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Can different groups within a single molecule photoisomerize at different wavelengths?
Azobenzene undergoes trans → cis photoisomerization when irradiated with light with a wavelength of 300–400 nm.
By adding electron withdrawing groups, the wavelength can be red-shifted.
If I were to make the molecule shown below, would it be possible to selectively photoisomerize around each of the $\ce{N=N}$ at different wavelengths?

Yes you can increase wavelength  by substituents and by increasing conjugation length also,  for example a simple polyene such as  butadiene  vs.  carotene or  a substituted polyene such as retinal. But no, you cannot just excite one bond  in the whole conjugated part of the molecule. Using tailored femtosecond pulses there has been much effort to do 'bond selective' chemistry but what the results mean is still hotly disputed.

I don't remember the reference any more, but there was a way to have selectivity that was between cis- and trans- double bonds. I think as long as the energy difference is big enough relative to the spread in your input light, you should be able to have some selectivity. The only thing to watch out for is that as you increase substitution, you're going to start disfavoring the cis- form, which I gather is what you want to make.

Related: http://chemistry.stackexchange.com/questions/31730/photoisomerization-of-azobenzene/31734

Thanks Porphyrin and Zhe. I am trying to design a photoswitchable ion binder for a supramolecular chemistry class. I am trying to design a way to 'lock' the molecule in a particular conformation using a thiol-disulfide dynamic interchange. I would like to be able to switch between which of the rings is locked to the other by using different wavelengths. Pretty far fetched idea, needs lots of refining, but it would be very nice to be able to rotate around a bond selectively.
Thanks again Klaus, I asked the azobenzene question the other time :) Still much to learn about photochemistry! :p

@zhe, yes its possible to selectively excite most cis and trans molecules as they generally have different absorption spectra. its doing that in the same molecule thas so far impossible.

cis and trans selectivity by wavelength, see for example hemithioindigo. Disclaimer: My former group.

@porphyrin I think I'm thinking too classically here. The entire conjugated system is going to be affected by substitution... :/

In theory: yes
As soon as you have two distinct groups, you will have two different, discrete energy differences. If you manage to irradiate at exactly the correct wavelength, only one of the two will be excited.
In practice: possibly, but it depends
Two diazo groups are still rather similar. You will either need a beam that has a good wavelength focus or you will have to make their wavelengths sufficiently different.
By the way, note that both your average IR spectrum and your average UV/Vis spectrum show this property. At a specific wavelength, exactly one vibration (IR) or electron (UV/Vis) will be excited. With sufficient resolution, you should be able to excite the different nitro group vibrations separately. If this applies to one light wavelength band, it will work in others, too.

I think that you are not correct here, in the proposed molecule both cis and trans chromophores are in the same conjugation so are part of the same delocalised MO. If the were not connected electronically, say widely separated by a  alkyl chain, then only if cis & trans absorb at different wavelengths would it be possible to change the one with the lower energy  or longer wavelength. At shorter wavelengths where both absorb no discrimination would be possible. In solution levels are so broadened and are so overlapped that specific narrow band excitation has no particular effect.


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Is blowing air (from lungs) into a water/soap solution changing the liquid into a gas contained in the bubbles?
I am teaching an art lesson to second graders. We are going to blow bubbles into a dyed soapy solution and then place the bubbles on paper.  When the dyed bubbles pop, it makes for a great bubbly background to our fish art.
To tie in science, specifically liquids and gas, I'd like to explain the reaction that is happening when we create a bubble.
Also, I'd like to explain what happens to the gas and liquid surrounding the bubble when it pops.
Any SUPER SIMPLE scientific explanations would be appreciated.

There's no reaction. It's a physical phenomenon. When the bubble pops, the gas inside starts to mix with the gas outside. The liquid splatters. Its movement is governed by air currents, gravity, and the initial trajectory as a result of the pop.

A bubble is like a balloon made from a soapy film.  A soapy film in a bubble is pressurized under tension like the rubber in an inflated balloon. Balloon rubber is different than soapy film because a soapy film is a liquid and rubber is a solid.  A solid balloon breaking makes a louder noise (pow) than a liquid bubble breaking (pop).          - Most 2nd graders cannot handle abstract concepts; however, they can handle similarities and differences.

The physics stack exchange may be a better place for this question.  There is a similar question there already.  http://physics.stackexchange.com/questions/167312/surface-tension-and-soap-bubbles

You don't have a chemical reaction happening there. What you have is they physical phenomenon of the gas from the lungs getting trapped inside of a film made of soap and water - it is like filling a balloon, but instead of rubber, you have soapy water.
And, as opposed from what you have with rubber, the soap+water film is NOT stable by itself, so after a while it collapses, hence the popping. What happens then is that the gas from the lungs then mixes with the air around.

The reason bubbles do not last long has mostly to do with weather and climate.  If the climate is foggy the soap film will dilute becoming more like water.  If it is hot and dry, the thin film will evaporate.  If it is windy, the wind will break the bubbles.  If it is after a rainstorm, the wind is still, and it is cool... then bubbles will last plenty long enough to have fun.    ... stability phenomena may be tough for second grader to grasp.

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Effects of inert gas on heterogenous equilibrium
What is the effect of the presence of an inert gas on a reversible reaction that was in more than one phase, such as a decomposition reaction?  If a solid or liquid or compound in aqueous solution decomposed in a reversible reaction, and one or more of the products was a gas, would the equilibrium shift if an inert gas was added and volume and temperature were kept constant (so total pressure would increase)?

Henry's Law would say that the inert gas does not affect the solubility of the gas that's involved in the reaction.

An inert gas (i.e.; unreactive gas) will not affect a stable equilibrium. Neither will adding a catalyst.

@Zhe I am not sure how applicable Henry's law is to this situation. In equilibrium, Le Chatelier's principle would suggest that the reaction would shift towards the side that produces less gases.

No, Le Châtelier's Principle only applies to the components involved in the equilibrium. If you add a gas that is not involved in the equilibrium, there is no shift. The reaction quotient $Q$ would have contributions from the partial pressure of the gas.

@Zhe The temperature and volume is kept constant. As a result, the reaction will shift its equilibrium to the side that reduces pressure. A similar thing happens for heat.

@QuantumCAPUCCINO I don't think that's correct. Consider the simple case of a closed container with just water and water vapor. If you inject helium into the closed container. The vapor pressure, which is just a function of temperature, is constant. You do not shift the equilibrium.

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Can two solids form in a double displacement reaction?
I am trying to react silver flouride with magnesium chloride. I know it is a double displacement reaction but so far the equation I have gotten is:
$$\ce{2AgF (aq) + MgCl2 (aq) -> MgF2 (s) + 2AgCl (s)}$$
Is this possible or am I doing something wrong? If my equation is correct would this mean that there is no reaction? I'm supposed to find the complete ionic equation and the net ionic equation but I'm not getting either with the above balanced equation. Can someone please tell me what I'm doing wrong?

This is surely possible, though uncommon; lithopone is a more widely known example. Also, welcome to Chem.SE, and pay attention to the formatting.

States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different.


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How are standard bond enthalpy values experimentally produced?
The question's essentially it. I've always wondered about how these values are produced, since we refer to them heaps in the lab to assess our experimental results, but I don't actually know how they're produced.

Probably via calorimetry?

As defined by the IUPAC Gold Book, mean bond energy (standard bond enthalpy) is defined as:

The average value of the gas-phase bond dissociation energies (usually at a temperature of $298\ \mathrm{K})$ for all bonds of the same type within the same chemical species.
Tabulated bond energies are generally values of bond energies averaged over a number of selected typical chemical species containing that type of bond.

These energies are calculated by studying the homolytic cleavage of a bond into its respective radicals, or the colligation of two radicals to form a particular bond. Though measuring the heat from the colligation of radicals might be the most direct method to determine these values, it is also the most difficult to perform with any accuracy, and is not the preferred method.
There are a variety of ways to measure bond dissociation energies through bond cleavage—by irradiating the molecule with electromagnetic waves, bombarding it with a beam of electrons, measuring the equilibrium constant of the reaction $\ce{A-B <=>A. + B \cdot}$, or by measuring the rate by which a radical reaction proceeds$^{[1]}$.

$[1]$ Szwarc, M. Proceedings of the Royal Society A: Mathematical, Physical and Engineering Sciences 1951, 207 (1088), 5–13.


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Imposing particle number subspace on second quantized Hamiltonians
Here is my question:
Second-quantized Hamiltonians act on any state, regardless of how many electrons the state has. They are also block diagonal on different particle-number subspaces, as they commute with the electron number operator.
Now, suppose that one writes down a molecular Hamiltonian in the second-quantized formalism. In order to get the ground state energy, does one need to project the Hamiltonian on the correct electron subspace and diagonalize?
Or is the bare diagonalization of the Hamiltonian (without projecting) sufficient to pick the subspace with the right number of electrons?

$\newcommand{\Ket}[1]{\left|#1\right>}$
$\newcommand{\Bra}[1]{\left<#1\right|}$
$\newcommand{\BraKet}[2]{{\left<#1}\left|#2\right>}$
It's an older question, but one worth answering!
For the uninitiated, second quantization is a bookkeeping technique that describes many-particle systems as excited states of a field, usually taken to be the reference field or vacuum state. I found the most clear description of it in atomic_rabbit's answer to this reddit post. The second-quantized operator is useful because this form allows it to be applied to an indeterminately sized system (the number of particles is not explicitly stated in the second-quantized operator).
Now, onto the question:
The "first-quantized" (electronic) Hamiltonian has the form (in atomic units):
$\hat{H} = \sum_i\frac{1}{2}\nabla^2_i - \sum_i\sum_A\frac{Z_A}{r_{iA}}+\frac{1}{2}\sum_i\sum_j\frac{1}{r_{ij}} = \sum_i\hat{h}(i)+\frac{1}{2}\sum_i\sum_j\frac{1}{r_{ij}}$
which, upon second quantization, goes to:
$\hat{H} = \sum_{p,q}\Bra{p}\hat{h}\Ket{q}a^+_pa_q + \frac{1}{4}\sum_{p,q,r,s}\Bra{pq}\Ket{rs}a^+_pa^+_qa_sa_r$
In the case of "first quantization", we construct the matrix form of the operator by projecting it onto whatever basis functions we're using. We then find the eigenvalues of the operator by diagonalizing its matrix form. Your question is an excellent one, because in second quantization, this projection seems to be already done within the operator, and so we should be able to diagonalize the operator outright. Right?
The catch is that the second-quantized operator still has the creation and annihilation operators that have not yet been projected onto any basis. This is equivalent to us not knowing whether there is a particle in the p$^{th}$ state. And what are the states? Those are properties of the field, which we must find out by projecting the second-quantized operator onto said field.

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what can 1 kJ/mol do in reality?
I am using an approximation of density matrix in a high order series expansion to predict the total internal energy of chemical systems. During my research,it was found out that the precision of such algorithm is accurate up to 0.00001 Hartree by optimising the independent parameters, which is roughly 0.026kJ/mol's error fluctuation. I wonder in reality, what can 1 kJ/mol do in our daily life, what phenomena can 1 kJ/mol of energy possibly trigger? I'm quite a mathematical person and haven't really studied enough chemistry nor physics. if someone could give me some physical intuitions that would be greatly appreciated!
And I meant to leave it as a broad question. Since we are not picking a specific chemical system to chew on here, any unbiased thoughts are welcomed and shall be considered in this post! A collective answer from the various point of view is probably a better way to get one's head around rather than a single aspect answer.

When I'm confronted with heat figures either from calorimetry work or molecular modelling studies I ask myself the same question.  What I've tried to do, over the years is to try and calibrate myself.  For example, what Wattage is your kettle?  How long does it take to heat up some water from room temp to boiling?  But not simply timing this process but getting a feel for it.

The average kinetic energy of molecules is $3RT/2$ which is $\approx 3.7 \pu{kJ mol^{-1}}$at $300$ K. There is also internal vibrational & rotational energy of molecules which adds $RT/2$  for  each squared term in the energy. So your $1 \pu{kJ mol^{-1}}$ does not account to much as its going to be overwhelmed by thermal energy. Of course at $30$ K its different.

@porphyrinas I agree with you. Lower the temperature, more quantum effects would be involved, the predicted result must certainly require a higher level of accuracy... that's totally reasonable to think for.

For this reaction
$$\ce{A -> B}$$
a $\Delta G$ value of $1\ \mathrm{kcal\,mol^{-1}}$ will result in an equilibrium value
$$K = e^{-\frac{\Delta G}{RT}}$$
of approximately 0.18 favoring reactant. Every additional increment or decrement by this value will change the relative ratio by roughly 5.


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How is phosphate stable disobeying octet rule?

Doesn't the phosphate now have 10 electrons mutually shared between the oxygen atoms? I am unable to understand how the phosphate bond would be formed if phosphate does not obey the octet rule.
Thank you!
Reference: http://www.chemguide.co.uk/organicprops/aminoacids/makedna.gif

The octet rule is just a rule. Once you leave the 2nd row of the periodic table, the octet rule doesn't apply. Further reading: https://en.wikipedia.org/wiki/Hypervalent_molecule

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Basicity of substituted pyridines

Which of the following pyridines are most basic?

I would've considered D, since chlorine is the most electronegative, and would therefore have the largest partial negative charge, making it more basic than the nitrogen substituent. The answer given, however, is B. How can I reason this for myself?

Is N,N-dimethylaniline a base? If so, is it stronger than pyridine? ;-)

The basicity of a compound is typically not directly proportional to the electronegativity of its substituents but indirectly proportional. That is, the more electronegative a substituent is, the less basic the entire compound is.
First let’s take a look at basicity and what factors into it. For a molecule or group to be basic, the following reaction (1) needs to occur adequately favourably:
$$\ce{B + H+ <=> BH+}\tag{1}$$
What is happening in this reaction is that we are giving the molecule $\ce{B}$ a positive charge. Positive charges are more stabilised if they are ‘balanced’ by negative charges — i.e. if the molecule $\ce{B}$ was already partially negatively charged before the proton came in.
Remember that electronegativity is the ability of an atom to draw bonding electrons towards it. That is exactly what you want to create a partial negative charge, so if the basic group of a molecule is more electronegative than its surroundings that is what you want. In the context of aromatic compounds, electronegativity is usually described as a $-I$ (negative inductive) or $+I$ positive inductive effect; the former being exercised by electronegative groups and reducing the electron density in an aromatic ring and the latter due to electropositive groups and increasing it.
If we want to create a pyridine that is as basic as possible, we want $+I$ effects, i.e. electron donating groups. If we rank the groups by their ability to donate electrons from $+I$ to $-I$, we get:

$\ce{CH3}$ — due to the electropositive hydrogens
$\ce{H}$ — defined as the standard ($0$)
$\ce{Cl}$ — mediumly deactivating
$\ce{Me2N}$ — strongly deactivating inductively.

However, inductive effects are not the only ones that can modify an aromatic system: mesomeric effects ($+M$ or $-M$) can do so, too. Mesomeric effects are exercised by groups that can either mesomericly donate electrons into the aromatic ring, increasing the electron density; or by those that can pull electrons out of the aromatic ring reducing its density. Mesomeric effects usually but not always outrank inductive effects, and they can have an effect across longer distances. Ranked by mesomeric effects from $+M$ to $-M$, the groups are:

$\ce{Me2N}$
$\ce{Cl}$
$\ce{CH3}$ — only very weakly; technically, this group can be explained almost solely by its $+I$ effect
$\ce{H}$

The dimethylamino group is a much better mesomeric electron donator than chlorine because nitrogen’s p-orbital is much more similar to carbon’s allowing for a stronger participation in the π system and because of nitrogen’s generally larger basicity than chlorine’s. If you want, you can visualise the $+M$ effect with the resonance structures Ringo’s answer shows.
Taking this all into account, we get the following final rating from more basic to less basic — and note that this corresponds nicely with the respective rates of electrophilic aromatic substitution:
$$\begin{array}{cccl}\hline
\text{Rank} & \text{Compound} & \mathrm{p}K_\mathrm{a} & \text{Source} \\\hline
1 & \ce{4{-}Cl-C5H4N} & 3.83 & \text{Nguyen}\\
2 & \ce{C5H5N} & 5.2 & \text{Bordwell}\\
3 & \ce{4{-}Me-C5H4N} & 6.03 & \text{Nguyen}\\
4 & \ce{4{-}Me2N-C5H4N} & 9.2 & \text{Evans}\\
\hline\end{array}$$
Compared to the ‘standard’ — pyridine — 4-chloropyridine is deactivated (strong $-I$) while 4-methylpyridine is mildly activated ($+I$ effect). 4-dimethylaminopyridine is, of course, strongly activated: the $+M$ effect strongly outweighs the $-I$ effect.

References:
Evans: the Evans $\mathrm{p}K_\mathrm{a}$ table
Bordwell: the Bordwell $\mathrm{p}K_\mathrm{a}$ data
Nguyen: N.-N. Pham-Tran, G. Bouchoux, D. Delaere, M. T. Nguyen, J. Phys. Chem. A 2005, 109, 2957. DOI: 10.1021/jp045339c.

Basicity has everything to do with electron density on the electron-pair donor.  Electronegative subsituents actually decrease basicity because they draw electrons towards themselves, though nitrogen and chlorine have comparable electronegativities (3.04 and 3.16 respectively), so I'm not sure you can use that argument here. Instead of inductive effects, let's consider the resonance structures of our molecules:
$\hspace{5.6cm}$
A positive charge on a nitrogen is always more favorable than a positive charge on a chlorine. From this we know nitrogen is donating its electrons into the ring to a much greater extent than chlorine. Because the electron density is much greater on the heterocyclic nitrogen in B, it will therefore be the more basic of the two.

I think N,N-Dimethylpyridine is the most basic compound due to + I efects of two methyl groups .

This answer is a bit short, but does not merit deletion. Please don't flag answers you think are wrong.


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The mass of carbon atom
I was reading Silberberg's principles of general chemistry when I noticed that it says that the mass of carbon12 is exactly 12 amu.
But when I calculated the total mass of its components    (6 protons : 1.00727 amu each , 6 neutrons : 1.00866 amu each and 6 electrons : 0.000548 amu each) it was obviously greater than 12. Why is that?
My explanation was that some of the mass is converted to energy but I don't know whether it was true or not.
And I'm sorry if my English is not good.

That's right: when you put together 6 protons, 6 neutrons, and 6 (not 12, mind you) electrons, some of the mass will be lost as energy. Also, your English is fine. Also, welcome to Chem. SE.

Sorry for the mistake (12 electrons) and thanks for your answer Ivan Neretin.


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Photocurable Polymer with no oxygen functional group or whose the structure is well known
I'm experimenting with the PDMS. I want to change the material with no oxygen containing functional group.
Is there a UV curable polymer with no oxygen atom?
Preferably cheap.
Or
Please recommend any UV curable or photoresist with well-known morphological property and molecular (polymer) structure.

I'm not sure if this addresses all your criteria,  but acrylonitrile is polymerisable by light and is oxygen free.

I will google it and check if it can be used in my system. Thanks.


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Calculation of equilibrium constant given the reaction temperature
I have only a chemical equation, like:
$$\ce{N2 + 3H2 <=> 2NH3}$$
How can I calculate the value of equilibrium constant from only this equation. Provided temperature is $\pu{298K}$. I can use other standard data. I am really stumped on this question.

You get the equilibrium constant by dividing the concentrations on the right side with the concentrations on the left side. Note that you should  account for the coefficients by using them as powers in your equilibrium equation.

Use $K= \frac{(NH_3)^2 }{ (N_2)  (H_2)^3}$ where the brackets indicate concentration.

there is no concentration given

Note that the standard Gibbs free energy of formation for everything on the left side is zero. The total change in free energy in this reaction is just double whatever the standard Gibbs free energy of formation of ammonia is.

@THELONEWOLF. The concentration is not given in the question.

By definition, the standard equilibrium constant is:
$K = e^{-\Delta_rG/RT}$
Where $\Delta_rG$ is the difference between the Gibbs Energies of the products and reactants.

I need the value of equilibrium constant.

Don't you want to calculate it from standard Gibbs energy tables?

Maybe I can use standard Gibbs energy table. I was looking for an direct method. But now it seems that there is no way of getting the value of  $K_c$ without knowing concentrations or Gibbs energy change.


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How is Oxygen Toxic?
We talk about divers using Scuba tanks that contain a only a small concentration of oxygen diluted in a mixture of other gasses such as nitrogen or helium.
From what I've read, this is because breathing pure oxygen, or a concentration that exceeds the normal partial-pressure of oxygen at sea level, when diving can cause major health problems.
Oxygen Toxicity is a phrase that keeps getting repeated, though I've found sparse few details surrounding what it actually is/means.
I ask therefore: How does oxygen become toxic to humans when breathed-in at high concentrations or underwater or wherever — on a chemical/biolgical level — and also why this is the case?
Because really, isn't good-old $\ce{O2}$ meant to be the gas that keeps us alive?

It' s a follow-up to http://chemistry.stackexchange.com/questions/61406/why-would-breathing-pure-oxygen-be-a-bad-idea?rq=1 I guess?

Well, what @Mith linked perfectly answers your question, I guess? If your question is any different, please edit and explain how.

Like I said before, everything is poisonous, except oxygen, nitrogen, and water. And even those may kill a man, if used improperly. We are accustomed to a very narrow window of possible conditions; one step outside and you are dead. Think of it this way: life on Earth never had a chance to get used to breathing pure oxygen.

The wikipedia page https://en.wikipedia.org/wiki/Oxygen_toxicity pretty much covers everything about this topic.

To get one thing out of the way first: toxic does not always mean the same thing. Some things are acutely toxic (e.g. carbon monoxide), some things are toxic upon accumulation (e.g. lead). Some things can get removed from the organism given enough time (e.g. methanol), some things will stay in your body ‘forever’ (e.g. lead). Some things kill you quickly (e.g. cyanide) while others will cause cancer that will kill you in maybe decades (e.g. benzene). And the organism can cope with some things given small enough concentrations (e.g. nitrite) while others are harmful at even the smallest concentrations (e.g. dimethyl mercury). Paracelsus is typically quoted in German with the saying:

Alle Dinge sind Gift, und nichts ist ohne Gift; allein die dosis machts, daß ein Ding kein Gift sei.
(All things are poison and nothing is without poison; only the dosis determines that a thing may not be poison.)

Thus, the question is only which category oxygen falls into: both in terms of acute/chronic/instant/accumulative/… and in terms of $\mathrm{LD_{50}}$ values.

Oxygen is, in principle, a very reactive gas. It is very electronegative and often reacts from its neutral $\pm 0$ oxidation state to compounds in which it features a $\mathrm{-II}$ oxidation state.
Life and living matter builds on being able to selectively use this type of oxidation enthalpy for its own uses, i.e. to build up complex structures (proteins, DNA) or to overcome entropic problems. From its beginning, ‘life’ was basically a method of using inspecific chemical energy for its own uses. Spontaneous reactions out of life’s control are typically suppressed as well as possible, workarounds made and so on. For example, aldehydes are useful, reactive chemical species that can be derivatised in may different reactions. Living organisms tend to make use of similar chemistry (as least partially) but it also tends to avoid the generation of free aldehydes as much as possible. Indeed, one way life keeps aldehydes ‘safe’ is by masking them as Schiff bases or as acetals (e.g. sugars). Life relies a lot on in situ generation of reactive species.
Indeed, imagining an evolution of the first living being in an oxygen-rich surrounding is very hard. When the first life forms came into being, the atmosphere and the conditions on Earth were generally much more reductive, much less oxidative than today. The first living organisms probably used chemosynthetic energy sources that did not rely on oxygen. It took a while for organisms to develop photosynthesis. This had quite a number of beneficial effects for those organisms: the light energy source is much more easily accessible, carbon dioxide to build up organic molecules was plentiful and these organic molecules could then be synthesised according to the organism’s needs either as building blocks or as an energy source. Photosynthesis only has one disadvantage: It has a strongly toxic by-product which can ruin the carefully-tuned enzymes your cell has by oxidising their metal cores, their sulfur atoms and more; and it can react with other molecules in the cell in unwanted ways. It had to be removed from the cell and the first photosynthetic life forms indeed proceeded to dump it into the environment.
At first, the dumped oxygen would react with reactive species and be chemically bound e.g. in rocks at the ocean floor. Partially, it was also absorbed and diluted in the oceans. After a certain oceanic concentration was reached, oxygen was released into the atmosphere and started to accumulate in land-side oxygen sinks by similar mechanisms. After all these oxygen sinks (‘dumps’) were filled, atmospheric oxygen levels began to increase dramatically within a few million years.
The effect that this had on existing life forms — even during the early evolution of oxygen-generating organisms — was catastrophic (hence the term Oxygen Catastrophy). Most life forms — obligate anaerobic ones — would have become extinct and paved the way for an entirely new generation of species: oxygen-tolerant and finally aerobic ones. The obligate anaerobes had no method to control cellular oxygen while later forms had these mechanisms in place.
The great advantage of having molecular, atmospheric or ocean-dissolved oxygen as an energy source is that it can supply a lot more energy than non-oxygenic processes: think of the amount of heat generated by burning and compare that to most other chemical processes to understand the difference. However, any organism relying on the use of oxygen must also cope with the by-products that can occur during oxidation — both desired oxidation procedures and undesired, spontaneous ones. Aerobic organisms tend to have a large battery of peroxidases and superoxide-dismutases for the controlled degradation of peroxides $\ce{R-O-O-H}$ and superoxides $\ce{H-O-O^.}$ — even though these species autodegrade almost at diffusion-controlled rates. The reason is that these reactive species, generated by the use of oxygen, would create too much havoc to allow them to degrade themselves; organisms must degrade them specifically and with minimal danger to the cell as quickly as possible.
And this is also the question why high oxygen levels can be considered toxic. Oxygen reacts with organic molecules present in the cell in an undesired fashion to generate the more reactive peroxides and superoxides. If they cannot be captured and destroyed quickly enough, they create more havoc by randomly oxidising further parts of the cell in undesired fashions potentially causing protein malfunction, DNA damage, cell wall damage and more. The higher the exterior oxygen concentration, the higher the lung oxygen concentration and the higher the blood oxygen level and cellular oxygen level. The higher the cellular oxygen concentration, the more damage it can perform to a cell and the harder it is for the peroxidases and superoxide-dismutases to keep the reactive species under control. Thus, more damage occurs leading, finally, to that bit of damage that can no longer be controlled.
A catch-it-all term for this is oxidative stress. All the different reactions are too numerous and too diverse to be presented here (or even on a medium as Wikipedia), but they all involve the reactive peroxides, superoxides and hydroxyl radicals $\ce{HO^.}$ (which I didn’t mention yet, but which are also terribly reactive).

Tl;dr: oxygen is great in principle, but too reactive. Cells can cope well with ambient oxygen levels and keep the reactive species under control. A rise in oxygen levels will generate too many harmful, reactive species that the cell can no longer control.


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Enthalpy of formation of ammonium bicarbonate
I'm helping with correcting lab reports, and one of the experiments the students did was heating of ammonium bicarbonate, $\ce{NH_3HCO_3}$ in powder form, over Bunsen burners, which turns the compound into water, carbon dioxide and ammonia (all gasses). They are supposed to say, in their report, whether this is an endothermic or exothermic reaction.
Now, I would be very surprised if this reaction turned out to be exothermic. An hour of googling seems to agree with me, but I can't find a definite answer. In the course we've talked about enthalpy of formation and Gibbs free energy, so it would fit into the curriculum if we could find the relevant data.
Unfortunately, it's apparently very difficult to find the enthalpy of formation of sodium bicarbonate through google. Many small tables do not list the chemical at all, and the other sources that I've found does not list enthalpy of formation. I've found one number, which is -32 kcal, without specifying whether it's per mole, g, kg, ounce or pound.
Does anyone have a source on the enthalpy of formation of solid $\ce{NH3HCO3}$ at room temperature?

−2580 cal/g.  Here is a reference that lists heats of formation:
https://engineering.purdue.edu/~propulsi/propulsion/comb/propellants.html

This becomes $-10.836 kJ/g$, and, without being too picky on atomic weights, $-845 kJ/mol$. Compared to the values for ammonia, water and carbon dioxide given on wikipedia, we can now conclude that the reaction is endothermic. Thank you.


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Deprotonation of 2- and 3-methylquinoline
Why are there many examples of 2-methylquinolines being treated with a base then an electrophile but none of the corresponding reaction with 3-methylquinolines? Does this have to do with the charge not being stabilised for the 3-methylquinoline?

what you want to say?

Yes, in contrast to 3-methylquinoline, the methyl group of 2-methylquinoline (1) is easily deprotonated by potassium ethoxide.

The anion has been examined by NMR spectroscopy (DOI), but its use in synthesis was reported long before. There's an article (in German) in Chem. Ber., 1909, 42, 1140 (DOI) where the reaction of 2-methylquinoline with diethyloxalate is described.


text:
Reason for decreased boiling point of hexafluoroisoproponal?
Ethanol and isopropyl alcohol have boiling points 78.37 °C and 82.6 °C respectively. The increase in the boiling point is obvious due to increase in carbon chain length which resulted in increase in boiling point of isopropyl alcohol.
But recently, I come across fluorinated ethanol and isopropanol, and the trend of boiling point is the reverse. Trifluoroethanol has boiling point 78 °C whereas hexafluoroisopropanol has boiling point 58.2 °C.
what is the reason for the decrease in boiling point of hexafluoroisopropanol? Is it due to increase in the number of fluorine atom, if so then why this resulted in decrease in the boiling point?

See https://en.wikipedia.org/wiki/Fluorocarbon particularly London dispersion forces

Since fluorine is so much more electronegative than hydrogen, its electrons are considerably less polarizable. This decreases the extent to which induced dipoles are able to form (London dispersion forces), which in turn lowers the energy necessary to break the intramolecular forces. This is what gives fluorocarbons lower boiling points than their hydrocarbon counterparts.


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Electrolysing vinegar
Yesterday I made a mixture of vinegar and salt and then electrolyzed it. I was just wondering if anyone could tell me what the process created. For instance, would it create ethane?

Vinegar in water is a slightly acidic solution. My guess would be evolution of hydrogen and oxygen gas at cathode and anode respectively. However, varying concentration and external potential might change the products

If you have table salt in the solution you will get Chlorine produced in preference to Oxygen at the anode.
Cathode still produces Hydrogen in this case.

As @Amritansh Singhal said, $\ce{H2}$ would be produced at the cathode from the electrolysis of both the water and some from the acetic acid (main ingredient of vinegar after water) and $\ce{O2}$ would accordingly be produced at the anode.
The question in your comment "Would it create ethane?" is where some more interesting chemistry might take place.  The short answer is "very little at the most" since in the presence of water, the cell potential will be limited by the electrolysis of water as described above.
However, if you could obtain pure acetic acid and produce the proper cell potential, then you could see the Kolbe electrolysis reaction, which would produce ethane.  According to excerpts from this Wikipedia article":

"The Kolbe reaction is formally a decarboxylative dimerisation of two
  carboxylic acids (or carboxylate ions)."

With respect to acetic acid specifically the above article states:

"The reaction mechanism involves a two-stage radical process:
  electrochemical decarboxylation gives a radical intermediate, then two
  such intermediates combine to form a covalent bond. As an example,
  electrolysis of acetic acid yields ethane and carbon dioxide:"
  $$\ce{CH3COOH → CH3COO− → CH3COO· → CH3· + CO2}$$
  followed by:
  $$\ce{2CH3· → CH3CH3}$$

Where the latter product is ethane.

Well $\ce{CH3}\cdot$ is too reactive to survive for any length of time in aqueous solution. I'd guess that you'd only get ethane in very concentrated solutions of acetic acid (e.g glacial acetic acid).

Yeah, it's going to react with most everything, though particularly another methyl radical.  Reactions with radical intermediates will just tend to have lots of unwanted byproducts unless you have pure reagents, the right conditions, etc.  That's part of the reason for my comment that you would get "very little at the most" unless you could get pure acetic acid, though I appreciate you mentioning that issue explicitly.


text:
Does organic chemistry somehow defy the periodic table?
My physics lecturer was introducing the idea of what a Theory of Everything was, and said something like this:

Other sciences have their own theories of everything too, fundamental axioms from which everything else can be derived from. In mathematics, we have ZFC, and well, the entire framework of abstract mathematics and axiomatic systems. Even in Chemistry, all of modern chemistry can be derived from the periodic table and the laws of atomic structure. Well ... except for two words: Organic Chem.

Is this true? Can organic chemistry somehow not be derived within the framework of the periodic table?

He/she was probably just kidding, it's very common for physical chemists (equivalently, physicists) to jokingly talk down towards organic chemists (and vice versa).

While some organic chemists are especially ignorant of anything beyond their art of synthetic work, actually nowhere in chemistry is it possible to accurately predict really new things. In hindsight, everything has however been explainable once the theoretical chemist (sometimes the same person) sat down and tried to. Good thing, too. Only logical knowledge can be handed down. Intuition and a fair amount of guesswork only work at the foremost front of science.

While all chemistry is unified because all chemical compounds are made from things in the periodic table, that isn't like mathematics and much of physics. In these sciences many important complexities are derivable from a simple set of axioms.
Some chemistry is simple to derive or deduce from the periodic table but not all of it. The electronic properties of iron atoms tell you a lot about many iron compounds; the location of alkali metals in the table tells us a lot about their common properties.
But carbon is odd. The properties of simple compounds follow simple periodic patterns. But there are more organic compounds (consisting of carbon skeletons, usually with C-C bonds) than all other chemical put together. The complexity and variety of carbon chemistry is almost unfathomably large. None of this is obvious or predictable from the periodic table. You can't deduce the enormous variety from the simple properties of carbon atoms or anything given in the periodic table.
That is probably the sort of distinction your lecturer was aiming at.

It is impossible to know exactly what your professor was implying, but one idea might be that life itself is a mystery. Generally organic chemistry is associated with living things that use a multitude of carbon compounds. Chemists know how organic reactions in general will proceed but we don't have a solid understanding as to how life itself can be created.


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Explaining The Different Order of Stability for singlet and triplet carbenes
Consider the following singlet carbenes:
$\ce{CH2}$ , $\ce{CF2}$, $\ce{CCl2}$
The order of stability expected considering back bonding is;
$\ce{CH2}$ < $\ce{CCl2}$ < $\ce{CF2}$
However the correct order is:
$\ce{CCl2}$ < $\ce{CH2}$ < $\ce{CF2}$
Can someone explain why dichlorocarbene(singlet) should be less stable than $\ce{CH2}$ even when back-bonding is present in $\ce{CCl2}$ (though it may not be strong).
Also, if we consider the stability order for triplet carbenes, it turns out that the order is just reversed. I assume that back-bonding doesn't work here due to non availability of empty p orbitals. But, if this is so, then what  factor(s) determines their stability?
Edit: There seems to be a confusion with the word stability. I would rather put it as in which species is carbon's electron hunger least/most satisfied(i.e. which are least and most electrophilic)

Stability with regard for what?

@Martin-マーチン  I'm interested in their relative stability. To put in simple words, their order of reactivity

Reactivity towards what? Nucleophilic or electrophilic substitution, for example, are two different shoes. Stability and in that regard reactivity are not absolute concepts, you need to specify what baseline you are comparing them to.

@Martin-マーチン I think the OP means relative stability when these species are formed as reaction intermediates.

@S007 you are implying that and that also does not define a proper baseline. In what reaction would they be formed, are the settings even comparable. There are just too many uncertainties to give a correct answer. These molecules don't even have the same ground state, significantly different geometries, etc. There is no universal and simple answer to this question and it needs more information.

@Martin-マーチン I'm sorry if my question appears uncertain. I dont know if it is the sole criteria for stability, but I wanted the species arranged in order of fulfilment of carbon's electron hunger(or electrophilic nature). Please check that I have now edited the post

Where did you get the "correct order" from?


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pH and solute solubility in waste mixed water
I was reading a module on waste management. There were two interesting points.

High pH results in higher solubility of $\ce{CN-}$. I understand that for HCN:
$$\ce{HCN ->H+ + CN-}$$
Higher pH means that equilibrium will be shifted rightwards and concentration of $\ce{CN-}$ will increase.
But will cyanide salts like KCN, NaCN have the same high solubility at higher pH? If so, how?
Low pH results in greater solubility of toxic gases. How?

HCN is a weak acid, so its salts get hydrolyzed to a significant extent, which makes the same argument applicable. 2. This is an example of hasty generalization. What toxic gases are you talking about?

Actually the book doesn't mention the gas names. Well, I don't think it will be SO2, CO2 because they will dissolve better in greater pH. Maybe alkaline toxic gases do the work

You think right. NH3 may behave like that.


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Side chain hydroxylation of pyrrole
What reagents are required to form 1-[1,5-dimethyl-3-phenyl-4-(propan-2-yl)-1H-pyrrol-2-yl]propan-1-ol with 1,2-dimethyl-4-phenyl-3-(propan-2-yl)-5-propyl-1H-pyrrole as starting material?

Is it possible for the starting material to react with a halogen gas e.g. bromine to form a bromine on the carbon and then react with e.g. NaOH to form the OH with Br as the leaving group?

If we consider that position allylic maybe it can be oxidized as such. The only problem is the methyl group on the other side of the nitrogen, whose reactivity will be similar. http://www.chemtube3d.com/ox_se_allylic.html

I'd like to propose a dark, non-radical pathway :)

Pyrroles, supposed that they are not fully substituted, undergo electrophilic aromatic substitutions with halogens.
For the given starting compound, a complete bromination isn't possible due to the lack of a proton that would have to be eliminated in the final step.
However, the addition of an electrophile is conceivable. Addition of "$\ce{Br+}$" at C-2 of the starting material, usually the prefered position in the reaction of unsubstituted pyrroles, is a dead end here. Addition at C-3 yields an $\alpha$-enamine cation intermediate from which a proton might be eliminated. The resulting species is both

a double enamine and
an allylic halide

This intermediate might rearrange upon elimination and addition of a nucleophile $\ce{Br-}$ in a way that the aromaticity of the pyrrole is reestablished and bromine is introduced in the side chain.

Edit 1
Supposed that you are mainly interested in the hydroxyalkyl-substituted pyrrole as the target molecule, you might want to consider an alternative strategy. Let's have a second look:

The target compound (1) has an $\ce{OH}$ group in a side chain. Alcohols may be prepared by different routes, such as

reduction of the corresponding ketone
reaction an aldehyde with carbon nucleophile

Let's focus on the second idea. This would leave us with the 2-formyl pyrrole (2), that could be transformed into (1) by a Grignard reaction with $\ce{EtMgBr}$.
Again, several methods are conceivable for the synthesis of the aldehyde (2), such as

selective reduction of a $\ce{-CO2R}$ substituent (or reduction to $\ce{-CH2OH}$ and subsequent oxidation to $\ce{-CHO}$)
direct introduction of a $\ce{-CHO}$ substituent at carbon 2 of  pyrrole (3)

The latter seems promising and you will probably realize that this an application of the Vilsmeier-Haack reaction, where the reactive intermediate, the $\ce{CHO}$ synthon, is prepared in situ from N,N-dimethylformamide and $\ce{POCl3}$.

This would leave you with a strategy for the synthesis of (3), but that's another question.

Though what you suggested may work to some extent (bromine is a liquid at room temperature, though, mind you), I feel it would be good to discuss the best conditions to maximize the yield of this synthesis and the reasons for them.
Let us first consider free-radical halogenation as a means to introducing a leaving group on the n-propyl side chain. Though we might first be worried that the isopropyl radical would propagate faster than the n-propyl radical, which would be unanimously true for benzylic radicals, the position of the substituents is much more important in pyrrole:
$\hspace{3.4cm}$
Hydrogen is abstracted preferentially from substituents at the 2 and 5 positions because the $\pi$ system the resulting unpaired electron occupies is more conjugated and thereby lower in energy. Given this, we need now only maximize abstraction from $n$-propyl over methyl.
Since more substituted radicals are more stable (due to hyperconjugative effects) and thereby react faster, by lowering the temperature of the reaction, the number of molecules with sufficient energy to form the methyl radical is decreased. Though this does the same for the n-propyl radical, the effect is less drastic due to the nature of the Maxwell-Boltzmann distribution, and a greater proportion of n-propyl halide is formed.
Lastly, the halide needs to be replaced with a hydroxyl group. Though you could try an $\mathrm{S_N2}$ using $\ce{NaOH}$ or $\ce{KOH}$, $\mathrm{E2}$ would be a major competitor, especially considering that its product only further extends the conjugation of the ring. Instead, just try microwaving the molecule in water for 3-5 minutes and you'll more than likely have converted it nearly quanitiatvely to the alcohol.
Using only water, we almost completely remove the possibility of elimination as it is such a weak base. Using the microwave allows us to thoroughly and quickly heat the solution to temperatures >$100º\mathrm{C}$ even, increasing the number of water molecules with sufficient energy to successfully collide with the reactant and the rate at which these collisions occur. Thus, one possible pathway would be:


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Is the Friedel-Crafts alkylation of benzene with cyclopropenyl chloride feasible?
Is the Friedel-Crafts alkylation of benzene with cyclopropenyl chloride according to the following scheme feasible?

Would the attack of the benzene towards the cycopropylium take place?
I am wondering about this, as the cyclopropyl carbocation is aromatic, which  implies that it is highly stable.
If it reacts with the benzene ring, it would lose the aromatic stabilisation, thus rendering it comparatively less stable.

If I were you I would go for PhLi + 3-chlorocyclopropene that seems to me to be more promising (and also far more convenient).

Brief Reaxys search shows some related Lewis-acid catalysed arylations, but they start with 3,3-dichlorocycloprop-1-ene and end up with 3-arylcyclopropenyl cations

@orthocresol That's pretty interesting. What happens to the other chloride?

@Zhe one chloride gets replaced with the aryl group, the other chloride is ionised

@orthocresol Sorry, I should've been clearer with my question: where does the hydride (or proton) come from that replaces that chloride?


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Classifying some molecular orbitals after Hartree-Fock calculation
I have done a Hartree-Fock (HF) calculation on a dioxetane molecule with one methyl group on one of the C-atoms (Fig 1), with minimal basis (using Molcas). I want to identify the active space that will be used in later CASSCF calculations on the same molecule. The interesting molecular orbitals (MO's) for this active space are the following ones: $\sigma_{CC}$, $\sigma_{CO}$, $\sigma_{CO}^{'}$, $\sigma_{OO}$, $\sigma_{CC}^*$, $\sigma_{CO}^*$, $\sigma_{CO}^{*'}$, $\sigma_{OO}^*$, $n_{OO}$, and $n_{OO}^{'}$. So basically all the ring bonding and anti-bonding MO's, plus the two non-bonding MO's on the oxygen atoms.
The CO MO's are not localized for just one CO pair, but rather they both exist at the same time in an "out-of-phase" and "in-phase" version. These are given in Figs 2-5. The same goes for the non-bonding MO's; they both exist in an out-of-phase and in-phase version, given in Figs 6 and 7. Then there are only the CC and OO MO's left. In the same calculation for 1,2-dioxetane (zero methyl groups, only H atoms on the carbons), I find separate CC and OO MO's. However, for this one-methyl dioxetane, it looks like the CC and OO MO's are combined, giving out-of-phase and in-phase version of both the bonding and anti-bonding. My issue is that the MO's look a bit ambiguous, in that some of the lobes are not where I expected them to be. These MO's are given Figs 8-10.
For it to be a $\sigma$ orbital, the electron density should be in the bond axis between the atoms. However, in Figure 8 this is not the case. It's like the same-phase lobes attract each other and overlap instead inside the ring. Is it correct to interpret this as the CC/OO in-phase bonding MO? Figure 10 looks more like the $\sigma_{OO}^*$ than the in-phase $\sigma_{CC/OO}^*$, as the orientation of the lobes on the C atoms seem strange to me. Especially the MO in Figure 11 is not obviously classified, in my opinion. The OO contribution seems very small, and the lobe orientations on the oxygens are not what I would expect for an anti-bonding $\sigma$ orbital.
However, there are no other candidates for these MO's, so I'm left with the current classification, even if it feels a bit incorrect. Perhaps this is a common issue in analyzing molecular orbitals in computational chemistry.
P.S. I apologize for the bad figure qualities, I don't know how to visualize molden files in any other way.

Figure 1. The molecule

Figure 2. CO bonding in phase

Figure 3. CO bonding out of phase

Figure 4. CO anti-bonding in phase

Figure 5. CO anti-bonding out of phase

Figure 6. Non-bonding in phase

Figure 7. Non-bonding out of phase

Figure 8. CC/OO bonding in phase

Figure 9. CC/OO bonding out of phase

Figure 10. CC/OO anti-bonding in phase

Figure 11. CC/OO anti-bonding out of phase

Generating neat orbitals/surfaces from molden/wfn-files

Avogadro does not render it correctly. The molecule is invisible, and weird layers appear. The orbitals themselves seem okay, though.

I know, that's also my problem with Avogadro. But there are other programs than avogadro. :)

Hmm, strange, but after re-installing Avogadro 1.1.1 everything works fine.

Actually, the rendering is fine, but the orbitals now look different from what they should.

That is because last time I checked Avogadro can't handle cube files where the origin isn't (0,0,0).

@Yoda As you use Molcas, you can use it’s own Orbital Viewer GV.

I quite like the figures, although I'd preferred it when the molecule rotation would have been the same in all of them. Since you are using molden (<3) you might already have a working gmolden on your system. Try that one... ;)

I found out that wxMacMolPlt worked quite well. But the software part is kind of beside the point of the question :P

@Martin-マーチン Oh Molden... Overlooked by more sophisticated programs but it will forever be a useful workhorse.

Natural bond orbital (NBO) methods may help you to localized the orbitals of your system after a post-HF calculation. I'm not familiar with Molcas but I know that several computational chemistry softwares have NBO methods implemented. Take a look at this simple explanation that uses MolPro:
https://sites.lsa.umich.edu/zimmerman-lab/tutorial/electronic-structure-calculation/casscf/


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Solvents adequate for dissolving overheated polyethylene terephthalate glycol-modified (PETG) residue?
I work in 3D printing. One of the materials we use is polyethylene terephthalate glycol-modified, or PETG. It is a pretty good material, both in strength and appearance. It is melted and pushed through a "hot end" nozzle, typically either brass or hardened steel, at about 230 degrees C.
Unfortunately, it is rather sticky, and tends to accumulate on the hot metal surfaces, where it slowly degrades due to the heat. The resulting residue is black, and somewhat hard to remove. I have tried blowtorching it, which works with polylactic acid (PLA), but that just results in the same black substance.
I have seen recommendations to use either acetone, dichloromethane, or phenol. Other substances like toluene are recommended - some say they work, some say they don't. However, all of these seem to be recommended for dissolving/welding the plastic in its original form. There doesn't seem to be any advice for dissolving the residue after it has degraded from heat, at which point its chemical composition has changed.
From my research, the residue may be composed primarily of acetaldehyde. I was not able to find information (that I can understand as a lay person) about how to dissolve that.
So far, I've tried putting a brass nozzle caked with PETG residue into a 50C acetone bath for an hour. A little of it came off, but not much. The ideal solvent would attack the residue so completely that it would come off without too much force. The inside bore of the nozzles (the melt zone) is typically on the order of 1-2 millimeters, so the best option is to use an abrasive cleaning filament. Something tougher, like a wire brush, is not possible in that confined space, as far as I know.
If anyone has a good answer, I'll spread it around the 3D printing community, so you won't just be helping me. Thanks in advance!

Do you know if the PETG is amorphous or crystalline? I was wondering if the heat exposure and rate of cool down might be changing the solid form on cooling. There are so many options to try it's hard to know where to start. For (fairly) innocuous substances, I would try DMSO, sulfolane mixed with about 5% water, silicone oil or a strong solution of lactic acid. I would try 350 cps grade for the silicone oil, it's cheap and not volatile.

According to Wikipedia, PETG is amorphous. The plastic is melted quickly (less than a second) in the hot end, may heat further from friction as it is pushed out the nozzle (typically 0.4 - 0.5mm), and then it cools (either from ambient air temperature, or from room air being blown over it by a cooling fan). This doesn't generally cause a problem. However, small amounts often cling to the hot end nozzle and may degrade. Additionally, if the hot end is left idle (heated up but not extruding), it could potentially sit there around 230C for hours.

If the residue comes from really long overheating, it might be partially carbonized. In this case it CANNOT be dissolved in anything except strong oxidizer that can eat metallic parts they come into contact with.

Anyway! I'd try toluene, xylene and similar aromatic solvents. They might work. If they don't, stick with mechanical cleaning, unless you can use ceramic/glass nozzle. In THIS case I would consider some aggressive oxidizing solutions.

Blowtorching is a nice option if your nozzle can survive it. If it can, use blowtorching without hesitation. It is easier and cleaner than any solution/solvent.

It is difficult to tell when you consider PETG overheated (how long? what temperature?). Thermal degradation likely produces several reaction products each of which having different solubility in different solvents. So I don't know whether I ever encountered "overheated" PETG during my own 3D printing experiments.
But I have had very good experience with Tetrahydrofuran (THF), which I happened to have in my basement for a long time (beware of peroxides!) and I have finally found an application for. It dissolves freshly printed PETG (and glue parts together) as well as the residues on the print nozzle.
Side notes: Acetone is wellknown for being absolutely useless for dissolving PETG and I can confirm this from my own experience. All you can expect is to roughen the surface a little bit.
You can read several reports that say dichloromethane is not working as a solvent for PETG. I have found a report that claims Cyclohexanone is working. But I haven't got personal experience with these both. Toluene neither.
But in general, the best method for cleaning nozzles is brute force, i.e. pyrolysis by bringing the (brass) nozzle to red heat with a gas torch. Fast (takes only a few minutes) and reliable (absolutely no residues, expecially at the nozzle hole). One thing I have noticed is that it is a good idea to polish the nozzle with finest steel wool after pyrolysis, especially the inside. On the inside this polish reduces friction for the filament (the brass gets "hazy" after heating it with the gas torch). On the outside, having a shiny surface may reduce emissivity/absorbtivity a little bit, which helps keeping set temperature during printing.

"Overheated" for PETG in 3D printing would typically mean hours to weeks at up to 260°C.

I see for melting point "melting point of brass (900 to 940 °C, 1,650 to 1,720 °F)" and the temperature of red hot brass "525C -- 975F -- Red heat, visible in the daylight,
581C -- 1077F - Red heat, visible in the sunlight,
700C -- 1292F - Dark red, 800C -- 1472F - Dull cherry-red,
900C -- 1652F - Cherry-red".  Presumably keep it down to dark red?

I admit that "red heat" was a little vague. What I meant was the one you barely see at the lower end of the temperature range. From the capabilities of my hot air gun (causing about the same 'redness' as my gas torch) I'd estimate that to something around 500°C, but of course not sure because I didn't check it with the thermal camera.

As to getting 'overheated': weeks is certainly not true for the sample of PETG filament I have had. Don't know anything about its purity of course. Say 1 hour seems realistic to me (gets slightly brownish after a decent print job).

Thanks for the temperature clarification!  Re the 1 hour time duration, please clarify, seems likely we are talking about different things.  Print duration often exceeds an hour (I did a 16 hour print recently, hardly exceptional).  It seems to me that at least the average 3D printer enthusiast is unlikely to remove and heat the nozzle to red hot after every print, or interrupt prints every hour, so I must have misunderstood.

@Technophile, I can only talk about my experience with the specific filament brand I have bought and with my specific 3D printer. Maybe that material variety of PETG is impure and degrades faster or at lower temp? Maybe my printer overheats?

what exactly is it that you are referring to with "Say, one hour"? Perhaps the time until you see the first residue on the nozzle?

If the removable nozzle is small enough, what about putting it in a test tube with potassium chlorate, sodium chlorate or magnesium nitrate and heating until the oxidizing salt melts? Basically just a fusion for maybe 30 minutes or so. Then dissolving off the crust and some light abrasion. Basically, the gummy bear method. Just a thought.

@Technophile: yes, until I see the first residue on the nozzle.

@oliver. OK, thanks.  I would expect to see the first bit of staining within the first hour.  My "hours to weeks" comment was about the interval I might expect between nozzle cleanings.  A useful cleaning method would have to be capable of removing filament residue that had been on the nozzle for hours to weeks.


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Two possible chains - isopropyl or ethyl and methyl?

I see two possibilities for choosing the longest branch, and they yield pretty different names:

3-isopropyl-3-hexene
3-ethyl-2-methyl-3-hexene

Is there a rule to determine which to choose? Does one of these prefixes take priority over the other, for example? Or is the second option preferable because of the lower number? I suspect this is a very straightforward rule...

You mean 3-(1-methylethyl)-3-hexene instead of isopropyl...

-isopropyl is not strictly a systematic name of a substituent. Use that chain which gives lowest position numbers.

The first important simplified criteria for the choice of a principal chain in such compounds are:

greater number of suffixes
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds

The corresponding wording of the nomenclature rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows:

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)
(j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);
(…)

Since the compound that is given in the question does not have any principal characteristic group (a characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name), the first criterion (greater number of suffixes) given in P-44.1.1 is not relevant in this case.
The second criterion given in P-44.3.2 is the longest chain. However, both possible names, (3Z)-3-ethyl-2-methylhex-3-ene and (3Z)-3-(propan-2-yl)hex-3-ene, indicate a principal chain with six skeletal atoms.
Also, both possible principal chains include the double bond. Therefore, the criteria given in P-44.4.1 (a) (greater number of multiple bonds) and (b) (greater number of double bonds) cannot be used to reach a decision.
The criterion given in P-44.4.1 (h) (lower locant for the suffix) is not relevant since the compound that is given in the question does not have any suffix.
Finally, the criterion given in P-44.4.1 (j) stipulates that the principal chain shall have the lower locant for the double bond, i.e., for the ‘ene’ ending. However, both names, (3Z)-3-ethyl-2-methylhex-3-ene and (3Z)-3-(propan-2-yl)hex-3-ene, have the locant 3 for the ‘ene’ ending.
Further criteria can be found in Section P-45:

P-45 SELECTION OF PREFERRED IUPAC NAMES
P-45.0 INTRODUCTION
Two or more names may result based on the same senior parent structure selected according to P-44, because of different substitution patterns or multiple occurrences of the same senior parent structure. (…) Application of
  the criteria in this section will generate the preferred IUPAC name.
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
(…)

In accordance with P-45.2.1, the preferred principal chain corresponds to (3Z)-3-ethyl-2-methylhex-3-ene, which has a principal chain with two substituents (3-ethyl, and 2-methyl), whereas the principal chain of (3Z)-3-(propan-2-yl)hex-3-ene has only one substituent (3-(propan-2-yl)).


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Why is the Jacobsen's catalyst with tert-Butyl groups a better catalyst for epoxidation?
In an experiment I synthesised the below Jacobsen's catalst:

I used it to create styrene oxide through epoxidation. Chiral GC analysis showed that the enantiometric excess was lower compared to using Jacobsen's catalyst with tert-Butyl groups (shown below). This leads me to believe that it is more enantioselective, but why is this?

I haven't worked on this stuff in a more than a decade, but if memory serves, having more electron donating groups in the aryl ring increases selectivity.

This type of oxygen transfer reaction consists of a two step
catalytic cycle, wherein, first an intermediate (salen)$\ce{Mn^V}$ oxo complex is generated, and then the activated oxygen is added to the olefinic double bond.
Electron withdrawing and electron donating group influence the strength of the $\ce{Mn=O}$ bond, with electron withdrawing substituents leading to a weaker $\ce{Mn=O}$, and electron donating substituents leadings to a stronger $\ce{Mn=O}$ bond.
An argument based on Hammond Postulate can be applied where in the case of electron withdrawing substituents (weak $\ce{Mn=O}$ bond), one expects an early transition state, and in the case of electron donating substituents (stronger $\ce{Mn=O}$) a late transition state is seen.
Enantioselectivity is governed by interactions of the salen ligand, which in the transition state geometry adapts a folded geometry forming a chiral pocket, and the incoming olefin.
A late transition state demands a certain proximity between the
olefin and the manganese oxo complex. This leads to a less  spatial separation between substrate and catalyst and consequently allows for better differentiation of stereoisomeric transition structures.
Thus the attenuation of the reactivity of the oxo-species by electro donating groups, leading to a comparatively late transition state is the reason for higher enantioselectivity.

References
This article looks at the electronic effects of 5,5' substituents using DFT calculations.
Cavallo, Luigi, and Heiko Jacobsen. "Electronic Effects in (salen)Mn-Based Epoxidation Catalysts." - The Journal of Organic Chemistry, 2008, 68(16), pp 6202–6207

Also, it was posited that some types of reactions involved two salen complexes, which is why the oligomeric catalysts were developed: http://www.sciencedirect.com/science/article/pii/S0040402014003779 (Yes, I'm the third author.) You can imagine that if this were the case, having t-butyl groups in the monomer would improve selectivity via steric factors...


text:
Transition state search for rotational motion
This question is regarding Gaussian Computational software package.
I have a system where one product forms from one conformational arrangement of molecular intermediate (conformational minimum) and another product forms from a second conformational arrangement of the same molecular intermediate (another conformational minimum). These two arrangements can be interconverted via a rotation about a single bond. In order to investigate the subsequent TS from each pathway, I also need to find the pathway that interconverts the two conformers.
How can I use Gaussian to find the rotational transition state that converts these two conformations? I have tried QST2 method using both conformers as input but did not work; I also tried doing a scan along the rotation about the single bond and do a subsequent TS search from the scan maxima but to no avail. Any other methods that I can use to find rotational transition states? I imagine the PES to be very flat, how can I improve my input for tighter convergence?

Could you be more specific on what your problems are? Namely, when you say that QST2 "input but did not work", what do you mean? What exactly happened? The same applies for PES scan.

Remember that for a QST2/3 scan to work, the atoms have to be ordered the same in each input block. Did it not work, because it aborted with an error, or did it not work because it did not lead to a stationary point? Please post an example of your input including keywords. Information on the version might also be necessary. In general such questions are not really on topic here though. I don't remember if gaussian has a user forum, but there would still be ccl.

Hi Wildcat and Martin, the qst2 outputs with normal termination; it is however not a product that i (or any chemistry) expect(s) as it forms strange ring fusions. It could have been that I did not order the atoms correctly in the same sequence in the two input structures. I will check this. Thanks!


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Does the mass of sulfur really decrease when dissolved in water and increase when burnt?
I was going through a bunch of interesting science 'facts' and one entry went this way:

Name an element whose mass decreases when it is dissolved in water and increases if it is burnt.

I tried guessing, but that didn't help. So I had a look at the answer: it was Sulfur
There are two questions that struck my mind after I saw the answer:

What does mass decrease/increase as stated here mean?(Does it mean that mass of sulfur along with other elements combined with it changes/or anything ambiguous such as change in actual mass is taking place) )
Is sulfur really the correct answer to that question? If yes, then why does sulfur show this anomalous property? Are there elements that display a similar property?

Any suggestion is welcome.

On combustion $\ce{SO2}$ is produced which has a greater mass than sulphur.

Shouldn't this question be true for more elements then??or sulphur is exceptional one.

Now this is one huge piece of bovine excrement. Sulfur does not dissolve in water, which means it does not change at all when you try to dissolve it in water, which in particular means its mass does not decrease. (It may dissolve in strongly alkaline solutions upon heating, but that's another story.) As for burning, sulfur forms gaseous products that fly away, which may be naively interpreted as "vanishing" or "decreasing mass", but hardly vice versa.

@IvanNeretin Isn't wettable sulphur available ? "Wettable sulfur is the commercial name for dusting sulfur formulated with additional ingredients to make it water miscible." --- available on Wikipedia https://en.wikipedia.org/wiki/Sulfur .... maybe the question isn't refer erring to that though. Because it has some impurities.

Yes this 'interesting science facts' article appears to be a pile of do-do! An atom i.e. element, cannot change its mass during a reaction, other than to gain or loose one or more electrons if ions are formed.  But as the electron mass is $\approx 1/1836$ that of the proton the effect is tiny.

@S007 You throw pure sulfur powder into water, and it just bluntly refuses to mix. It floats on top (despite having greater density), and if you try to mix it with a spoon, it forms ugly lumps, much like flour or starch, only worse. That's because it is not wettable. Now you add some chemicals and it becomes wettable. It starts to mix with water nicely, forming a suspension. Everything I said before is still true. Sulfur still does not dissolve in water, nor does it react.

@porphyrin If the reaction releases energy and you let that energy leave the system you're looking at, the mass of the system decreases by a tiny bit.

Except for tricks, matter does not change mass in the circumstances mentioned. It reminds me of: "Matter cannot be created or destroyed, nor can it be returned without a receipt"

@THELONEWOLF of course. Hydrogen (H2), when burned, produces water (H2O), which has a greater mass than H2. Magnesium (Mg), when burned, produces Magnesium Oxide (MgO), which has a greater mass than Mg. This is not a unique property of Sulfur at all.

@RobertColumbia what about decrease in mass when dissolved in water???

@CodesInChaos how do you envisage the energy is being carried away, i.e. what particles are going to do this?

Upon reading the answers on Quora (thanks S007 for pointing that out) I realized this trick question is a lousy play upon two somewhat peculiar features of sulfur:

When submerged in water (not "dissolved", mind you, for sulfur does not dissolve), its apparent weight becomes less, thanks to Archimedes and the buoyancy force. This is indeed true of any element and compound, but sulfur has density of about 2, so its apparent weight decreases about twofold.
When burnt, sulfur forms a compound with oxygen which (containing all that sulfur plus oxygen) naturally weighs more than sulfur alone. Again, this is true of any element or compound that can be burnt, which are numerous. If anything, sulfur is "special" in that its atomic weight is 32, and the molecular weight of $\ce{SO2}$ is 64, so the increase is exactly twofold.

The way it stands now, though, the question is quite meaningless. Mass of any element never changes (barring nuclear reactions), that's the law of conservation. If you count compounds, then the mass would increase in any reaction (presuming we started from the pure element) and stay unchanged otherwise. If you count all compounds except gases, then the mass would sometimes increase and sometimes decrease, so there might be an answer, but then sulfur fails both conditions: it does not dissolve in water at all, and it all transforms to gases (that is, "vanishes") when burnt.

The "decrease" in weight isn't Archimedes' buoyancy force but surface tension. If it was really buoyancy then large particles would float in water too.

I think I addressed this properly: I said "when submerged", which pretty much rules out the surface effects (strong as they are).

Wouldn't the mass of the element decrease in most reactions (or at least thermodynamically favourable ones)? Sure, the difference between S bound in an allotrope and S bound in SO2 isn't large, but it's not nonexistent, when you factor in the energy it sheds in forming the more stable compound; right?

You mean, because of $E=mc^2$? Yes, it would; I ignored that.

@IvanNeretin Einstein must be spinning in his grave. ;) It's still a stupid question relying on intentional ambiguity, of course (obligatory xkcd: https://xkcd.com/169/), but it still presents a good opportunity to remind readers of such things, I should think.

@WillihamTotland So true, man. So true.

sulfur has density of about 2 2 g/cm^2?

@cat g/cm^3, but yes. Depending on the allotrope it's more like 1.8, though, which for any appreciable quantity is a ways out.

@cat And probably due to the simplicity of the unit calculation, densities are generally given as the raw number, at least in my experience, unless dimensional analysis is in play.

@WillihamTotland et al. ya the fractional change "due to Einstein" is small, 3 eV vs 60 GeV (roughly) but I was hung up trying to figure out what fraction of the mass loss to assign to the Sulfur, and what to the Oxygen, since the energy loss is (roughly speaking) in the electrons. So I really misunderstood the question!!

Sulfur is not "buoyant." I'll guarantee that a fist sized chunk of solid sulfur dropped into a five gallon bucket of water will sink and not float.  As I said you can "float" small particles of sulfur on water due to surface tension.

The decrease in weight is actually due to buoyancy force and not due to any chemical reaction as such. Though hydrogen sulphide gas may be formed but it will be in negligible quantities. The relative density of sulphur is about 2.
On combustion, $\ce{SO2}$ is produced which has a greater mass than sulphur.

P.S: Did you know that "magic" is science in disguise ?

Apparently the question seems to be quite popular.
https://www.quora.com/Does-sulfur-really-weigh-two-kilograms-when-dry-one-kilogram-when-wet-and-three-kilograms-when-burnt

A.C. Clarke: "Any sufficiently advanced technology is indistinguishable from magic." (emphasis from mine) I don'tknow about science. ;-)

@Karl "Word Play" =)

upvoted for including the title of the quora question, which is the only thing that makes any of this make sense.  What's interesting is the 1:2:3 ratio, not just decrease/increase, so it makes sense that someone would try to point this out in a clumsy way, and that further distortion could lead to the nonsense that the OP saw.

If you want to say that buoyancy forces "decrease weight" then you're also saying that you're weightless when you sit on a chair, since the chair provides an upward force that "decreases your weight" to zero.

This is, mostly, nonsense. Your question didn't speak to weight, it spoke of mass. It is generally agreed that elemental sulfur exists as an S8 ring. (cyclo-S8). (really we know it exists in a number of allotropes at STP) It is quite water insoluble, (with a solubility of 2½ ppm).
It is a basic assumption of chemistry (with the exception of radiochemistry) that mass is fixed (conserved) and does not change in any chemical process (including phase changes and dissolution). It is possible that when S dissolves in water, the S ring is broken up so that the molecular mass is decreased - I don't know. When it is burnt (oxidized with O2), the product's molecular mass is quite a bit less than the 256 of S8, meaning the "answer" is nonsense.
However. As any physicist knows E=mc². So an increase in an atom's ENERGY increases mass by E/c². Of course the energy change will be small in burning it, and c is a very large number, so that the tiny mass increase is not significant for a chemist.
Similarly, if S8 is dissolved in water cooler than it is, energy will be lost, but the same "insignificance" argument can be made. For a chemist (as I said, except for radioactive elements (or those in high energy environments like fission reactors, fusion experiments, or particle colliders (or when exposed to cosmic rays))) Mass is conserved, it does not change upon dissolution, burning, heating or cooling. The is no chemical element which changes mass upon dissolution or burning, although there are plenty of chemical compounds, including molecules of just one element, that may change molecular mass upon a phase change or dissolution, or that react to form products of more or less molecular mass.

Could you stop answering as unregistered? You already have an account.

When sulfur is burnt, energy is taken out of it, so its mass decreases (by a tiny amount, as you said, but it's worth getting the direction right).

Theoretically yes, if something loses energy it also loses mass (seriously no joke). Also if it gains energy it gains mass(By Einstein's mass energy equivalence)
Of course the energies involved in chemical reactions are in no way affecting mass of the molecule atom in any measurable way.
This applies to all forms of energy, potential, chemical, doesn't matter.
Example: the planet Saturn would be pulling harder at my toenail then any change might due to the potential energy of being in a gravity well (Earth).
What is the mass gain loss? We use Einstein's equation: $$\Delta M=\frac{\Delta E}{c^2}$$
So you see the  $c^{-2}=1.1126501 \times 10^{-17} $ part will slap the mass change to something almost nonexistent.
So yes,the mass will change,it depends on you how do you measure such a small change.

I don’t see how this answers the question.

You should write your answer to actually answer the question asked. I see no reference to sulphur in your answer at all.

Okay, let me state my criticism in a different way. Yes, you are technically correct about energy and mass and losing one means losing the other. To answer the question asked (‘(Why) Does sulphur’s mass decrease when dissolved and increase when burnt?’) requires relating what you posted in your answer to sulphur, the process of mixing sulphur and water, and the process of reacting sulphur with oxygen in a strongly exothermic way. That exactly is what your answer is failing to do and therefore it does not answer the question.

hey look,..let smake Sulphur heavier S+O{2} ==> (SO{2} after burning).__
So how does this logic work? we get to add the mass of 2 Oxygen atoms to every Sulphur atom and presto "sulphur just got heavier by burning"
Seriously? .....But hey you can actually drop words like  "exothermic", cool

Words have meaning, heck if SOx is sulphur why not call everything a frog?

Congratulations. You found the flaw in the question. That should be the answer. See the other answers.

You should re-read the previous answers, I expanded on the mass-defect answer already given. Maybe you missed that. You also dont understand why type of reaction (your qoute)"exotermic" or not is not relevant in the least here.

Answers on SE should be self-containing. Every ‘answer’ that provides additional information without answering the question is not an answer.

I answered the question I clearly stated it is not possible (except when you consider the exotic case of mass defect). People initially not getting what that means or going off tangents into "exothermic reactions" is sad, but not really my fault.


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Order of reaction / Rate-determining step in enzyme catalysed reactions
According to my chemistry book, there are 3 simplified steps within an enzyme catalysed reaction:

$\displaystyle\ce{S + E -> ES}$
$\displaystyle\ce{ES -> EP}$
$\displaystyle\ce{EP -> E + P}$

Where S is substrate, E is enzyme, ES is enzyme substrate complex, EP is enzyme product complex and P is product.
It is my understanding that at first, when substrate concentration is low the rate determining step is step 1, as this is the slowest step in the mechanism.
When you increase substrate concentration to a certain level, where there are no free enzymes remaining to bond with them, adding more substrate will not make the reaction faster, so the order has become 0 with respect to substrate conc.
Is the rate determining step now 2 or 3? I thought it would be 3 as that is the step that frees up more enzymes which could go on to catalyse more reactions. My book says that it is step 2, as "it depends on how fast the ES can convert into EP". But surely this step won't free up more enzymes to catalyse reaction further?
Is the book correct? Can anyone help me understand this?

The rate determining step doesn't have anything to do with any of the factors you mentioned. It's not even true that the rate determining step can be determined directly from the rate of $$\ce{\mathrm{ES}->\mathrm{EP}}$$
The correct definition is that it is the biggest barrier from a stable species (reactant or intermediate) to any subsequent transition state along the reaction coordinate.
Your comment about the apparent rate of the reaction is unrelated. The rate-determining step is an intrinsic property of the reaction and its mechanism. It is unaffected by concentration, though concentration effects may change which mechanism out of several is predominant.

To add onto Zhe's answer, consider the Michaelis-Menten model of enzyme kinetics.$$v=\frac{k_{cat}[E_0][S]}{K_M+[S]}$$The $k_{cat}[E_0]$ term is sometimes given as $V_{max}$ which is the maximum speed of the reaction while $k_{cat}$ itself is the rate at which the enzyme produces product molecules. $K_M$ is known as the Michaelis constant and behaves like an equilibrium constant. Thus, enzymes with a small $K_M$ are said to have a strong affinity for their substrate because at very low concentrations of substrate, the enzyme is bound.
To answer the question you ask about a large concentration of substrate, this is the case where,
$[S]>>K_M:$
In this case, we can rewrite our equation as,
$$v=k_{cat}[E_0]=V_{max}.$$Notice the cancellation of substrate concentration when $K_M$ is very small.
This means, as you point out, the reaction goes no faster for large concentrations of substrate because the enzyme is always completely saturated and able to find another substrate molecule immediately. Notice that what exactly is a large concentration, however, depends on the size of $K_M.$
The Michaelis-Menten equation does not make any assumptions about what is the rate determining step when being derived, so there is no reason to worry about that. There are limiting cases to think about though.
It's worth pointing out that in the other limiting case, we get a linear equation.


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Shallenberger Acree example
At the moment I'm writing a paper about sweeteners and want to compare saccharides, polyoles and synthetic sweeteners like cyclamate.
I'm having some problems trying to apply the Shallenberger–Acree model and would like an explanation for Glucose/Sucrose and Maltitol since I already figured it out for Cyclamate.
Please keep your English simple since it isn't my native language.


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Can a first order reaction change to a second order reaction under specific temperatures?
I'm studying basic reaction behavior. My question is:
Consider a Continuous Stirred Tank Reactor (CSTR) where an exothermic, first-order, irreversible reaction takes place:
$$
\ce{A ->[k_1] B}
$$
Is there any possibility or (does it make sense in chemistry) that under high and/or low reaction temperature the reaction will switch to a second order reaction? i.e.
$$
\ce{A + A ->[k_2] C}
$$
If yes, can you give an example? Please consider answering in layman's terms.

An example in time would be triplet-triplet annihilation leading to delayed fluorescence, bi-molecular at short times just after light absorption and then first order at longer times. With temperature one can imagine first order in a  viscous solvent at low temp. and bimolecular at higher temp when viscosity is reduced.

It's theoretically possible, though I don't know if there exists specific examples.
The first thing to recognize is that reactions don't really "switch" in an on/off fashion (at least not based on temperature). If there's two physically possible reactions, they both occur under all conditions ... theoretically. The main reason you see one and not the other is due to the comparative rates of reaction - one reaction might be so slow that no appreciable amount of product forms in reasonable timeframes.
So while the rates of reactions change based on temperature, as Zhe mentions, the way that rate changes in most cases (the Arrhenius equation) means that the two reactions aren't going to swap positions based on those considerations alone.
However, kinetics is not the only knob we're playing with here, when we change temperatures. We're also messing with the thermodynamics. Specifically, temperature is a factor in the free energy equation
$$\ce{\Delta G = \Delta H - T \Delta S }$$
If the two reactions have different $\Delta H$ (enthalpy) and $\Delta S$ (entropy) values, then we can change the spontaneity of the reactions. At low temperatures, something with a very favorable enthalpy will be spontaneous, even with a disfavorable entropy. Conversely, at high temperatures, something with a very favorable entropy will be spontaneous, even with a disfavorable enthalpy.
So if you have a system which can undergo first-order and second order reactions (for example, a molecule which has ) and choose your reactant such such that the first order reaction has a more favorable enthalpy and a less favorable entropy than the second order reaction (or vice versa), there will be a temperature where the thermodynamically favored product will go from that of the first-order reaction to that of the second order reaction. Whether this would be enough to manifest itself as an appreciable difference in product depends on the particular numbers.
While we're on the discussion, there is another technique related to controlling product outcome, that exploits the balance between the two factors already discussed. That's using thermodynamic versus kinetic reaction control. In this setup, you have one reaction which provides a more stable product, but due to things like steric hindrance the reaction rate is lower.  The other product forms faster, but the product is less stable. If you set up reaction conditions such as short reaction times at high temperature where the speed of the reaction is the dominant factor, the faster-made "kinetically controlled" product dominates. However, if you use conditions such as long reaction times at low temperatures, where the stability of the products is the dominant factor, the more stable "thermodynamicly controlled" product dominates.

I think the short answer is no.
The reason being that the reaction coordinates are intrinsic properties of the reactions. They define the rate determining steps that give rise to the two rate laws. These barriers are temperature dependent, but they are temperature dependent in similar ways, so I do not think it is possible to have a temperature based switch over in this specific case.
I purposefully omitted an mathematical descriptions because you asked for an answer in laymen terms, but please comment if more details are required...

Thanks for your swift answer, but can I add, is there any other, controllable or uncontrollable , factor which can cause a switch in the reaction ?

You can have a catalyst that works for reaction 2. If it's bound to a solid surface, you'd able to control easily the second pathway by adding or removing the catalyst.


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Can the colour change in litmus paper be explained by conjugated systems?
I have been trying to understand the colour change in litmus paper on a more detailed level than "the chemical identity changes, and therefore also physical properties". I have knowledge of the particle in a box model of quantum mechanics, and i would like to apply this to the conjugated system in the active indicator-component of litmus dye.
In other indicators I have looked at, you can clearly see that when the dye molecule dissociates or is protonated, there is change to the length of the conjugated system (number of alternating double bonds in a row), which explains the colour change. An example is given below for BTB (bromo thymol blue)

However, i can't find any explanation like this as to why the indicator dye in litmus paper (7-hydroxyphenoxazone) changes colour when protonated/deprotonated. From what I can see, the length of the conjugated system is the same in both versions of the molecule.

Edit: If someone knows of a model that more easily explains this colour change than conjugation, feel free to share that as well.

Did you search for the reaction of litmus/other indicators in different mediums on the Net ? Usually a chemical reaction takes place according to medium.

I think that's already addressed in the question. The conjugated system extends throughout the molecule regardless of whether or not you protonate on the carbonyl...

You have shown the structural change in bromothymol blue upon deprotonation.
For the anion of 7-hydroxyphenoxazone, the following resonance structures involving a p-quinone imine are conceivable:

The assumption that the $\pi$-system of the neutral species and the corresponding anion are identical is not justified.
A smaller system, in which a similar effect can be observed, is 4-nitrophenol. A solution of the phenol is colourless, whereas the solution of the phenolate is yellow ($\lambda_{\mathrm{max}}$ = 405 nm).
Again, loss of a proton leads to tautomerism with contribution of a quinoid structure.

I think the OP's question is that this system is just as conjugated as the original.

Yes, these are the resonance structures I thought would arise from the deprotonation of 7-hydroxyphenoxazone. But, as @Zhe pointed out, I dont understand why this structure gives rise to a different colour than neutral 7-hydroxyphenoxazone. From what i can see the conjugated system is similar in both.

@Adroit Similar, but not identical. Change in uv-vis absorption due to protonation/deprotonation without breaking further bonds is a rather common effect. Think in methyl red, methyl violet, or the anthocyane dyes. This is about electron distribution in the HOMO, isn't it?

@KlausWarzecha Hmm, not sure  I understand. Does the dissociation add another electron to the same conjugated system, and so the energy of the HOMO is higher => HOMO-LUMO gap corresponds to lower energy light => more red light absorbed => appears blue? If this is correct, can you explain why the dissociation adds one electron to the conjugated system?

@Adroit In my (somewhat limited) experience, deprotonated oxygens (for example) with lone pairs and a negative formal charge have a tendency to  withdraw a much larger fraction of electron density from the structure they're attached to than they do when they're protonated. So, even though the structure remains conjugated, the bond system has a different density distribution -- different enough to have distinguishable absorbance behavior.

How would i og about predicting the colour of the above two examples? From the provided answers it seems to me that the particle in a box model is not sufficient.

Klaus, is the resonance/tautomerism critical to the absorbance in the visible? Are there examples where simple deprotonation shifts the (ground and/or excited) electronic structure sufficiently to induce a measurable change?

The resonance/tautomerism view probably is more like a guideline for a practical organic chemist to rationalize the observation. In the end, it can't replace a real calculation. As for the second part, I may not have fully understood the the question. Aren't the cases where where a change in pH results in a colour change (the two examples here or anthocyanes) already examples (or at least indications) for a change in the electronic structure?

The answer @Kalus Warzecha seems to make sense. (a) the particle in a box or on a ring is a very crude model and only a 'rule of thumb' method (b) the molecule is not symmetric wrt C=O & OH so changes in energy are to be expected if the swap around, plus the change in energy to cause a colour change is quite small (c) the molecule is very simple , what else can cause a colour change?

@KlausWarzecha See my new question here. Also, I would argue these are not tautomerizations, since the graph of the molecular connectivity doesn't change in either case -- just the bond orders and the location of the formal charge.


text:
How to measure the amount of calcium in a supplement?
In chemistry lab, my group was assigned to determine the actual amount of calcium in a supplement.  We did this successfully using HCl to dissolve the tablet, vacuum filtration, and then the addition of sodium carbonate.  Next week, it has been suggested that we attempt it again using a different method.  Thus, we are seeking the most interesting way, albeit using elementary tools and chemicals, that one might do this.

For the determination of calcium (at least) two methods are conceivable:

gravimetric analysis
Sulfate is a rather bad choice here. As compared to the carbonate, the sulfate has a much higher solubility. A better choice would be the oxalate with a solubility around $6\,\mathrm{mg\,l^{-1}}$.
photometry
Calcium forms a purple complex with the Arsenazo III dye that can be detected at wavelengths of 660 and 700 nm.
Alternatively, you can use o-cresolphthalein and detect the complex at 545 nm.
The photometric methods are typically used in automated high-throughput clinical analysis of calcium in serum, but given that a photometer isn't available in the students' lab, Karl's comment on EDTA is exactly what you might be looking for. Complexometric titration is the keyword you might want to look for. Here, an indicator that forms a coloured complex with $\ce{Ca}$ is replaced by EDTA. The liberated indicator then shows a different colour.

Titration with EDTA?

@Karl Yes, that's definitely the classic. But there was a twist, wasn't it? $\ce{Mg}$ could be determined directly with Eriochrome Black T as the indicator, while $\ce{Ca}$ was run in a repacement titration  because the binding to the indicator was somewhat sluggish.

I also remember something like that. But second semester is too long ago. I'd have to dig up Jander-Blasius  in my basement. ;-)

@Karl Same to me ;) You probably had the Merck broschure on complexometric titrations too :)

Right. Where'd you get your indoctrination?

Let us continue this discussion in chat.

If I remember correctly, the real problem about the experiment of complexometric titration was that you had a mixture of Mg/Ca (at least at my school). If you only have Ca present there should not be any twist to it.