaction coordinate is the progress of a reaction from reactants to products with various intermediates and transition states in between. It is an abstraction. It has no relation to time. Rather it is the progress of bond-forming and bond-breaking reaction steps. The free energy change of partially formed and partially broken bonds cannot be measured. The free energies of transition states (local maxima) cannot be directly measured. These states are transient and cannot be directly observed. For a simple reaction with few mechanistic steps, the activation energy (transition state energy) can be calculated from kinetics measurements. The free energy of intermediates can often be determined directly or indirectly if they are long-lived. Most of these diagrams are assembled from a few measured data points and either extrapolation or theoretical computation.

If two reaction coordinate axes are presented, this often means that there are two separate pathways the reaction can take, and each pathway is presented on one of the two reaction coordinate axes. Perhaps the reaction involves two different steps, but those steps need not happen in the same order every time. As a generic example, consider nucleophilic substitution reactions in organic chemistry: A generic nucleophile ($\ce{Nu-}$) reacts with an organic molecule R-X, where X is a leaving group, and R is the rest of the molecule (which is unchanged or nearly unchanged). The overall reaction would be

$$
\ce{Nu- + R-X -> Nu-R + X-}
$$

This process involves two steps: forming the bond between Nu and R, and breaking the bond between R and X. These two steps can happen in three different orders: 1) the R-X bond breaks first and the R-Nu bond forms second (this is the dissociative or $\mathrm{S_{N}1}$ mechanism); 2) The R-Nu bond forms first and the R-X bond breaks second (this is the associative or $\mathrm{S_{N}Ac}$ mechanism); and 3) The R-X bond breaks and the R-Nu bond forms at the same time (this is the intermediate or $\mathrm{S_{N}2}$ mechanism). All three of these cases occur in reactions that are covered in any undergraduate organic chemistry text. These three processes would look like the following. The items in parentheses are transition states with partially formed/broken bonds ($\ce{\bond{....}}$).

Dissociative:

Step 1: $\ce{Nu- + R-X -> (Nu- + R\bond{....}X) -> Nu- + R+ + X-}$
Step 2: $\ce{Nu- + R+ + X- -> (Nu\bond{....}R + X-) -> Nu-R + X-}$

Associative:

Step 1: $\ce{Nu- + R-X -> (Nu\bond{....}R-X) -> [Nu-R-X]-}$
Step 2: $\ce{[Nu-R-X]- -> (Nu-R\bond{....}X) -> Nu-R + X-}$

Intermediate:

Only step: $\ce{Nu- + R-X -> (Nu\bond{...}R\bond{....}X) -> Nu-R + X-}$

A reaction coordinate energy diagram for a substitution process could then be graphed over two different reaction coordinates: breaking the R-X bond and forming the R-Nu bond.  For ease of two dimensional representation, these are often presented as contour diagrams. Various trajectories could be taken through this space based on what real species R, Nu, and X represent. For example, if R+ is a stabilized cation, then the dissociative pathway becomes more likely. These plots are called More O'Ferrall-Jencks plots. This link will take you to the wikipedia article, which is not a very good article. The references in that article to the papers by R. A. More O'Ferrall and W. P. Jencks are probably better places to go to find out more. Most advanced undergraduate or graduate level textbooks of physical organic chemistry, particularly Advanced Organic Chemistry, Part A: Structure and Mechanisms by Carey and Sundberg, will cover this topic, at least as it relates to organic chemistry.


Sodium azide to Nitrogen ratio in airbags I've been given this question and I don't know how to solve it.

Airbags are designed to protect the occupants of a car during a collision. They use the decomposition of sodium azide, $\ce{NaN3}$. Three chemical reactions occur rapidly after impact and produce nitrogen gas ($\ce{N2}$) to fill the airbag.
The equations summarise the overall process that takes place.

$\ce{2 NaN3 -> 2 Na + 3 N_2}$
$\ce{10 Na + 2 KNO3 -> K2O + 5 Na2O + N2}$
$\ce{K2O + Na2O + 2 SiO2 -> K2O3Si + Na2O3Si}$

What is the ratio between the number of sodium azide ($\ce{NaN3}$) molecules used and the number of nitrogen ($\ce{N2}$) molecules produced in the overall process?

I initially just used the first equation to get a ratio of 2:3, but obviously that's not right. I have no clue where to start. How do you do this? (How this can be considered a Year 10 question baffles me)

Assuming you have an excess of $\ce{KNO3}$, the ratio of $\ce{NaN3}$ consumed to $\ce{N2}$ produced would be 5:8.

Let's renormalize the first two equations, dividing the first through by 2, and the second through by 10.  By doing this we can see, directly, that in the overall reaction, 3/2 + 1/10 = 8/5 = 1.6 molecules of $\ce{N_2}$ are produced for every molecule of  $\ce{NaN3}$ consumed.  If you prefer an integer ratio, it's 1:(8/5) = 5:8.

  $\ce{NaN3 -> Na + 3/2 N_2}$
  $\ce{Na + 1/5 KNO3 -> 1/10K2O + 1/2 Na2O + 1/10 N2}$


Why aren't resorcinol and catechol classified as dihydric alcohols? Why aren't resorcinol and catechol classified as dihydric alcohols? They contain two alcohol groups.

As @Loong pointed out in his comment, resorcinol and catechol are phenols, not alcohols. Although both classes of compounds contain an O-H group, they are different in that the oxygen in phenols is bonded to an arene ring such as benzene or napthalene.
Importantly, a lone pair on the phenol oxygen is able to interact with the aromatic $\pi$ system via resonance, which gives phenols characteristically different properties to normal alcohols.

The lone pair delocalisation increases the electron density in the ring dramatically at the ortho and para positions, as indicated by the negative charge at these positions in the resonance structures. This makes these positions much more reactive towards electrophiles than they are in unsubstituted benzene.
The lone pair delocalisation also occurs in the phenoxide ion, the conjugate base of phenol, which makes it much more stable than a regular alkoxide ion, and therefore makes phenols much more acidic than alcohols.


Does the "Bat Gen" claim to produce more energy than it uses? The following links claim that inventor Arturo Solis Herrera has created an "everlasting battery".

http://www.mexiconewsnetwork.com/news/bat-gen/
http://www.faqs.org/patents/app/20090134007#ixzz10Sh9iLQM

The article claims that:

  Bat gen is the name of this ever-lasting battery. It consists of a biochemical process triggered by mixing water and melanin, which results in a substance capable of separating oxygen from hydrogen (the components of a water molecule), therefore liberating energy.

  The process continues as the same molecule brings together both elements turning them into water again; as a result one more energy load is triggered.

  According to Solis’ research, once this process is reached, it can continue for 100 years! Moreover, there are several ways of artificially producing the melanin found in human nails, hair and retina; two of them, which he already patented, are based on vegetables and oil.

Although they do not claim that the battery is perpetual, they seem to imply that mixing water and melanin can give an (oscillating?) reaction that can last for 100 years, providing cheap power.

Does this have any chemical basis? What matter is consumed?
Would that work without the user needing to add anything to the battery or is the article just describing the process in a misleading way?

I'm not sufficiently familiar with the chemistry involved to state authoritatively that this guy is full of bologna, but there are some pretty fantastic claims made in the second link.  Also, if that's a US patent application, it's a disaster.

The device is able to use any light from 900-200nm, mechanical agitation, sound, or magnetic fields (and unspecified "other") as sources of power.
He has also made all of his cells inside Tupperware containers sealed by silicone caulking, so the 100 year claim is a bit sketchy.
The inventor's testing methodology is pretty dubious as well.  The device was placed in sunlight for 30 minutes and it experienced a change of pH from 7.3 to 7.1, therefore it works.  Also, in his own words:

  It is worth saying that such variation from 0.2 to 1.0 pH unit; and the reverse reaction that occurred once the containers were placed in a dark place, was anticipated by our theoric background, when we performed the experiments, we known already what was going to happen, as the matter of fact, we didn't do many experiments, we only made two or three times, obtaining the results we expected.  {All grammar and spelling mistakes are those of the original author(s).}

That said, the idea of a cell that can catalytically break down water into hydrogen and oxygen using sunlight and then recombine them into water, extracting the energy, is not impossible by any stretch of the imagination.  Designing a cell that is economically viable compared with silicon based devices has yet to be accomplished, to the best of my knowledge.


How can I safely store highly outgassing self-pressurizing chemicals? I have a standard copper etching solution of 1 part $\pu{10M}$ $\ce{HCL}$ and 2 parts 3% $\ce{H2O2}$ (as described in numerous places online).

I previously had this stored in a tightly-sealed glass bottle, as described in the instructions I followed and general acid storage instructions.

However, my solution seems to be outgassing hydrogen or chlorine gas quite significantly, and when I opened the sealed container a few days later it violently liberated much of its contents into my face. (I was wearing a partial face shield and immediately went under the shower, so no ill effects were observed).

I know that for some chemicals that self-pressurize, such as cryogenic nitrogen, the storage method is essentially constant venting - is that what I'll have to resort to?

Bottles for 30% or stronger $\ce{H2O2}$ usually have vented caps to prevent sudden out-gassing, splashing and boil-over that you experienced, as well as to prevent bursting the container. The vent may be as simple as thinned spot, which may be scored or slitted, or it may have a polythelene, PTFE or other sponge or "paper" separator that helps retain liquids while allowing gas to escape.

In either case, I would not suggest storing a mix of $\ce{H2O2}$ and $\ce{HCl}$ for more than a few hours for a few reasons.

As gases evolve, the solution becomes weaker and less effective an etchant.
Stored on a shelf, the gases will corrode the shelving surprisingly quickly. You would not want shelving with caustic or toxic chemicals to collapse.
If gases given off are other than $\ce{O2}$, you probably don't want to breathe them for long.

Ferric chloride is a slow etchant at room temperature, but gentle warming speeds etching greatly (PDF). At 54 °C, $\ce{FeCl3}$ removes ~35 um of copper per minute, so standard PC boards ("0.5 oz" - "1 oz") should be etched in ~1 - 2 minutes.

The biggest nuisance I find using $\ce{FeCl3}$ is that it tends to stain containers, tables and hands, but Iron Out and similar bisulfite or hydrosulfite stain removers can be used on cleanup.


How do I synthesize 3-hydroxydihydrofuran-2-one from formaldehyde? I want to synthesize 3-hydroxydihydrofuran-2-one from formaldehyde but I do not know of a method. I tried a Mannich reaction and lactonization after dehydration.

Starting from pyruvic acid which is readily available and not very expensive: react pyruvic acid (only one enolizable side) with a Mannich base, then react the resulting amide with $\ce{MeI}$ and base to obtain the exo-methylene product. Brominate the exo-methylene. Now cyclization will prefer the five-membered ring over the strained four-membered ring. Reduce the ketone with a reletively mild reducing agent like $\ce{NaBH4}$ and then eliminate $\ce{HBr}$ in a E1cB type reaction, yielding 3-hydroxydihydrofuran-2-one.

This synthesis involves a Mannich reaction as you asked, but it can be done much simpler without it I think.

Note: I'm not an expert, just an enthusiastic amateur. Before trying anything in the lab  make sure your synthesis is checked by a supervisor.


Why is cyclooctatetraene non planar but the cyclooctatetraenide anion planar? I was reading some organometallics and I came upon the preparation of uranocene, $\ce{U(C8H8)2}$. It has two cyclooctatetraenide anions, $\ce{C8H8^2-}$, as ligands, with a uranium atom between them.

It was mentioned that the neutral cyclooctatetraene molecule $\ce{C8H8}$ is non aromatic and has a tub-shaped conformation, but the dianion $\ce{C8H8^2-}$ is planar but aromatic. Why is this so? Can fulfilling Hückel's rule force a cyclic molecule to become planar?

Fulfilling Hückel's rule means that you get to be aromatic.

The reason cyclooctotetraene is non-planar is because its planar arrangement would be antiaromatic. It twists to avoid it, because antiaromatic compounds are unstable with respect to localisation of the π-electrons (see here: What is the justification for Hückel's Rule?)

The veritable nanosecond that a molecule has the potential to become aromatic though, it will leap at the opportunity. Any angle strain from being planar is so quickly overtaken by the huge jump in stability that aromaticity brings that the molecule will twist to form an aromatic structure in a blink.


What is the molecular geometry of NHF₂? I had to draw the molecular shape for $\ce{NHF2}$ as a Trigonal Planar like this

But when I check to make sure it was right, I came across a source saying

  NHF2-difluoroammonia--tetrahedral, almost trigonal
  (Source 1)

After looking some more I found a source that now said it was

  Trigonal pyramidal, Bond angle of 107 degrees Bond angle of F-N-F > F-N-H
  (Source 2)

Can someone clarify which one it is so I can understand.

You can look up the molecule on chemspider, where you have a little applet for the 3D structure. Or you can download a coordinate file from NIST and view it in a molecular viewer, like Avogadro. Or keep on reading for some deeper insight.

As Klaus already pointed out, if VSEPR is a valid concept, one would arrive at the conclusion, that the molecule is trigonal pyramidal, like ammonia. I performed a DF-BP86/def2-SVP calculation on this to analyse the electronic structure a bit more.

One reason why VSEPR does fail (a lot) is that lone pairs are more stable in orbitals with high s character. Another point that is often observed is that orbitals with higher p character are directed to more electronegative elements (Bent's Rule).[1,2] As a result of that, the bond angle between electronegative elements gets smaller. This can be observed here, too. From the ideal sp³ arrangement we would expect an 109.5° angle, but here it is about 5° smaller. As a consequence also the hydrogen-nitrogen-fluorine angle has to decrease. In this case the bond angle becomes much smaller because of an intramolecular hydrogen bond.[3]
The NBO analysis consequently builds the same picture:
$$\begin{array}{llll}\hline
\text{type}&\text{main contribution}&\text{other contributions}\\\hline
\sigma\text{-Bd}~\ce{N-F}& 64\%~\ce{F}~(14\% s + 86\% p) & 36\%~\ce{N}~(15\% s + 85\% p) \\
\sigma\text{-Bd}~\ce{N-H}& 68\%~\ce{N}~(25\% s + 75\% p) & 32\%~\ce{H}~(100\% s) \\
\sigma\text{-LP}~\ce{N} & 100\%~\ce{N}~(52\% s + 48\% p) &  \\
\sigma\text{-LP}~\ce{F} & 100\%~\ce{F}~(75\% s + 25\% p) &  \\
\sigma\text{-LP}~\ce{F} & 100\%~\ce{F}~(10\% s + 90\% p) &  \\
\sigma\text{-LP}~\ce{F} & 97\%~\ce{F}~(100\% p) &  2\%~\ce{N}~(7\% s + 93\% p) & 1\%~\ce{F2}~(15\% s + 85\% p)\\\hline
\end{array}$$

The planar (trigonal) structure on this level of theory is about 33 kcal/mol higher in energy. It is the transition state for the nitrogen inversion, which is therefore very slow.

Notes:

Bent's rule on wikipedia
Bent's rule on our network: Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?
The bond length is much smaller than the sum of the corresponding van der Waals radii, suggesting a hydrogen bond. Compare A. Bondi J. Phys. Chem., 1964, 68 (3), pp 441–451.
$r_{\mathrm{vdW}}(\ce{HF})=r_{\mathrm{vdW}}(\ce{H})+r_{\mathrm{vdW}}(\ce{F})= (106 + 140)~\mathrm{pm}= 246~\mathrm{pm}$


What will happen when a metal reacts with an acid which is neither dilute nor concentrated? We have always studied about metals reacting with dilute and concentrated acids, but I do not understand what will happen if the same metal reacts with an acid that is neither concentrated nor dilute?

This question is a bit vague, but briefly, the reason some acids behave differently at high and low concentration is because changing the activities of components in the solution changes the redox potential of the various possible reactions.

Let's take the reaction of iron with sulfuric acid as an example. This is not necessarily the only factor and I'll make a few assumptions that are likely quite inaccurate to simplify the problem, but it will illustrate the situation:

Let's assume that the half-reaction with iron is always the same—iron is being oxidized:

$\ce{Fe_{(s)}<=> Fe^2+_{(aq)} +2e-}\ E^\circ=+0.44\ \mathrm{V}$

Now, if you look at a table of standard reduction potentials, you'll see that there are a couple of potential half-reactions that could be reduced to drive the iron oxidation:

$\ce{2H+_{(aq)} +2e- <=> H2_{(g)}}\ E^\circ=0\ \mathrm{V}$

$\ce{SO4^{2-}_{(aq)} +4H+_{(aq)} + 2e- <=> SO2_{(g)} +2H2O_{(l)}}\ E^\circ=+0.17\ \mathrm{V}$

If all these reagents were at unity activity, then the second reaction would be more favoured and we would see mostly $\ce{SO2}$, but the activities of the substances in solution are also important. By the Nernst equation, $E=E^\circ -\frac{RT}{zF}\ln Q$, we see that the potential of a complete cell or a half-cell reaction is dependant on the reaction quotient $Q$. Because we're assuming that only the one reaction happens to the iron, we can just look at the reduction half-cell. If we assume that the sulfuric acid dissociates completely to $\ce{SO4^2-}$, then $[\ce{H+}]\approx 2\cdot[\ce{SO4^2-}]$. If we assume the gaseous products of the reaction don't affect the equilibrium, for our two reactions, $Q$ is $\frac{1}{[\ce{H+}]^2}=\frac{1}{(2[\ce{H2SO4}])^2}$ and $\frac{1}{[\ce{H+}]^4[\ce{SO4^{2-}}]}=\frac{1}{(2\cdot[\ce{H2SO4}])^4[\ce{H2SO4}]}=\frac{1}{16\cdot[\ce{H2SO4}]^5}$, respectively.

Now, if we substitute these into the Nernst equation for different concentrations of acid, we can see which reaction is more favourable: In dilute acid (1 mM), we get −160 mV for hydrogen formation and −238 mV for sulfur dioxide. The picture changes as the acid gets more concentrated. In 1 M acid, the hydrogen formation half-cell has a potential of 18 mV, but sulfur dioxide is now up to 206 mV.

So you can see how the concentration of acid changes which reaction is more favourable. To answer your question about what happens in the middle, one need only do the calculation for a moderate concentration and find that at some point, the two half-reactions become close in potential and we may see a significant mixture of both reactions occurring.

This is a very simplified picture and the Nernst equation is only technically valid under conditions of no current flow, meaning that other factors may influence what is observed experimentally, but I hope it's apparent that there's not really a solid line between dilute and concentrated. Many different reactions can occur simultaneously, even if a single reaction is dominant under certain conditions.


Percentage of carbon in stainless steel to avoid brittleness What is the maximum percentage of carbon allowed in stainless steel 202 so that it does not cause brittleness?

All  standardised grades of alloys have definite limits of composition.
If its common, then a quick search will find it. 0.15% is the specified limit for SS202.

http://www.azom.com/article.aspx?ArticleID=8209

Whether its brittle or not depends on what you mean by brittle and its thermal processing. Its a martensitic stainless steel so note the details in azom and that it will be more brittle when hardened as described.


Why are the ethyl ester methylene protons not diastereotopic in ethyl (+-)-2-methylbutyrate? In the following compound, there is one chiral center and two methylene groups.  Why are the protons on the ethyl ester methylene not diastereotopic?

You are correct to conclude that they are diastereotopic. There appears to be a bit of splitting evident if you look closely at the NMR, but it might be expected to be very small given the distance between the ethyl group methylene protons and the chiral center. But even if the distance were greater and the resulting splitting too small to be observed, the protons would still be technically diastereotopic.


What happens if I draw too much current from a home made battery? If I try to make a battery like one of these: https://www.wikihow.com/Make-a-Homemade-Battery And then I short the two terminals, what will happen?

Disregarding anything getting hot and burning up, I'm curious what happens to the chemical reaction. I assume there is a max speed at which the reaction can occur because it seems unlikely that shorting the wires would give "infinite" current and use up the reaction in an instant.

My assumption is that the current will max out and the voltage will drop to basically nothing. Is this true? And if so what is actually limiting the rate of the reaction? Is this the same chemical mechanism as a store bought battery or are they different somehow?

Yes, you're right. There will a maximum speed that is determined by the kinetics of several parts of the system.

Electronic Resistance: Pretty straight-forward. All the parts have some inherent resistance. Though this tends to be minial compared to some of the other causes.

Migration: At high currents, the ions become depleted (or in excess) at the electrode's surface. This means that the reaction will not run at the same potential until new ions diffuse from the bulk.

Adsorption: In order for the redox reaction to occur, there is often a step needed where the ions adsorbs to the metallic surface. This is a sort of chemical reaction, so has a rate constant associated with it.

Chemical Reaction: Some batteries can also involve a chemical reaction to form an activated reactant.

Electron Transfer: This is the actual redox part itself. Transferring an electron from the metal to the reductant, or vice-versa.

All of this will play a role, but I mostly think of migration effects when I think of kinetic limitations. Once that surface layer is depleted, all of the other processes stop too.

This is most likely to be worse for a home-made battery. The electrical resistance will probably be higher because there are longer wires, less reliable construction, etc. Distances between the electrodes is longer, so in the long run there's more distance for ions to travel to equalize a charge imbalance. Your electrode surfaces are likely to be dirtier, causing slower absorption.

The book Electrochemical Methods by Alan Bard and Larry Faulkner is a good resource if you want to do a deep dive.


Total pressure v/s Partial pressure If I have 2 containers, something like in the diagram. One has gas A, while other has gas B. Both containers have equal volume and at same temp. Then when I remove the corks, and let the gases flow, until when will they flow, and will I finally have a half of Gas A in one container and rest half in the other, and similarly for gas B, or will only one of them flow until total pressure in the 2 containers is same?

In other words, what is the necessary condition for flow of gas - Difference in its partial pressure, or difference in total pressure?

If the membrane is permeable to both gases, then both gases will become uniformly distributed between the two containers, and the final pressure will be uniform.  So, if the initial pressures in the two containers are $p_1$ and $p_2$, the initial number of moles in the two containers are $$n_1=\frac{p_1V}{RT}$$ and $$n_2=\frac{p_2V}{RT}$$  The total number of moles does not change, so the final pressure is $$p=\frac{(n_1+n_2)RT}{2V}=\frac{p_1+p_2}{2}$$The final mole fractions in the combined container are $p_1/(p_1+p_2)$, and $p_2/(p_1+p_2)$.  The final partial pressures are $p_1/2$ and $p_2/2$.


How significant is air decomposition on formation of nitrogen oxides? How significant is $\ce{NO$_x$}$ formation from $\ce{O2}$ (reactant) and $\ce{N2}$ decomposition from simply heating the two as opposed to from $\ce{CO2}$, $\ce{H2O}$, (and $\ce{N2}$) dissociation? In other words, which of the two pathways for $\ce{NO$_x$}$ formation is more significant (I imagine this will vary with equivalence ratio):

$\ce{O}$ and $\ce{N}$ radical from $\ce{N2}$ and $\ce{O2}$ dissociation
$\ce{O}$ radical from $\ce{CO2}$ and $\ce{H2O}$ dissociation with $\ce{N}$ radical

My feeling is that the second becomes more significant as equivalence ratio increases (more fuel means more $\ce{O2}$ is consumed in combustion).

In situations where NOx originiates form $\ce{N2}$, such as lightning or a high temperature furnace, the Zeldovich mechanism dominates:

$$\ce{O2 -> O +O}$$

$$\ce{O + N2 -> NO +N}$$

$$\ce{O2 + N-> NO +O}$$

If a combusting fuel actual contains nitrogen, other mechanisms can be involved.


Difference between freezing point and melting point I do not understand why for mixtures freezing point is lower than melting point. Are not the two representing the same equilibrium between solid and liquid which we are approaching from two different sides, so why should the temperature at which this happens be different for the two cases?

The Bodner research group at Purdue University has posted this excellent description of the difference between melting point and freezing point:

Pure, crystalline solids have a characteristic melting point, the temperature at which the solid melts to become a liquid. The transition between the solid and the liquid is so sharp for small samples of a pure substance that melting points can be measured to $\pu{0.1^\circ C}$. The melting point of solid oxygen, for example, is $\pu{-218.4^\circ C}$.
Liquids have a characteristic temperature at which they turn into solids, known as their freezing point. In theory, the melting point of a solid should be the same as the freezing point of the liquid. In practice, small differences between these quantities can be observed.
It is difficult, if not impossible, to heat a solid above its melting point because the heat that enters the solid at its melting point is used to convert the solid into a liquid. It is possible, however, to cool some liquids to temperatures below their freezing points without forming a solid. When this is done, the liquid is said to be supercooled.
An example of a supercooled liquid can be made by heating solid sodium acetate trihydrate ($\ce{NaCH3CO2 . 3H2O}$). When this solid melts, the sodium acetate dissolves in the water that was trapped in the crystal to form a solution. When the solution cools to room temperature, it should solidify. But it often doesn't. If a small crystal of sodium acetate trihydrate is added to the liquid, however, the contents of the flask solidify within seconds.
A liquid can become supercooled because the particles in a solid are packed in a regular structure that is characteristic of that particular substance. Some of these solids form very easily; others do not. Some need a particle of dust, or a seed crystal, to act as a site on which the crystal can grow. In order to form crystals of sodium acetate trihydrate, $\ce{Na+}$ ions, $\ce{CH3CO2-}$ ions, and water molecules must come together in the proper orientation. It is difficult for these particles to organize themselves, but a seed crystal can provide the framework on which the proper arrangement of ions and water molecules can grow.
Because it is difficult to heat solids to temperatures above their melting points, and because pure solids tend to melt over a very small temperature range, melting points are often used to help identify compounds. We can distinguish between the three sugars known as glucose ($\vartheta_\mathrm m = \pu{150^\circ C}$), fructose ($\vartheta_\mathrm m = 103{-}\pu{105^\circ C}$), and sucrose ($\vartheta_\mathrm m = 185{-}\pu{186^\circ C}$), for example, by determining the melting point of a small sample.
Measurements of the melting point of a solid can also provide information about the purity of the substance. Pure, crystalline solids melt over a very narrow range of temperatures, whereas mixtures melt over a broad temperature range. Mixtures also tend to melt at temperatures below the melting points of the pure solids.


Antoine law and ideal gas law I have the following question:

  A $100\ \mathrm{ml}$ sealed flask with $60\ \mathrm{ml}$ acetonitrile is heated to $140\ \mathrm{^\circ C}$. The boiling point of acetonitrile is $82\ \mathrm{^\circ C}$. What will be the pressure in the flask?

1) I understand I can use the Antoine equation, $\log{p} = A-B/(C+T)$, to calculate the vapor pressure of acetonitrile, but that equation only applies if there is some liquid acetonitrile left in the flask. How do I know if this is the case, or instead all acetonitrile has boiled and is in the vapor phase?

2) The total pressure is the sum of the partial pressures of acetonitrile and air. I can calculate the air pressure inside the flask with the ideal gas law, PV=nRT. However, I'm not sure what is the volume? Is it the total volume minus the acetonitrile volume ($100\ \mathrm{ml}-60\ \mathrm{ml}=40\ \mathrm{ml}$), or is it $40\ \mathrm{ml}+x$, where $x$ is the additional volume due to the conversion of liquid acetonitrile into vapor?

Any ideas are welcome.

Assume that the initial temperature is room temperature and neglect the initial amount of acetonitrile in the gas phase.  Calculate the equilibrium vapor pressure of acetonitrile at 140 C and compare it with the partial pressure you would calculate from the ideal gas law if all the acetonitrile had evaporated (so that its volume is 100 cc).  If the latter is greater than the former, then there will still be some acetonitrile remaining as liquid.  The partial pressure of the acetonitrile in the gas phase will then be equal to the equilibrium vapor pressure.  Assuming that x gm of acetonitrile have evaporated, you can then calculate, in terms of x, the new mass and volume of the liquid and the new volume of the head space.  You can then determine x from the requirement that the total amount of acetonitrile has not changed.


How does common salt speed up the setting of the plaster of Paris? We use plaster of Paris (POP) to cast and lift foot prints/tyre prints from the crime scene for evidence purpose. While preparing cast, we follow standardized protocol and add a pinch of common salt to the POP solution. It catalyzes thickening of the POP. I am keen to know chemistry behind it. How does salt react with POP?

My guess would be that it absorbs the water in the plaster of paris. Plaster of paris normally has what's called 'water of crystallization' in it. However, it sets too fast with just this, so extra water is added. The salt absorbs that extra water, thereby making it set faster.
Here's a website with details about plaster of paris chemically: http://collections.infocollections.org/ukedu/en/d/Jg16foe/7.4.html
and a book called 'Science in the Real World: A simplified story of how technology using chemistry and physics is used in the real world of industry' has good information on it as well.


Alkali and alkali earth metal carbonates and bicarbonates decomposition Which of the following can not decompose on heating to give $\ce{CO_2}$?

$\ce{Li_2CO_3}$
$\ce{Na_2CO_3}$
$\ce{KHCO_3}$
$\ce{BaCO_3}$

What I have read till now: Lithium carbonate is not so stable to heat and bicarbonates of alkali metals decompose at approximately $\pu{300 ^\circ C}$. The stability of alkaline earth metals increases down the group and they decompose on red heating.
This only leaves us with one option as correct (2nd), but what is the reaction of decomposition of $\ce{Na_2CO_3}$?

From your question, the heating temperature is unclear. See these links for sodium and lithium carbonates, the hydrocarbonate or barium.

They are all also mentioned on Wikipedia.

In other words, they all seem to decompose at some point, so what is meant by heating?

The reaction for sodium carbonate decompositon at melting point should be:

$$\ce{Na2CO3 -> Na2O + CO2}$$


How does the addition of NaCl to the cathode side effect the potential of a voltaic cell: Zn(s) | Zn2+(aq) 0.1M || Cu2+(aq) 0.1M | Cu(s) By addition of $\ce{NaCl}$ to the $\ce{Zn}$ side of the cell, I observed precipitation of $\ce{ZnCl2}$.
I measured the voltage to $\pu{1.05 V}$ before adding $\ce{NaCl}$ to the cathode side, I then observed the voltage go down by addition of the salt.
According to the Nernst equation: as the $\ce{Zn^2+}$ ion concentration reduces with precipitation, there should be an increase in voltage.
$$E = 1.05 - \frac{RT}{nF} \ln Q$$
$$E= 1.05 - \frac{RT}{nF} \ln \left(\frac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]}\right)$$
since $\ln Q \lt 0$ for $[\ce{Zn^2+}] < [\ce{Cu^2+}]$
"$-\frac{RT}{nF} \ln Q$" term would be positive. Therefore the voltage should go up. But the opposite was observed and I don't know how to explain it.

@JR19. If you modify the anode compartment, and if you then observe that this change produces a decrease in the cell voltage, it means that [$\ce{Zn^{2+}}$] has increased. It is probably an effect of "activity". At high concentration, concentration must be replaced by "activity". The voltage delivered by the electrodes is not really proportional to the log of the concentration, but it is proportional to the log of the so called "activity". I explain.
Let's consider the case of $\ce{HCl}$, which is best known. If you measure the pH of $0.01$ M HCl, you find pH = $2.0$. If now you add and dissolve some $\ce{NaCl}$ to this solution, the pH goes down to $1.5$ and even to $1.2$, as if the solution gets more and more concentrated. It is hard to believe, but easy to check. Do it in the lab, if you can. Here is an explanation taken from an old issue of J. Chem. Educ. (I don't remember which one). The measured "concentration" given by an electrode is the number of moles divided NOT by the volume of the solution, but divided by the volume of the "free water", namely the volume of water not engaged in adsorption around the ions. This sort of "new concentration" is called activity. By adding more $\ce{NaCl}$, the volume of free water decreases, so that the activity of $\ce{HCl}$ increases. In $1$ liter concentrated solution of $\ce{NaCl}$, there is only about $0.2$ liter of free water. The activity is much higher than the concentration. It must be the same phenomena in your cell.