What exactly happens when vegetable oil is partially hydrogenated? My current assumption is that partial hydrogenation requires a polyunsaturated fat, because partial implies at least one but not all, which means there must be at least two double bonds in the fat, hence polyunsaturated, right?

Assuming that is correct:

My current understanding is that when you bubble hydrogen (using some heavy metal catalyst) through a vegetable oil, somehow this results in transforming polyunsaturated fats to unhealthy forms of unsaturated fats called trans fats.

What exactly is happening, here? I initially assumed that adding hydrogen to the fat would do little more than making double-bonded carbons into single bonds and then attaching two hydrogen atoms to the now-free valence electrons.

But apparently it also removes cis-kinks and makes them into straight trans-lines, which doesn't change the number of hydrogens but rather changes the shape.

So what's going on here exactly? What do we say partial hydrogenation is actually doing to the polyunsaturated fat?

You actually don't need a polyunsaturated fat to be partially hydrogenated. The "partial" refers to the macroscopic view of the fat, rather than a per-molecule basis. That is, if you have, say, one mole of monounsaturated fats and hydrogenate the double bond in half a mole of them, then you've partially hydrogenated the original mole of fat. Taken individually, each molecule has either been hydrogenated or it's still monounsaturated - no single molecule has been "partially hydrogenated". However, the mole taken as a whole has only been partially hydrogenated - 50% of the double bonds that were originally present have been hydrogenated.

Regarding trans fats, the important thing to realize is that (parts of) the hydrogenation reaction is reversible. The addition of molecular hydrogen is not concerted. Each hydrogen is added in a step-wise fashion (roughly). It's in that one-hydrogen added, one-to-go stage where the reaction can reverse course and go back to a double bond. But prior to reforming, the catalyst has broken the double bond and you have free rotation around it. Which way it's put back together is a matter of chance.

So you start with a fat which has all cis bonds. (Because the stereo and regio-selective enzymes that make them only make cis fats.) You then start the hydrogenation process. Each time you add hydrogens some goes to completion, and you end up with saturated bonds. But a fraction of the reactions get part way and then reverse. But as the hydrogenation catalyst is not stereo- and regio-selective, it's just as happy reversing to a trans bond as it is to a cis bond. So it's those bonds - the ones which go part way and then reverse - which end up as the trans fat in the final product.

The reason why fully hydrogenated fats don't have this issue is because they're produced with an excess of hydrogen. Yes, some of the double bonds react part way and then reverse to trans fats, but the catalyst is not stereo- and regio-selective, and it's able to convert the trans bonds to saturated bonds just as easily as the cis bonds. All the trans fats that get formed in the reaction are subsequently destroyed by the complete hydrogenation.


When two containers at different T and P are kept together, will pressure become same or temperature? Two containers, kept at different temperatures are connected through a stopcock and two different ideal gases at different pressures are filled inside them. Initially the stopcock is closed. When the stopcock is opened, will the pressures become same or will thermal equilibrium be established first? Why?

Here's an extract from an online source about thermal equilibrium-"As we have seen in the zeroth law of thermodynamics, when two objects are placed in contact heat (energy) is transferred from one to the other until they reach the same temperature (are in thermal equilibrium). When the objects are at the same temperature there is no heat transfer."

We are taught in class that after the stopcock is opened, gaseous molecules travel from container at higher pressure to the container with lower pressure. There wasn't any discussion on thermal equilibrium. If the temperature of two gases kept at different temperatures don't become equal, isn't that a violation of zeroth law?

When the stopcock is opened, the gas molecules at different temperatures do come in contact. So, why thermal equilibrium is not established and instead there is a greater driving force to make pressure in both the containers same?

The situation that is being covered in your class is where the two containers are thermally insulated from one another.  In this case, the only place where heat can be transferred from one container to the other is through the gas in the valve (by conduction).  But the assumption is being made in your class problem that heat conduction through the gas in the valve is negligible.  In any event, even if the valve were left open until the pressures effectively equilibrated, the temperatures in the two containers would require a much larger time to equilibrate.  So, if the assumption is made that, once the pressures effectively equilibrate, the valve is closed again (and does not allow heat transfer between the containers), then at final thermodynamic equilibrium, the temperatures of the gas in the two containers will be different.


How can I get this glass stopper out? Need troubleshooting! I placed a glass stopper in a $\pu{100 mL}$ round bottom flask while my solution was still warm and let it cool overnight. Now I think I created a vacuum seal and I can’t force the stopper out to filter my product. I’ve tried heating it up again but to no avail. I don’t want to slip oil in there and risk getting it in.

Anyone have a good idea?

I found gently taping a glass stopper with another glass stopper sometimes loosens it(maybe the shock is allowing air to pass through), but you could always put ice on the stopper and surround the top of the flask with warm water and wait a minute or two. This allows the top of the flask to expand while the stopper shrinks just a bit. Then, just take the stopper out. If you lose all hope (I hope that doesn't happen) and don't care about about your solution I would just use WD-40.


Order of reaction for hydrogen ions being ignored I have been reading through a case study about calculating the order of reaction for the following reaction:

$$\ce{H2O2 + 3 I- + 2 H+ -> I3- + 2 H2O}$$

The paper uses an iodine clock reaction with vitamin C to calculate its order of reactions, experimentally. However, when writing out the rate law, the author writes it as such:

$$\mathrm{rate} = k[\ce{H2O2}]^x[\ce{I-}]^y$$

How come it is not written as follows:

$$\mathrm{rate} = k[\ce{H2O2}]^x[\ce{I-}]^y[\ce{H+}]^z\;?$$

For what reason is the order of reaction of hydrogen omitted? My thinking was that it has an order of reaction of $z = 0$ and therefore it does not affect the rate at all. However, I was unable to find a source to verify this.

Imagine you are doing a process which takes 'n' steps. Label the steps : 1,2,3..n, now consider the time it takes to do step 1 , step -2 and so on. Now, wouldn't you agree that the time taken to complete process would be entirely dependent on the step which takes the most time(slowest step) to complete?
So, in your example the hydrogen ions are in one of the faster steps. So, if the concentration of hydrogen is 'enough' such that once you come to the step that their ample hydrogens to fuel the reaction, then it would seem that rate is independent of hydrogen concentration ( I say seem because there may be minute effects of it on the rate)


What does it mean when something is said to "contain ions"? (I'm looking for a very basic level explanation because my only chemistry experience is one fast-paced high school course.)

So, according to the professor of that course, ions are never found in nature as ions; everything would be neutrally charged. My question is, how can something have ions in it, without having them be neutralized? Doesn't nature want to lower potential energy, which would mean bonding the ions asap? Or am I misunderstanding things and said ions are already bonded to other ions, but they are still called ions even though the charge of the whole molecule should be neutral?

Your teacher is almost right but some terminology needs to be clarified.

Pure substances will almost always be electrically neutral (and the exceptions involve temporary local transfer of electrons from one substance to another as when you rub a rubber balloon on a wall surface). Beyond a certain level of charge separation you get enough electrical potential to cause sparks which allows the charges to equalise again. So persistent charge imbalance isn't easy to get. I think that is what your teacher means.

But if you look inside some pure substances, they do contain ions, just an equal number of each to avoid the bulk material having an overall charge. Common salt, for example, consists of ions (Na+ and Cl- in equal numbers). The substance is neutral, but the individual atoms making it up are all ions.

So your teacher is wrong if she meant that ions don't exist in nature but correct if she meant that bulk substances are electrically neutral.


Why do diacidic and triacidic alkalis dissociate in one step? My book states that dibasic and tribasic acids dissociate in water in two and three steps respectively but for diacidic and triacidic bases it states that they dissociate in one step. I do not know why.

For example:

$\ce{H2SO4}$ dissociates in water in two steps as follows:
$$\ce{H2SO4(aq) <=> HSO4^{-} + H+}$$
$$\ce{HSO4^{-}(aq) <=> SO4^{2-} + H+}$$

but

$\ce{Cu(OH)2}$ dissociates in water in one step as follows:

$$\ce{Cu(OH)2(aq) <=> Cu^{2+} + 2OH-}$$

Why is this so. Why does not $\ce{Cu(OH)2}$ dissociate in two steps to form $\ce{CuOH+}$ in the first step and $\ce{Cu^{2+}}$ in the second step?

Polyprotic acids

In most acids, even the strong acids, the bond to the acidic hydrogen atom is covalent (though frequently very polar). Thus, for acids, the more correct description of what happens in solution is ionization - a covalent bond is breaking heterolytically to form ions. The solvent is involved in solvating the proton, which is why we usually write $\ce{H3O+}$ for protons in water and not $\ce{H+}$.

Polyprotic acids ionize in separate steps because after the first ionization, the new species has a different strength covalent bond to break. Consider your example of sulfuric acid. Sulfuric acid is a strong acid that ionizes readily and (nearly) completely in water:

$$\ce{H2SO4 + H2O -> HSO4- + H3O+}$$

The resulting hydrogen sulfate anion is a different chemical species. Its bonds have different strengths. The second proton is not as easily ionized. In fact, the hydrogen sulfate anion is considered a weak acid. It does not completely ionize in water. An equilibrium is established with an equilibrium constant of $K_a = 1.02 \times 10^{-2}$. This means that at $\mathrm{pH = 2}$, the hydrogen sulfate anion is 50% ionized: half is hydrogen sulfate and half has ben converted to sulfate.

$$\ce{HSO4- + H2O <=> SO4^2- + H3O+}$$

What evidence do we have for this step-wise ionization? We have titrations! If we add small volumes of a solution of a stong base (say $\ce{NaOH}$) of known concentration to a solution of sulfuric acid and monitor the pH, we can see two distinct equivalence points (the inflection points on the graph where the pH change is very large over a small volume of base added) corresponding to the two ionization events. However, because the first ionization of sulfuric acid is complete before the base is added, a titration  curve of sulfuric acid will only show one equivalence point when two equivalents of base are added (though there is a blip at 1 equivalent).

If we consider maleic acid $(\ce{H2C4H2O4})$, for which both ionizations are "weak" $(K_1=1.26\times 10^{-2},\ K_2=8.51\times 10^{-7})$, we get a different looking titration curve with inflections at 1 and 2 equivalents of base. If the ionization of both protons occurred simultaneously, we would not see a curve like this.

$$\ce{H2C4H2O4 +H2O <=> HC4H2O4- + H3O+}\\
\ce{HC4H2O4- + H2O <=> C4H2O4^2- + H3O+}$$

  Bases

Bases that are the hydroxide salts of metals are ionic compounds. Though many of them have poor solubility in water, what is dissolved is completely ionized. Thus, for copper (II) hydroxide in water at equilibrium, there is only solid copper (II) hydroxide ad solubilized copper(II) and hydroxide ions:

$$\ce{Cu(OH)2 (S) <=> Cu^2+ (aq) + 2OH- (aq)}$$

A titration curve for copper (II) hydroxide with a strong acid like hydrochloric acid would show only one equivalence point at two equivalents of acid:

All ionic bases will behave similarly.

However, some bases are covalent and ionize by reacting with water:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

If there was to be a base of this type that had two ionizable groups, we would see two equivalence points.


How to find the molecular weight of a given gas based on its density? The density of a gas $\ce{X}$ is $10$ times that of hydrogen. In that case, what is the molecular weight of gas $\ce{X}$?

Well what I've done up to now is this:

The molecular mass of hydrogen is $M(\ce{H2})=2$, the density of hydrogen is $\rho(\ce{H2})= \pu{0.089 kg/m3}$.
Assuming that equal volumes $V$ of both gases are taken, the mass of hydrogen $m(\ce{H2})= (0.089 \cdot v) \pu{kg}$ and the mass of the unknown gas $m(\ce{X}) = (0.89 \cdot v) \pu{kg}$.

Although this hasn't really gotten me anywhere. How will it be done?

Indeed you can use Avogadro's law here, but I guess it is useful to give a little bit more insight in Avogadro's Law. What you can do is apply the ideal gas law:

$$p V = n R T $$

Here $n$ is the amount of substance of gas, which you can also write as:
$$ n=\frac{m}{M},$$
where $m$ is the total mass of the gas and $M$ is the molecular mass.
You can rewrite the ideal gas law now to:
$$p M = \frac{m}{V} R T,$$
where you should recognize that $\frac{m}{V}=\rho$.

The gas constant $R$ is (as the name shows) a constant so you can write
\begin{align}
  R &=\frac{p M}{\rho T} &
  \rightarrow            &&
  \frac{p_1 M_1}{\rho_1 T_1}&=\frac{p_2 M_2}{\rho_2 T_2}
\end{align}

And therefore, for equal pressure and temperature you will find that
$$\frac{M_1}{\rho_1}=\frac{M_2}{\rho_2},$$
which means that if the density of gas 2 (the unknown gas $\ce{X}$) is $10$ times as high, the molecular mass should be as well.

What is interesting to add is that this assumes the ideal gas law to be applicable. If, for example, your gases would be at high pressure (e.g. $\pu{200 bar}$) this is typically not a good assumption so the molecular mass that you would find for a $10$ times as dense gas will not be equal to $10$ times the molecular mass of hydrogen.


Thermal stability of charged and uncharged alpha helices (the concept of alpha-helix capping) I just learned that charged alpha helices are less thermally stable than uncharged ones. The only difference I see between them would be a larger dipole moment in the charged helix, but how would this contribute to stability?

Here a picture for clarification:

This might relate to the concept of 'capping' alpha-helices, where the placement of beneficial charges at either end of a helix is shown to increase the overall stability of the protein.  See the article1 published in Nature.
In general, by introducing preferential charges at the ends of alpha-helices you stabilize the inherent dipole moment created by the alignment of amide groups in the protein backbone.

Here is a different citation (found here):

In 1988, Richardson and Richardson observed that certain amino acid
residues are stastically favored as the amino-terminal caps for the
alpha helices in 45 globular proteins. The introduction of a
negatively charged side chain at the N-cap, which can neutralize the
partial dipole created by the unpaired amide protons, has been shown
to increase stability in T4 lysozyme.

I also believe the figure in the question is a bit misleading - as the illustrated charges may be intended on side-chain groups. I therefore assume that this helix is a part of a protein, and that the illustration simply wants to highlight the fact that you place charged side-chain groups on the ends of the helix (i.e. capping) within a protein structure.
Reference

Serrano, L., Fersht, A. Capping and α-helix stability. Nature 342, 296–299 (1989). DOI: 10.1038/342296a0


Why do elements in columns 6 and 11 assume 'abnormal' electron configurations? When I look around for why copper and chromium only have one electron in their outermost s orbital and 5/10 in their outermost d orbital, I'm bombarded with the fact that they are more stable with a half or completely filled d orbital, so the final electron enters that orbital instead of the 'usual' s orbital.

What I'm really looking for is why the d orbital is more stable this way. I assume it has to do with distributing the negative charge of the electrons as evenly as possible around the nucleus since each orbital of the d subshell is in a slightly different location, leading to a more positive charge in the last empty or half-filled d orbital. Putting the final electron in the s orbital would create a more negative charge around the atom as a whole, but still leave that positive spot empty.

Why does this not happen with the other columns as well? Does this extra stability work with all half or completely filled orbitals, except columns 6 and 11 are the only cases where the difference is strong enough to 'pull' an electron from the s orbital? It seems like fluorine would have a tendency to do do this as well, so I suppose the positive gap left in the unfilled p orbital isn't strong enough to remove an electron from the lower 2s orbital.

As I understand this, there are basically two effects at work here.

When you populate an $\mathrm{s}$-orbital, you add a significant amount of electron density close to the nucleus. This screens the attractive charge of the nucleus from the $\mathrm{d}$-orbitals, making them higher in energy (and more radially diffuse). The difference in energy between putting all the electrons in $\mathrm{d}$-orbitals and putting one in an $\mathrm{s}$-orbital increases as you fill the $\mathrm{d}$-orbitals.

Additionally, pairing electrons in one orbital (so, adding the second $\mathrm{s}$ electron) carries a significant energy cost in terms of Coulombic repulsion because you're adding an electron essentially in exactly the same space as there's already an electron.

I'm assuming that the effect isn't strong enough to avert fluorine having a $\mathrm{2s^2}$ occupation, and if you look at gadolinium, the effect there isn't strong enough to stop the $\mathrm{s}$ from filling (large nuclear charge and orbital extent at the nucleus is a good combination energy-wise), it does manage to make it more favourable to add the electron into the $\mathrm{5d}$ instead of the $\mathrm{4f}$ orbitals.

Also, if you take a look at tungsten vs gold, there the effect isn't strong enough for tungsten to avoid a $\mathrm{6s^2}$ occupation, but is for gold - more $\mathrm{d}$ electrons making the screening effect overcome the strong nuclear charge and enhanced nuclear penetration of an $\mathrm{s}$-orbital.


Avogadro Molecule Editor: How to render image from command line? Given a molecular geometry in the .xyz file format, how can I use Avogadro on the command line to render an image of it?

The website states that there is some command line API available, but I can't seem to find any documentation/information about it. Alternatively, are there any other molecular visualization tools that provide a convenient command line interface for image rendering of .xyz files?

There is no way to currently do this using Avogadro, although Python bindings have been "in the works" for a while. The problem is, with volunteer developers, there's not always enough free time to tackle all features.

You can create pseudo-3D images using Open Babel. I developed this for the Pitt Quantum Repository to give SVG and PNG images in the search interface. It's not perfect, but I like the effect.

obabel beta-carotene.xyz -O beta-carotene-2d.svg -xS -xd -xb none

The -xS indicates this 3D "ball and stick" style, -xd prevents the molecule name from being included, and -xb none gives  transparent background to the SVG (useful for presentations).


Why is a alkyne anion more stable than a benzene anion? I would think the benzene anion (A) is more stable b/c the compound aromatic. Yes I understand that sp hybridized carbons are more stable than sp2 for negative charges, however I thought that aromaticity is a more important factor in determining stability.

Aromaticity enters the reckoning only if the excess electron associated with the negative charge enters the conjugated pi electron system.

In your case that does not happen.  The negative charge is on a formally nonbinding pair on the carbon like other carbanions, but in a conjugated ring when you do not have a ligand on that carbon the electron pair takes the place of the ligand and isn't in the pi system.  So not stabilized by aromaticity.  Here you don't have the ligand, the top carbon in A where they put the negative charge has no hydrogen atom.  If a hydrogen atom were there the molecule would be neutral benzene.

You can have a negative charge stabilized by aromaticity in two ways represented by the benzyl anion and the cyclopentadienyl anion.

Benzyl anion:  This has the formula $\ce{C6H5CH2^-}$, a phenyl ring attached to a methylene group.  The methylene group acts as the necessary ligand that's missing in your $\ce{C6H5^-}$, so the excess electron pair enters the pi system and conjugates with the ring.  Note that the $4n+2$ rule for aromaticity does not work because with that extra pi-bonded carbon you don't have all the conjugated electrons in a ring anymore.  But you do have a highly de-localized structure that inherits the stability of the ring.

Cyclopentadienyl anion:  this ion, $\ce{C5H5^-}$, has the negative charge and all the ligands on the ring atoms, so the extra electron pair and negative charge are already in the pi system without having to conjugate with anything else like the case above.  You have a pure aromatic ring with the negative charge built in.  Pyrrole, furan, and thiophene are a variation on this theme where the ring is negative but neutralized by a positive formal charge on the nitrogen, oxygen or sulfur atom.


Defining a stereospecific reaction My textbook says that a reaction which gives different stereoisomeric products when the different stereoisomers of starting material are used in a stereospecific reaction.

Is it necessary that all different stereoisomers will give different stereoisomeric products? For example, in bimolecular elimination diastereomers give different geometric isomers while enantiomers don't. So, do we say that this is an example of stereospecific reaction?

A stereospecific reaction is one in which there is only one possible stereochemical outcome for the product and that outcome is controlled by the stereochemistry of the reactant.

One of the earliest examples studied by most students of organic chemistry is nucleophilic substitution by the $\mathrm{S_N 2}$ mechanism. This reaction in its simplest form has a concerted mechanism. Reactions with concerted mechanisms often have stereospecific outcomes because the concerted mechanism demands a certain geometric arrangement of the reactants. In the case of the $\mathrm{S_N 2}$ mechanism, the nucleophile must approach from the opposite side of the molecule from the leaving group for steric and electronic reasons.

Likewise, the $\mathrm{E2}$ elimination mechanism leads to a stereospecific outcome. Stereospecificity does not require a one-to-one mapping of stereochemical inputs to outcomes. Rather, it requires a predictable mapping of inputs to outcomes.

For example, consider the $\mathrm{E2}$ elimination of 2-chloro-3-methylpentane with hydroxide. The reactant has four possible stereoisomers and the major Zaitsev product has two possible stereoisomers. Each stereoisomer leads to only one possible stereoisomeric product even though each stereoisomer of the product can be prepared from two stereoisomers of the reactant.


Deriving fluorescence intensity equations I've been having trouble with deriving the equations in the following problem.

  The interaction between DNA and AO to form the AO–DNA complex can be expressed by the following reaction:

  $$\ce{AO + DNA <=> AO-DNA},$$

  whose equilibrium constant is
  $$K = \frac{\ce{[AO-DNA]}}{\ce{[AO][DNA]}}  \tag{1}$$

  Consider that initially there is only AO in the measuring cell displaying an emission at $\lambda_{em} = \pu{520 nm}$, and finally at equilibrium both AO and AO–DNA complex have emission at the same wavelength.

  Additionally, the binding equilibrium constant for AO intercalation to DNA (ignore AO self-aggregation and dimerization) can be determined from the equation:

  $$\frac{C_\ce{AO}}{ΔF} = \frac{1}{Δϕ}+\frac{1}{ΔϕK}\frac{1}{\ce{[DNA]}} \tag{2}$$

  $F - ϕ_\ce{AO} \cdot C_\ce{AO} = ΔF$, $F = ϕ_i \cdot C_i$, and $ϕ_\ce{AO} - ϕ_\ce{AO-DNA} = Δϕ$ is given.

  $F$ is the overall intensity, $ϕ$ is the fluorescence constant and $i$ denotes a particular component.

  1) Show that $ΔF = [\ce{AO-DNA}]Δϕ$

  2) Derive equation (2) starting from equation (1)

In terms of the first part, I tried working out formulas for both initial and final fluorescence intensities $F$ using $F = ϕ_i \cdot C_i$ but didn't really get anywhere after that. In terms of the second part, I tried substituting $ΔF = [\ce{AO-DNA}]Δϕ$ into the equilibrium equation but didn't really get further from that either. I would really like to get some tips on where to get started.

Concentrating on fluorescence intensity for the moment the equation becomes $I_O = I_D $ where $I_O$ is that for the free dye and $I_D$ for that bound to the DNA. If $\alpha$ dissociates the equilibrium is then

$$\begin{align}  &I_O \quad \rightleftharpoons  &I_D \\ &1-\alpha &\alpha \end{align}$$

and the equilibrium constant $K=I_D/I_O=\alpha/(1-\alpha)$.

The total emission is $\displaystyle F=I_O(1-\alpha)+I_D\alpha=I_O\frac{1}{1+K}+I_D\frac{K}{1+K}$

Rearranging this gives $\displaystyle \frac{1}{I_O+KI_D}+\frac{1}{I_D}=\frac{1}{F}$

Replacing intensity with product of fluorescence yield and concentration and using the rather odd definitions given  should give the equation you seek. Note that $F-\phi_{AO}C_{AO}$ is the intensity of the bound dye.

The first question states that the intensity due to the bound dye $F-\phi_{AO}C_{AO}$ is equal to the amount bound times the difference in yields, this because the dye fluoresces at the same wavelength whether bound or not.


Process to separate Lauric Acid from Coconut Oil I am trying to find a (relatively) simple process to separate lauric acid from other MCTs contained within coconut oil.   I am wanting to obtain a separation of +-5% purity.

One process I am trying is heating the coconut oil until it is completely liquid (~50'C), and then slowly cooling it to ~22'C using a fridge I modified with a digital temperature controller. This process sometimes works in making "solid clouds" of lauric acid in the oil, but not sufficient to separate easily from the remainder of the oil.

Is there something I can try in order to separate lauric acid from the coconut oil, using relatively simple processes?

Any comments welcome.

With respect to economy, waste management and regulations concerning food processing, your current approach, a solvent-free crystallization, is probably the most reasonable approach.

There's a Unilever patent which claims that the crystallization, both stirred and quiescent, is significantly improved by addition of sucrose laureates.

The patent holders attribute this to the observation that the additive leads to the formation of

  large and non-porous spherulites, which highly increases separation efficiency


Is there a reason for the mathematical form of the equilibrium constant? Why are the two molarities multiplied and not added, and why is each raised to the power of the coefficient rather than multiplied by it? What is the reasoning behind this form? Was it simply determined that the probability of reaction was proportionate to both and thus they are multiplied together? I'd like to know whether its form is purely empirical or has a theoretical backing.

Equilibrium is also the state in which the forward and reverse reactions are proceeding at equal rate. Consider a hypothetical reaction

$$\ce{aA + bB} \overset{k_1}{\underset{k_{-1}}{\ce{<=>}}}\ce{ cC + dD}$$

If we assume that the forward and reverse processes are elementary reactions (a big assumption), then:

$$rate_1=k_1 [\ce{A}]^a [\ce{B}]^b$$
$$rate_{-1}=k_{-1} [\ce{C}]^c [\ce{D}]^d$$

At equilibrium:
$$rate_1=rate_{-1}$$
$$k_1 [\ce{A}]^a [\ce{B}]^b=k_{-1} [\ce{C}]^c [\ce{D}]^d$$
$$\dfrac{k_1}{k_{-1}}=\dfrac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}$$

The ratio of the rate constants is the equilibrium constant:

$$K=\dfrac{k_1}{k_{-1}}=\dfrac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}$$


In a two electrode system, is the applied current density referring to just the working electrode or to the auxiliary electrode as well? I'm doing an experiment electrosynthesizing a polymer using a two electrode cell and it says to do it under 0.5 mA/cm^2. However, I am unsure if that means 0.5 mA per square centimeter of surface area on the working electrode or for both electrodes. If anyone could clarify this for me, it would be most appreciated.

It is normal in many laboratory electrochemistry experiments for the working electrode to have a much smaller surface area than the counter (auxiliary) electrode. This when the current density at the working electrode is 0.5 mA per square cm, the current density on the counter electrode is much smaller.

It is more common that someone will express the potential and the current density on the working electrode.

If for example we have a reference electrode which is at 0 volts (standard hydrogen electrode) and the potential of the working electrode is + 400 mV. We could have a flow of 1 mA out of the working electrode. As no current flows through the reference electrode all of this current flows through the counter electrode.

The electrical potential between the counter and the reference electrodes is known as the compliance voltage. This is the voltage required to create a flow of current which is just right to make the right amount of current flow through the working electrode to keep its potential at + 400 mV.

This is done using a potentistat, a range of different designs of such devices exist. I think that the most simple design and one of the best is the general design used by PAR (Princetown Applied Research) in the classic 173. It has been joked that at one point almost all electrochemists have used a 173, I rather like the 173 it is a nice machine. But we are not here to discuss classic electrochem machines.

Back to the design of a machine like a 173, If we set the drive at -400 mV then the top op amp will impose a current on the cell via the secondary electrode which will bring the potential of the reference electrode to -400 mV.

The bottom op amp which is a I to V converter will hold the working electrode at zero volts (ground or GND) as far as the electronics is concerned. Thus for the chemist if they hold the view that the refernece electrode is the zero volts point then the working electrode will be at +400 mV.

If 1 mA flows into the working electrode and the value of resistor R is 1000 ohms then the output of the I to V converter (bottom op amp) will have to be -1 volt for the sum of the currents coming to the inverting (-) input of the op amp to be zero. We assume that the input resistance of an op amp is infinity. Thus if we attach a x-y plotter to the drive terminals (x to E or potential axis) and the (y axis to the current indication terminal) we will get the plotter to plot -400 mV, -1 volt.

A word of warning the Europeans and the Americans often follow different conventions on what sign to give a reducing current (cathodic) and what sign to give a reducing potential.

If I recall rightly the European convention is for bottom left to be reducing current and reducing potential.


How does "breathable" siloxane water sealant work? (This is an interdisciplinary question, between molecular chemistry, materials science & construction. I think Chemistry SE is the right fit, but please move it if needed.)

Silane/siloxane solutions are applied to porous building materials (e.g. brick or concrete) to repel water. Unlike a solid coating, these sealants are "breathable", allowing water vapor to exit instead of being trapped, which would cause slow structural damage. How does that work, molecularly?

I presume these compounds affect surface tension in some way, reducing the ability of liquid water to permeate, without completely sealing the pores, but I don't understand it.

Optional: also, how does this sealant affects the permeability of other liquids (polar vs non?) and other gases?

According to How silicone water repellents work,
"Because silicones have a very open molecular structure, water vapor trapped in the substrate can easily slip through. Water molecules, however, are too large to enter."

If I remember correctly, siloxanes' Si-O bonds are not intrinsically hydrophobic, but the long molecular chains with outward-facing hydrophobic methyl groups, and thermal molecular motion, prevent liquid water from seeping through, which presents less of a barrier to water vapor and other low-molecular-weight gases.


Distinguish between symmetric and asymmetric molecules with IR I get increasingly nervous as I get my organic lab course reports. For, I often get asked whether the peak in the IR comes from an asymmetric or symmetric stretch.

I've looked at about ten books and turned google upside down. But I still can't answer the question in every case, though our molecules are quite simple.

Can you help me out with a table or something equivalent?

The most urgent example is a $\ce{C-H}$ streching at $3006~ \mathrm{cm}^{-1}$ from p-tbutyl-(2-ethanol)-benzene (don't mind the nomenclature, please).

Whether the stretch of a bond in a molecule is asymmetric or symmetric depends on the relative movement of the atoms connected by the bond. Lets take a $\ce{CH2}$ group as an example. If the two hydrogen atoms connected to the carbon move in the same direction during stretching, it s a symmetric stretch, if they are going in opposite directions, the stretch is asymmetric (see also here):

Left: symmetric stretch, right: asymmetric stretch

A bond vibration like stretching will only be IR-active (i.e. give a band in the IR spectrum) if it is accompanied by a change of dipole moment. In the case of the $\ce{C-H}$ bond, carbon is slightly negatively and hydrogen slightly positively polarized. A symmetric stretch increases and decreases the distance between these partial charges periodically, and therefore also changes the dipole moment, because the latter is dependent from the distance between charges (source). During an asymmetric stretch, the elongation of one $\ce{C-H}$ bond is compensated by the shortening of the other, so the overall change of dipole moment should be weaker.

In the case of 2-(4-tert-butylphenyl)ethanol (the correct name for your compond), the band at 3006 $\mathrm{cm}^{-1}$ can either be assigned to the $\ce{CH2}$ groups of the substituted ethanol or to $\ce{C-H}$ bonds in the benzene ring. In the first case, you can distinguish between symmetrical and asymmetrical stretch (symmetrical seems more likely to me), while in the latter case, a single hydrogen is attached to a carbon, and a distinction of the stretching, like in the case of two substituents at the same carbon, cannot be made.


Do symmetric hydrogen bonds in neutral molecules exist? As far as I know, there have only few truly symmetric hydrogen bonds been observed. Unquestionable is the existence of it in the bifluoride ion, $\ce{[F-H-F]-}$, see also here. There are a couple of more, like in a water hydroxyl complex $\ce{[HO\bond{~}H2O]- @Cu($110$)}$,[1] or in the hydrogen disuperoxide anion $\ce{[O2-H-O2]-}$.[2]

Long time it was believed that malonaldehyde and acetylacetone[3] form keto-enol tautomers with symmetric hydrogen bonds. The reason for this is the lack of resolution of the experiments.

I recently came across the article about substituted malonaldehydes.[4] According to this, if you chose R1 to be an amine and R2 a nitro group, you could end up with a single well on the potential energy surface.

However, there is one response, that left me a bit puzzled. Henry Rzepa wrote:

  Searches of the Cambridge crystal structure database will reveal quite a number of instances of symmetrical hydrogen bonds. These are mostly homonuclear, ie N…H…N or O…H…O. Years ago, we came across a more unusual example, of an (almost) symmetrical N….H….O heteronuclear example (http://dx.doi.org/10.1039/C39890001722 ).

So apparently there exist many. Unfortunately, he doesn't extend us the courtesy of even a single example. If you look on Google Scholar, you do indeed find a few examples of symmetric hydrogen bonds. However, most of them are either "crystallograpically" symmetric, for charged species, or both.

Also Evangelista et al. conclude their paper about malonaldehydes with the following statement, which leads me to believe, that there is no such thing as a symmetrical hydrogen bond in a neutral molecule:

  A remaining goal is the identification of a substituted malonaldehyde with no barrier at all, i.e., a C2v equilibrium geometry.

Since then a few more years have passed, but my search did not turn up any more results.
Do symmetric hydrogen bonds in neutral molecules exist?

T. Kumagai, M. Kaizu, H. Okuyama, S. Hatta, T. Aruga, I. Hamada, and Y. Morikawa, Phys. Rev. B 2010, 81, 045402
S.J Knak Jensen , I.G Csizmadia, Chem. Phys. Lett. 2000, 319 (3-4), 220-222.
Daryl L. Howard, Henrik G. Kjaergaard, Jing Huang, and Markus Meuwly, J. Phys. Chem. A 2015, 119 (29), 7980–7990.
Steven Bachrach, 03 Feb 2009, Malonaldehydes: searching for short hydrogen bonds Also: Jacqueline C. Hargis, Francesco A. Evangelista, Justin B. Ingels and Henry F. Schaefer III, J. Am. Chem. Soc. 2008, 130 (51), 17471–17478.

It appears that perhaps water may have symmetric hydrogen bonds between oxygen atoms, but it would take at least 60 GPa of pressure to make water to bond like that.

Article named very descriptively "Compression of Ice to 210 gigapascals: Infrared Evidence for a Symmetric Hydrogen-Bonded Phase" proves it:

  Protonated and deuterated ices ($\ce{H2O}$ and $\ce{D2O}$) compressed to a maximum pressure of 210 gigapascals at 85 to 300 kelvin exhibit a phase transition at 60 gigapascals in $\ce{H2O}$ ice (70 gigapascals in $\ce{D2O}$ ice) on the basis of their infrared reflectance spectra determined with synchrotron radiation. The transition is characterized by soft-mode behavior of the $\nu_3$ $\ce{O-H}$ or $\ce{O-D}$ stretch below the transition, followed by a hardening (positive pressure shift) above it. This behavior is interpreted as the transformation of ice phase VII to a structure with symmetric hydrogen bonds.

On the other hand, article "The Hydrogen Bond in the Solid State" mentions:

  First example of an $\ce{O-H-N}$ hydrogen bond with a centered position of
  the proton: pentachlorophenol/4-methylpyridine at 100 K characterized
  by neutron diffraction studies (...)

Unfortunately:

  Even very small chemical changes in the system lead to loss of the
  symmetry (...)

so symmetric hydrogen bonds can probably exist also in very low temperatures.

References

Goncharov, A F; Struzhkin, V V; Somayazulu, M S; Hemley, R J; Mao, H K. Science 273.5272

Steiner, Thomas Angew. Chem. Int. Ed. 41 1 71


Why are solid phase organics often conjugated to HCl? Many uncharged organic compounds are sold in solid phase with HCl conjugate - e.g. the antihistamine diphenhydramine.HCl or the derivitizing agent pentafluorobenzyl hydroxylamine.HCl, among many others. Assuming the compound will be dissolved at relatively low concentrations in a buffered medium (for instance when a drug enters the bloodstream) the HCl shouldn't have much chemical effect. Why is it there - does it aid crystallization, or perhaps it's leftover from synthesis and is difficult to separate?

Does it aid crystallization?

Yes.

All of the drugs that you see as "hydrochlorides" are alkaloids - a class of basic natural products containing one or more amine functional groups. The dimethylamino group in the upper right of the structure of diphenhydramine (below) is a prime example.

Many of these compounds are not crystalline (at best amphorous, at worst waxy or liquid). These compounds are expensive to purify, since the inexpensive method of recrystallization is not an option. Chromatography is an expensive and wasteful process on the industrial scale.

Reacting alkaloids with HCl (or other acids - for example dextromethorphan HBr) usually produces a crystalline compound. These crystalline compounds are easier to purify.

There is a second more important benefit associated with the hydrochloride salts - they have higher bioavailability than the freebases. The freebase is nonpolar and poorly soluble in water. The hydrochloride salt has an ionic group, which increases water solubility.


Why isn't my electrochemical cell producing its potential voltage For a school project, I am testing how concentration affects the voltage output of three metals. So I am building voltaic cells with 0.1 M copper nitrate and 0.1 M, 0.01 M, and 0.001 M of zinc, aluminum and nickel nitrates as the electrolyte solutions. For the cells with zinc and copper, I found that the voltage produced was generally fairly close to their theoretical value of 1.10 V with the lowest voltage produced being 0.95. However, for aluminum and nickel, I have found that they are very far off from their theoretical values of 2.00 and 0.57 respectively as I have only reached a maximum voltage of 0.50 for aluminum and 0.20 for nickel.
I am using a porous cup and have made sure all of the electrodes are thoroughly sanded before each trial. I am fairly confident the issue is not related to the concentrations, or to the porous cup. At this point, I just have no idea why it isn't working. I read a similar question on here suggesting that the issue might be that aluminum does not easily dissolve in NO3 and I was wondering if the same was true for nickel.

Aluminum redox potential has nothing to do with the voltage of the cell you have build with it. Let me explain why, because it is not obvious.
Electrolytic cells made with aluminum anodes always yield rather low voltages. The reason is that usual pieces of aluminum are always covered by a thin, continuous and colorless layer of aluminum oxide $\ce{Al2O3}$. This oxide is an electrical insulator which makes a barrier between the metal and the solution. Aluminum ions cannot cross this barrier. There is no way of preventing this aluminum oxide formation. If you rub the aluminum surface to remove it, the pure metal so exposed will immediately react with air to remake this layer.
The only way to suppress this layer is to dip aluminum in a mercury chloride solution. Of course today this operation is forbidden by law. But this experiment was done before this legal decision. So let's see what was happening. Some mercury ions may cross the barrier and react with aluminum to produce an amalgam $\ce{Al(Hg)}$ plus some $\ce{Al^{3+}}$ ions getting in solution$$\ce{5 Al + 3 HgCl2 -> 3 Al(Hg) + 2 Al^{3+} + 6 Cl^-}$$ The formula Al(Hg) is not sure, as it is an alloy. Maybe the ratio Al:Hg is different from $1:1$. It does not matter for the present study. The only important thing now is that the aluminum oxide layer does not perfectly cover the amalgamated metal any more, and is partly destroyed. Therefore aluminum atoms are suddenly in contact with water, and they quickly react with it according to :  $$\ce{2 Al(Hg) + 6 H2O -> 2 Al(OH)3 + 3 H2 + 2Hg}$$ or$$\ce{2Al + 6 H2O ->  2 Al(OH)3 + 3 H2}$$This shows that unprotected aluminum metal quickly reacts with water, producing $\ce{H2}$ bubbles, and these $\ce{H2}$ bubbles are responsible of the real anodic half cell according to :
$$\ce{H2  -> 2 H^+ + 2 e-}$$ As a consequence, this "aluminum" electrode would have, at maybe $p$H$ 5$, a redox potential $\ce{E(H2/H+)}$ given by Nernst law : $$\ce{E(H_2/H^+) = E°(H_2/H^+) + 0.0296 V·log[H+]= 0 V + 0.0296 V·log 10^{-5} = - 0.148 V}$$ In your setup, this anode is coupled with an ordinary copper half-cell as cathode, whose standard redox potential is + $0.34$ V. The voltage of this cell is $$\ce{E(Cu/Cu^{2+}) - E(H2/H+) = 0.34 V -(-0.148 V) = 0.488 V}$$ This value is nearly the voltage that you have obtained in your cell ($0.50$ V).


Why are enantiomeric excess % and specific rotation considered? I was wondering, when working in a lab in an area related to organic compounds, when and why would a chemist consider the enantiomeric excess % and how would a chemist used that information?? Same Thing goes with specific rotation, why, how, and when would a chemist use this? I was just curious because I saw this in my organic chemistry book and they relate the problems to lab work and I want to get a good grasp of tis context and it made me wondered.

Enantiomeric excess is very important concept in organic chemistry because it clearly states the purity of one isomer over the other. For example, have a look at the structure of Thalidomide which is racemic (50:50 —R:S isomers) in nature. Between the two form, one is having very important medical property where as another form is sedative. In these cases, synthesizing enantiomerically pure (>99%) form is mandatory and in January 1996, the FDA announced it would consider further incentives for developing single isomer drugs, owing to their better pharmacokinetics prosperity, safety, and tolerability.
http://en.wikipedia.org/wiki/Thalidomide

These are the reason which makes chemists to target enantimerically pure form using latest available technologies like, Chiral GC or HPLC.

Specific optical rotation is very old concept to judge the enatiopurity of the compounds but its not accurate because it depends on various factors like concentration of the sample, solvent which we use, temperature, etc. Even small impurity in the sample gives drastic change in the rotation value. Thats why chemists adapt latest technique like HPLC of GC, nevertheless, chemists still carry out specific optical rotation to make sure that the enantiomerically excess compound has some rotation.


What does nutella add to the Coke+Mentos reaction? Inspired by this question on SFF.SE.

The Diet Coke + Mentos experiment has been a popular one for a while now, and I've read a little bit about the science behind it.

I recently discovered this video, where an Italian man adds Nutella to the mix1, to his great excitement.

However, I'm not clear on why he did this; I couldn't see a significant difference in the reaction (although my view was somewhat obscured by the condom). What does the Nutella add to this reaction?

1 Do not watch this video in a public place without headphones on (or the sound turned way down); he gets very, very excited when his abomination experiment (literally) blows up in his face

Nutella was used just to protect cola from incidental reaction with Mentos. And it makes more effective insertion - all mentos tabs are inside at the same time.

The reaction is completely the same with simple Mentos. (this video also is quite noisy)

This is not chemical reaction, is carbon dioxide evaporation.


Will acetone damage/attack anti UV coating on glasses? My glasses collect a lot of grime in hard to reach spots. I want to dunk them in acetone to loosen the grime and clean them.

Does this put me at risk of losing the anti-UV coating? Sorry I was unable to find out what the coating chemical was.

Without knowing the material your glasses are made of, I would discourage cleaning glasses with acetone.

Often, glasses use no longer use glass as lens material -- polymers like polycarbonate represent an alternative, especially if a high degree of correction is intended, or simply because of weight concerns.  Polycarbonates however withstand ketones (and acetone is an example of ketones) poorly (listing).  As a result, contact with this solvent may blind your glasses permanently.  Happens to safety goggles worn in the chem lab, too -- yet better the goggles, than damage to the eyes.

In addition, what is the material of the frame?  Hence, I recommend a mild detergent; and if this is not working, consult your optician -- they have an ultrasound bath.


Identification of solutions by solubility when mixed with each other $\ce{Pb^{2+} (aq), Cl- (aq), Ag+ (aq), (SO4)^{2-} (aq)}$.  You number the four bottles, 1,2,3,4and while numbering them, you notice that bottles 1 and 3 have the word "cation" on their lids while bottles 2 and 4 have "anion" on their lids. You decide to test them by mixing them together in pairs, but you have only three empty beakers. When the solutions are mixed in pairs.
Three combination give white precipitates;
Bottle 1 + bottle 2. bottle 1 + bottle 4, bottle 2 + bottle 3. ive done each combination and theyre all insoluble? but my chart says there has to be 1 combination thats soluble and 3 that arent ?

Since It is a Homework Question I would Like to add a hint instead of giving full solution.

You might not be able to Watch the difference Carefully.

For Your information

$\ce{PbCl2}$ = insoluble

$\ce{PbSO4}$ = insoluble

$\ce{AgCl}$ = insoluble

$\ce{Ag2SO4}$ = slightly soluble

Try doing it again and watch the difference carefully afterall now you know which one is soluble.

Hope it helps.


How to convert benzamide to benzylamine? Question
  Benzamide can be converted to benzylamine using

  A) $\ce{Br2,\ KOH}$
   B) $\ce{PCl5}$
   C) $\ce{LiAlH4}$
   D) $\ce{NaBH4}$

My answer:

Reducing agent is used so C and D options both must be correct but since its single choice answer, I do not know which is more appropriate. Any help?
Also, what does option A and B reagents do to amides?

The more reactive $\ce{LiAlH4}$ is required for the reduction of the amide to the corresponding amine (source). The carbonyl carbon of an amide is less electrophilic compared to a ketone, because electron density is donated by the amide nitrogen via resonance.

Bromine and hydroxide react with an carboxylic acid amide to yield an amine with one fewer carbon atom (Hofmann degradation). $\ce{Br2}$ and $\ce{KOH}$ react to form $\ce{KOBr}$ in situ, which transforms the amide into an intermediate isocyanate ($\ce{R-N=C=O}$). Hydrolysis of the isocyanate yields a carbamic acid $\ce{R-NH-COOH}$, which decomposes to the amine $\ce{R-NH2}$ and $\ce{CO2}$. As described in this paper, the reaction of $\ce{PCl5}$ with carboxylic acid amides produces amidochlorides $\ce{RCCl2NH2}$ as intermediates, which break down to the corresponding nitrile $\ce{RC#N}$ and $\ce{HCl}$.


Why do halogens have odd numbers as oxidation number? Halogens like $\ce{Cl}$ always exhibit $+7,+5,+3,+1,-1$ as their oxidation numbers. I found the following answer after checking several sources:

  Halogen atoms have $7$ electrons in their valence shell, so $1$ of those electrons is spin unpaired, which is an unstable situation. If an even number oxidation occurs, said even number of electrons are taken away, leaving still an odd number of electrons with at least $1$ unpaired. Taking away an odd number of electrons leaves an even number of electrons with the opportunity of them to be all paired.

These oxidation numbers are observed in compounds like $\ce{HClO4}$ and $\ce{HClO3}$. We know that, before forming these compounds, the valence electrons of $\ce{Cl}$ are excited into the d orbitals. Thus now, they are not in ground state, but in the excited state instead.

The quoted explanation is giving the reason for my question on the basis of the ground state of halogens. (I especially consider $\ce{Cl}$). But we know that, before formation of compounds, the halogen atoms are in their excited state instead.

So is the quoted explanation correctly justifying why $\ce{Cl}$ exhibits odd oxidation states only?

Because of the way we construct oxidation numbers.  Put an atom into a molecule with all electrons paired and all the bonds heteronuclear, and by assigning all the valence electrons to the more electronegative atom of each bond we render even number on each atom.  Combine that with an odd number required for a neutral atom and you count an odd difference, thus an odd oxidation number.

Note that I have assumed all electrons to be paired and all bonds as heteronuclear.  But, it is not always the case.  There could be homonuclear bonds as in the halogen element itself ($\ce{X_2}$), with an oxidation number = $0$.  Or there could be unpaired electrons on the halogen so we count an odd rather than even number of valence electrons per halogen atom.  $\ce{ClO}$ (chlorine oxidation number $+2$) and $\ce{ClO_2}$ ($+4$) are such examples.


Chemical hardener to cast breastmilk into gemstones I'm looking for a chemical hardener that I can add to breastmilk to turn it into a stone. I have seen others use one but they will not share what they are using. In their video they add about 1ml of secret solvent to 15ml of breastmilk and heat it on low heat for a couple minutes. It hardens completely within a few days. Any help would be greatly appreciated.

Referring to real milk (bovine, human breast milk; but not coconut milk) contains casein which may be polymerized.  You may end up with edible cheese, or green adhesives, or gemstone-like galalith, for example.

Frankly, I have no hands-on experience to experiment with breast milk.  And I recommend to refrain from experimenting with formaline at home.  However, the procedure "Sculpted Science: Turn Milk into Plastic!" in the open access section Bring Science to Home by the Scientific American, which triggers the process just by vinegar and gentle heat, sounds much safer and may be worth to follow.


What happens in an acid and metal reaction where the acid is twice as concentrated and the metal twice as much in quantity? Experiment A

3cm metal

0.5M acid

5 degrees temperature change

Experiment B

6cm metal

1M acid

How much would the temperature change be, and how much faster or slower would the reaction take place in?

I thought the answer would be either "the same" or "4x as much". I don't exactly understand the reaction yet; would the doubling of each material make the reaction 4x stronger (2x2), or would they cancel each other out?

Of course, it might've been something else I haven't thought of, too.

Without providing more detailed specifications of the metal such as: state of the metal - powder or solid, shape of the solid form etc., the rate of the reaction will be the same, but the second reaction will be a bit faster because of the bigger temperature change (as it affects reaction rate coefficient).

I assume that you meant 3 and 6 $cm^3$ and not $cm$.
The same metal reacts in both reactions. So we can assume that the density of both metals is also the same ($\rho_1=\rho_2=\rho$). We can further assume that metals were dissolved completely and that the temperature change is actually temperature increase.

$\rho=\frac{m_1}{V_1}$, $\rho=\frac{m_2}{V_2}$

$\frac{m_1}{V_1}=\frac{m_2}{V_2}$

$\frac{m_1}{3}=\frac{m_2}{6}$

$\frac{m_1}{1}=\frac{m_2}{2}$

The bigger piece of metal weights 2 times more than the smaller metal. We can convert weights to moles (molar mass of both metals is the same):

$n=\frac{m}{M}$

$\frac{n_1}{1}=\frac{n_2}{2}$
so we can say $2*n_1=n_2$

The 1st reaction:

$n_1$ reacts with 0,5M of acid.

The 2nd reaction:

2*$n_1$ react with 1M solution and it is the same as $n_1$ reacts with 0,5M

The volume of the solution is the same and the specific heat capacity will be nearly the same. The reaction enthalpy corresponds to the amount of reacted material. Because two times more material reacts in the second reaction,  the temperature change will be nearly two times bigger (nearly - because of the change in specific heat capacity).


Predicting rate equations from data I am having trouble wit the following question on my problem sheet:

"The reaction $\ce{Br- + ClO^- \to BrO^- + Cl^-}$ occurs with a rate constant of $\text{0.35 dm}^3 \text{mol}^{-1}
\text{s}^{-1}$

at
298 K. For the following initial conditions calculate the time taken for the Br^-
concentration to fall to half its original value:

i) $[\ce{Br-}]=[\ce{ClO-}]=0.0017$

ii) $[\ce{Br-}]=0.0017$ and $[\ce{ClO-}]=0.084$"

I am sure I could do this question if I knew the rate equation but it isn't given. From the units of the rate constant the reaction must be second order but how am I supposed to know whether it's second order with respect to one of the reactants (and thus zeroth order w.r.t the other) or first order with respect to both? Also, what's the significance of the concentrations being equal (there's another question that stresses that the concentrations of both the reactants are equal).

You have to assume it is first order in each reactant.

Zero-order kinetics is always an artifact of the conditions under which the reaction is carried out.

The reaction can't really be zero-order in a reactant, otherwise it would imply the reaction continues after there is no reactant remaining.


Existence of Deuterium peroxide Does deuterium peroxide exist? Wikipedia has no reference. Also, the sites for delivery of chemicals like "Pubchem" or "Chemspider" have no reference.

In my book, a preparation of $\ce{D2O2}$ is given,

$$\ce{K2S2O8 + 2D2O -> 2KDSO4 +D2O2}$$

If $\ce{D2O2}$ exist, what are its physical properties (structure and appearance etc.)?

It does exist. At least Google shows one supplier situated in New Jersey.

  ID 1130   Deuterium Peroxide (30% in D2O) D2O2    98  5   5.0g / 10.0g

Though for some reason I could not find any reputable supplier like Cambridge Isotopes or Sigma-Aldrich (strange!). As far as physical properties are concerned, as pointed out by @snurden there exists a significant amount of literature. Here is one article from the Canadian Journal of Chemistry that includes study of many properties (behind a pay wall though).

Abstract:

And to answer final part of your question:


How is the hydrolysis of amides done in a lab? I have seen this video about the hydrolysis of amides in order to convert them to a carboxylic acid. I understood how it works but I want to know how would someone physically do it in a lab.

I need to know the temperature needed, the time required for the reaction to happen, the reactants ratio and the process itself.

The answer, as with so much chemistry, is it depends. What substrate are you hydrolysing, is it a primary, secondary or tertiary amide? What else is in the molecule that you want to preserve?

Both acid and base hydrolysis require concentrated conditions and elevated temperature. There is a lab procedure here that uses sodium hydroxide in water at reflux to hydrolyse benzamide to benzoic acid. The standard cleavage of an acetamide protecting group to give amine plus acetic acid is overnight reflux in 2N HCl or KOH in MeOH at reflux ref here, for a benzamide protecting group it is 48h in 6N HCl or HBr in Acetic Acid  at rt.

In my own experience a particularly resistant acetamide required 3 equivalents of KOH in refluxing ethylene glycol to hydrolyse.


Organic Structure Challenge Recently we had this challenge in our Organic Chemistry Department.
I was not involved in it but I thought it might be interesting for the SE Chemistry site, so here it is.

Propose a structure for C20H18 with the longest IUPAC name.
Following rules apply:

Names must be generated with MarvinSketch: Structure>Generate Name…>Preferred IUPAC Name (https://www.chemaxon.com/products/marvin).
No isotopic markings can be used.
The structures must be possible (no cycles with too much unsaturations nor systems with impossible tension).
All the characters will be counted: numbers, letters, points, colons, dashes, everything.
Longest name wins.

Here is the example that was given:

and the name provided by MarvinSketch:

(1'R,4'Z,6'Z,8'Z,10'Z,12'S,15'R,16'R,17'R)-spiro[cyclopropane-1,18'-tetracyclo[13.2.1.04,17.012,16]octadecane]-2,2',4',6',8',10',13'-heptaene

141 characters

Good luck and prosperous year 2018!

EDIT

Well, what is a challenge without a prize?
So, I have set a bounty worth +50 reputation for the longest name.
Sorry, I can not spend much more on it.
Remember that you can post more than one structure.
If you have proposed one and after some days you find a structure with a longer name, you can post it too.


How does the reaction of HCl and KMnO4 proceed? I don't get how does this reaction of hydrochloric acid and potassium permanganate proceeds:

$$\ce{HCl + KMnO4 -> KCl + MnCl2 + Cl2 + H2O}$$

What I know:

$\ce{HCl}$ dissociates into $\ce{H+}$ and $\ce{Cl-}$.
$\ce{KMnO4}$ dissociates into $\ce{K+}$ and $\ce{MnO4^{2-}}$

What I don't get: How $\ce{MnCl2}$ and $\ce{Cl2}$ can appear there?

Could someone, please, guide me through this reaction, step by step?

As others have pointed out, you don't have a "normal" reaction, you have a oxidation-reduction reaction (if you want to sound especially geeky, calls them RedOx (pronounced "ree-dox") reactions.

The electrons (which we don't normally balance in a equation) feature prominently because there are elements changing oxidation state.

First, split your reaction into its half reactions. Bring the "spectator ions" along for the ride and make sure they balance:

Oxidation: 2 $\ce{Cl-}$ gets oxidized to $\ce{Cl2}$ and gives up 2 electrons. $\ce{H+}$ is your spectator ion. $$\ce{2 H+ + 2 Cl- -> Cl2 + 2e- + 2H+}$$
Reduction: $\ce{Mn^7+}$ needs 5 electrons to reduce it to $\ce{Mn^2+}$. $\ce{K+}$ and $\ce{O^2-}$ are your spectator ions. $$\ce{K+ + MnO4- + 5e- -> Mn^2+ + K+ + 4O^2-}$$
Now, just add water. Oxidation reactions get $\ce{OH-}$; reduction reactions get $\ce{H+}$:
$$\begin{align}
\ce{2OH- + 2 H+ + 2 Cl- &-> Cl2 + 2e- + 2H2O}\\
\ce{8H+ + K+ + MnO4- + 5e- &-> Mn^2+ + 4 H2O + K+}
\end{align}$$
Now we balance the electrons (remember least-common-multiple from math class? You thought you'd never use that, didn't you?)

LCM for 2 and 5 is 10, so multiply the Cl equation by 5 and the Mn reaction by 2.

$$\begin{align}
\ce{10OH- + 10 H+ + 10 Cl- &-> 5Cl2 + 10e- + 10H2O}\\
\ce{16H+ + 2K+ + 2MnO4- + 10e- &-> 2Mn^2+ + 8 H2O + 2K+}
\end{align}$$

Add everybody back together, make more water, and cancel out the electrons:
$$\ce{10H2O + 10 HCl + 6H+ + 2KMnO4 -> 5Cl2 + 10H2O + 2Mn^2+ + 8 H2O + 2K+}$$
Take out the excess water:
$$\ce{10 HCl + 6H+ + 2KMnO4 -> 5Cl2 + 2Mn^2+ + 8 H2O + 2K+}$$
Looks like we're short something to balance out the charges. We can add more chloride spectator ions to both sides of the equation. They're not part of the redox since they are $\ce{Cl-}$ on both sides of the equation:
$$\ce{10 HCl + 6HCl + 2KMnO4 -> 5Cl2 + 2MnCl2 + 8 H2O + 2KCl}$$
Collect terms one last time and we're done:
$$\ce{16HCl + 2KMnO4 -> 5Cl2 + 2MnCl2 + 8 H2O + 2KCl}$$


Why is Ma*Va=Mb*Vb still valid for weak acids/bases? I asked my chemistry teacher this today and didn't get an answer. It would seem to me that the equation for determining the morality of the unknown in a titration, MaVa = MbVb, would only work if the acid and the base were both strong.

I always interpreted the equation as a way to say (moles H+)=(moles OH-), but if the acid was weak, the moles of H+ would not simply equal MaVa, because the acid wouldn't fully dissociate, and the same for a base and OH-.

Yet the formula still works for both strong and weak acids and bases. Why?

We normally titrate a weak acid by a strong base, or  weak base by a strong acid, to ensure that the reaction of titration is total (quantitative). We don't titrate weak acid by a weak base (and vice-versa).
I will calculate the equilibrium constant of the titration of  weak acid $\ce{HA}$ by a strong base $\ce{OH-}$:
 $$\ce{HA + OH- <=>A- + H2O}$$ The equilibrium constant is $$K=10^{(\mathrm{p}K_w -\mathrm{p}K_a)}$$ For common weak acids (acetic acid $\mathrm{p}K_a=4.7$ ): $$K>> 10^3$$
So, the equation of titration (neutralization) must be written:$$\ce{HA + OH- ->A- + H2O}$$
At the equivalence point: $$n(\ce{HA})=n(\ce{OH-})$$


Are there types of glue that are designed to be dissolved easily? I've heard that some types of glue are easily dissolved.  For example, wood glue dissolves in vinegar and super glue dissolves in acetone.  This doesn't mean that wood glue or super glue were necessarily intended to be dissolved.

Are there types of glue/adhesive that are pretty strong and have a cheap solvent that is intended to remove the glue/adhesive later?

"Wood glues" can have several different compositions, chances are if the ones you are working with respond to vinegar they are either protein based or PVA.  In the protein case the solvation is likely accomplished by the vinegar protonating the polymer chains, and therefore changing their attraction to each other. In the case of PVA based glues, vinegar is likely contributing to acid catalyzed hydrolysis.  The softening of cyanoacrylate (super glue) with acetone is a more traditional dissolving mechanic, as no chemical changes are being made.

That being said, all three mechanisms mentioned above can be pretty well represented in the glue world with different polymers.  Based on what you consider cheap solvents, you have a few options. Cyanoacrylate (superglues), some UV-cured adhesives (which are frequently very well crosslinked and strong), PVC, and many others are all softenable and removable by acetone

Additionally, anything that can be taken out by acetone will be solvated in MEK (methyl ethyl ketone).  MEK is less water miscible and more aggressive though, which may be relevant to your needs.

Silicone adhesives are great for flexible applications, but frequently need stronger organic solvents (like toluene) to swell it and then it can be mechanically removed.  There are also commercially available silicone removers that appear to use acidic or basic bond cleavage to break up the adhesive.

Rubber cement is another flexible and relatively strong bonding adhesive mainly composed of solvated polymers, drying and tackifying by solvent evaporation.  Reintroducing the solvents (many commercially available paint thinners should work) will soften the adhesive.

As an aside note, depending on the materials you are bonding together, heat and cold can also be used to weaken joints and destabilize the adhesive bond separately or in tandem with chemical treatments.


Low-cost nontoxic corrosion inhibitor for steel exposed to hydrochloric acid It seems that many commercial descaling solutions designed to remove scale (calcium carbonates, etc that form on water heaters) are a simple combination of an acid and an acid corrosion inhibitor for metals.

The acid renders the scale soluble in water, while (some) inhibitors coat metal surfaces preventing the acid from corroding the metal.

The problem is that these descaling solutions, despite their simplicity, are often \$30-\$100 per gallon (at just 5% HCl concentration!). Furthermore, most corrosion inhibitors are toxic.

A smart cost-sensitive consumer would obtain hydrochloric acid at a low cost (\$10 for 1 gallon of 35% HCl), and obtain or make a non-toxic, low-cost corrosion inhibitor.

Is there a non-toxic, low-cost, and reasonably well-established corrosion inhibitor that can be reliably obtained by a consumer for use as a descaling agent in combination with HCl?

An extremely wide variety of plant extracts have been shown to serve as corrosion inhibitors, including black tea, chamomile, orange peels, black pepper, mustard seed, neem, the list goes on an on.

I have skimmed almost 100 papers so far and handfuls of review papers, and it is not clear to me that any of these "results" can be trusted. Most of these papers (all?) are from "PhD Factories" in developing countries, and zero of them give actionable insights. However, apparently almost every plant extract any author has ever tested offers protection against corrosion by a factor of roughly 5x in grams-per-liter concentrations.

A plant extract is generally made by taking about 10 grams of dried plant, and heating it in 0.5-1 liter of water (or some other solvent), and removing all but ~20ml/20g of liquid, leaving a very concentrated solution of plant compounds ("plant tea"). Or you can mix the dried material directly with the hydrochloric acid if you can then filter it.

Thus a food-grade hydrochloric-acid / "plant-tea" mixture might be an effective and cheap alternative to extremely expensive commercial products. 1 gallon of muriatic acid is \$10, producing 5 gallons of plant-extract/HCl descaling solution at a cost of \$2/gal. It may have comparable performance to commercial descaling solutions at more than \$50/gal.

However, these extracts are poorly understood and are extremely complicated bio-chemical solutions. My search continues for a single well-understood compound with low-toxicity and good corrosion inhibiting properties for steel in acids like HCl.


Grignards do transmetallation? Grignards don't add to alkyl halides to create linear carbon extensions because Grignards do transmetallation rather than SN2 displacement of halide.

Is this correct?

I thought transmetallation a) involved two metals and b) involved ligand exchange between the two metals … correct me if I'm wrong, but if I have a Grignard and an alkyl halide, I only have one metal.

You'd think that reaction of an alkyl grignard with a (primary) alkyl bromide/iodide gave all sorts of grignards via metal/halogen exchange.  I guess that's what your professor means by transmetallation.  You'd also get some elimination products.  Certainly that's what would happen with alkyllithiums.

That's why for carbon-carbon bond formation reactions you'd use cuprates.  Reduced basicity, increased nucleophilicity, all that.


Metal nail reaction in wall filler Please help explain what is happening to this nail. It has been in this wall filler for a good few months.

The substance being excreted was rising up forming these stalagmites, with a liquid forming on them.

Removing the nail, there were also excretions at the part that was sunken.

It is such a beautiful picture, perhaps good for a chemistry book cover. It is a similar phenomenon as what happens in a so-called chemical garden. Water seems to be creeping up to the nail head, and helping in oxidation of iron salts to insoluble hydroxides. Finally the gelatinous membranes act like semi-permeable membranc in the osmosis like experiment.

The figure shows several salt crystals soaked in a slightly alkaline medium.

Ref: From Chemical Gardens to Chemobrionics,Chemical Reviews, 2015, 115(16)
DOI: 10.1021/acs.chemrev.5b00014


Can a substituted cyclohexane progress through SN1? I have read that when the leaving group substituent is stuck in the equatorial position for some reason (like from a t-butyl group or a trans decalin), SN2 is either very slow or just doesn't happen at all because the rear attack is hindered by axial hydrogens/other substituents. I haven't been able to get a straight answer as to whether it can instead progress through SN1 at an acceptable rate, since the attached carbon will at least be secondary.

The only thing I can think of which would prevent this happening is the difficulty of forming a planar sp2 carbocation in a such a rigid system.

Any help is appreciated, thanks.

I think it's difficult to say for sure without actually making experiments. It should depend a lot on the conformation of the reactant, the substituents, the leaving group and also on the solvent. Nevertheless, I would imagine that it should be possible in some cases if you have a good leaving group.

In the case of a substituted cis-decalin, for example (I suppose you actually meant to refer to cis because in the trans isomer both hydrogens are stuck in axial positions), the elimination of the leaving group must indeed induce some strain, but maybe not that much. The structure could adapt with small changes in angles all over the molecule and both cycles would keep chair-like conformations. A tertiary carbocation would also be quite favorable.

So yes, in my opinion it should be possible in some cases, specially if heating is employed or you have a good leaving group. However, I should reinforce the fact that it really depends on many factors. This type of system can get quite complex if you consider the influence of the solvent, the influence of the substituents and the possibility of side reactions, including eliminations reactions, which commonly compete with substitution (on the presence of bases).


What is an example of chemical reaction that can be assisted by both an inorganic catalyst and an enzyme? I have been researching chemical reactions of inorganic catalysts and enzymes and cannot find a chemical process where an inorganic compound can be replaced by an enzyme (or vice versa) and have the same products. I have looked at article after article with no luck.
What is an example of chemical reaction that involves an enzyme that can be replaced by an inorganic catalyst?

A good example is the conversion of hydrogen peroxide to oxygen and water
$$\ce{2H2O2 ->2H2O + O2}$$
It can be catalyzed by catalase, which has one of the highest known turnover numbers, or by an inorganic catalyst such as manganese(IV) oxide or lead(IV) oxide.
This has been suggested for instance as an educational project by the University of York (see references).
References

Decomposing hydrogen peroxide. Chemistry Review (1995), Volume 5, Number 1, p. 30. https://www.york.ac.uk/chemistry/schools/chemrev/


How to calculate concentration from reaction quotient and initial concentration? Say I know that the reaction quotient $\left(Q = \dfrac{[B]}{[A]}\right)$ for a simple reaction $\ce{A\to B}$, at $t = 0$, $[A] = 1$ and $[B] = 0$, at some time in the future is $Q = 1$ (or any number really).

Is there a way to calculate the concentrations of A and B, or is more information required?

You have everything you need.

$$\ce{A -> B}$$

From the initial condition, stoichiometry and the conservation of mass, always:

$$\ce{[A] + [B]=1}$$

At time $t$:

$$\ce{\frac{[B]}{[A]}=1\to [B]=[A]}$$

Substitute into the first equation:

$$\ce{\to [A] + [A]=1\to 2\times [A]=1\to [A]=\frac12=[B]}$$

For a more general case:

$$\ce{[A] + [B]=n}$$

At time $t$:

$$\ce{\frac{[B]}{[A]}=Q\to [B]=Q\times [A]}$$

Substituting as above, we get:

$$\ce{[A]=\frac{n}{1+Q}}$$

$$\ce{[B]=\frac{n\times Q}{1+Q}}$$


Why is equilibrium achieved at different stages of a reaction? In other words, I want to know why some reactions attain equilibrium early in the reaction while some reactions obtain equilibrium at the end of the reaction.

Why is this the case?

If you define early or end of reaction by the how much the concentration of reactants change from initial reaction to once equilibrium is reached, it is because the equilibrium constant itself is essentially a ratio of the forward and reverse rate constants (can be approximated by the Arrhenius equation posted by t.c.). A high ratio means "more" products (at least a higher concentration) are present at equilibrium than reactants (so in a sense near the "end of the reaction"), since the forward rate is much higher than the reverse, a lower concentration of reactants compared to products is needed to maintain equilibrium. A low ratio means the opposite.


Structure of phosphate ion I understand that the hybridization in the phosphate ion is $\mathrm{sp^3}$ since phosphorous forms $4$ sigma bonds with $4$ of the oxygens and there are no lone pairs on the phosphorous atom.

The outer electronic configuration of phosphorous is $\mathrm{3s^2 3p^3}$. In the case of $\mathrm{sp^3}$ hybridisation, the $\mathrm{3s}$ and the all of the $\mathrm{3p}$ orbitals are involved. Hence, among the $4$ equivalent orbitals now obtained, $1$ of them contains a pair of electrons, while the other $3$ are singly occupied. So shouldn't phosphorous be able to form sigma bonds with only $3$ of the oxygen atoms since it has only $3$ unpaired electrons?

I am trying to compare this case with that of carbon tetrachloride. There, by $\mathrm{sp^3}$ hybridization, we were able to get $4$ equivalent orbitals each with an unpaired electron, and therefore explain its tetravalent nature. In the case of phosphate ion, one of the $4$ orbitals obtained already has a pair of electrons. So, could anyone please tell me where my understanding is flawed?

Disclaimer: hybridization is a mathematical concept and fails at higher levels.  As per the linked answer, it is clear that the participation of $d$-orbitals in phosphate ion is minimum. But this is just a high-school level, layman's explanation for your question.

  So, could anyone please tell me where my understanding is flawed?

Your reasoning is flawed in assuming that carbon atom has four unpaired electrons in its orbitals after hybridisaton, while phosphorus does not. Actually, both of them do.

In fact, carbon atom in its ground state does have a lone pair of electrons in its $\mathrm{2s}$ orbital. It is only when it gets into its excited state, that that lone pair is broken up, yielding four unpaired electrons. The excited carbon atom now undergoes hybridization as depicted in the figure below, yielding four equivalent covalent bonds as in carbon tetrachloride.

(source)

As you noted, phosphorus also has a lone pair of electrons in its $\mathrm{3s}$ orbital. But, that is its ground state electronic configuration. Once it gets excited, it will have five unpaired electrons, the fifth electron going into its $\mathrm{3d}$ orbital. Now, phosphorus will undergo $\mathrm{sp^3}$ hybridization, to form $4$ sigma bonds. The fifth electron helps form the $d\pi-p\pi$ bond with the neutral oxygen atom.

I hope this helps!


Why are X-rays used in crystallography? Can any other wavelength of electromagnetic radiation be used?
Why can't gamma radiation be used?

There are a few reasons, but the most direct answer is that the wavelength of X-ray photons is on the order of the distance between atomic nuclei in solids, e.g. ~ 4 ångströms (bonds are roughly 1.5-2.5 Å). You can think of it like the waves fit nice and snugly between the atoms and "fill" the crystal and thereby give us information about where the "cavities" are, and thus by reciprocal analysis (inverse Fourier transform) we can determine where the "rigid" stuff is (the atoms).

Other kinds of incident radiation are used e.g. neutron beams (for crystallography) and even electron beams (for surface spectroscopy), but X-rays are a more readily available source of radiation. In other words, it's easier/cheaper to produce X-rays than a beam of pure neutrons.

Gamma rays are about 100 times smaller in wavelength than X-rays (so instead of 4 Å, something like 0.04 Å, and so they generally penetrate through any solid material and don't produce a clear diffraction pattern like X-rays. Also if I'm not mistaken, it is much more difficult to produce gamma rays.

It's important to realize that diffraction is not the same as reflection and refraction. The latter two are surface phenomena - diffraction is a bulk phenomena completely unrelated to surfaces.

Electrons interact strongly with the surface of solids not only because of their charge but their energetic wavelength - gamma rays interact very weakly with a bulk crystal and go right through it. X-rays and neutrons (bearing no charge) are in the "happy medium" and produce a sufficient interaction with crystals as to give information about the arrangement of particles.

Picture taken from wikipedia's X-ray article:

More useful info in Economou's book "The Physics of Solids" can be found.


Why would I not see a reaction between aluminum and tin(IV) chloride? I was performing an experiment where I was supposed to construct a modified activity series between certain elements. I decided to use chlorides with the other metals for single replacement reactions so I could compare them all in relation to one another for my modified activity series.

When I placed about one gram of aluminum into about five milliliters of tin(IV) chloride, I was not able to see any reaction taking place. However, aluminum is higher in reactivity on the activity series so I know the aluminum had to react to replace the tin, but why would I not be able to see it?

You need to activate the $\ce{Al}$ to remove the tenacious oxide layer that protects it from reaction. This can be done by a quick wash with dil $\ce{HCl}$, rinse with $\ce{EtOH}$ and finally $\ce{Et2O}$ before adding to the tin chloride.


Initial rate of reaction from the graph? I tried finding the slope between two points but I keep getting wrong values, how does one find the initial rate by using such graphs?

So I started by trying to find a slope of two points, $(0,0.4)$ and $(10,0.2)$ after calculations I got  $-0.02$, which is the opposite of the right answer in sign. I tried to check myself by trying to find the slope for other points like: $(10,0.2)$ and $(20,0.1)$ which gives me $-0.01$, I'm confused as whether my initial idea is right or not.

It seems like you have the right idea to me,

For the "Initial Rate," you're taking a curve and doing a straight approximation of the curve over a small portion, as you appear to be doing.

Your logic also appears correct to me, a decrease in concentration should correspond to a negative rate.

My guess is that the question, (a poorly worded one?) is actually referring to the rate of product creation, and asking you to assume that there is a 1-1 relationship between the decrease in reactant, and the appearance of product. Thus if your reactant is disappearing at $0.02~M/s$ (as indicated by $rate= -0.02 ~M/s$) , then your product should be appearing at $0.02~M/s$, making A the correct answer.


Can a very high concentration of hydrogen peroxide, inhibit or denature the enzyme catalase? My experiment, where I am investigating the effect of increasing substrate concentration of hydrogen peroxide on enzyme activity hasn't worked out as I hoped. The last result, highest concentration of hydrogen peroxide, shows a decrease in enzyme activity compared to the lower substrate concentration and I don't understand why?

20% hydrogen peroxide sounds HUGE. But something can happen that is called substrate inhibition. It can be detected if the substrate acts as a non-competitive reversible inhibitor (i.e. binds only to the active ES Michaelis complex). In the case I studied during my PhD, the inhibition constant ($K_\mathrm{i}$) was roughly 1000 times higher that the Michaelis constant ($K_\mathrm{M}$).
The steady-state rate was well fit with :
$$r = \frac{r_\infty}{1+\frac{K_\mathrm{M}}{s} + \frac{s}{K_\mathrm{i}}}$$
So the activity would go through a maximum ($s$ is substrate concentration).

Tip : if it is possible, do you assays with substrate concentrations growing exponentially, like $1\,\mathrm{\mu M}$, then 2, 5, 10, 20, etc.

A good way to decide between enzyme denaturation and substrate inhibition would be to know if it is reversible : denaturation must not be reversible. You could first make a baseline assay (at low conc.), expose the enzyme to a high concentration, then change the composition of the solution (eg dialysis, dilution) and run another assay at low concentration.

A "volume of bubbles" does not sound very accurate. Could you use a Clark electrode to measure the activity instead ?


How many covalent bonds are there in carbon suboxide? Calculate the number of covalent bonds in $\ce{C3O2}$.

Structure:

I am confused. I can count covalent bonds as 8 by counting single-single bonds, but also I can say that there are 4 "double covalent bonds". Which is correct?

Widely used terminology might not match what an examination wants
It is possible that your question wants the answer 8 in this context, but whether this answer would be widely accepted by anyone familiar with bonding theory is doubtful.
In the early days of theories about bonding Lewis described bonds as consisting of two shared electrons. You are probably familiar with "Lewis Structures" which are simple pictures of bonding where electrons are represented with two dots between atoms. This is also how the theory of bonding is initially introduced to students. In this view the number of bonds in the molecule is, indeed, 8. But this–vastly over simplistic theory–is not good for explaining the bonds in many real compounds. Hence the dispute about the correct terminology.
A good theory of bonding needs to take quantum mechanics (and the shape of electron orbitals) into account and those were not understood in the early 20th century when Lewis developed his explanation of bonding.
But we don't need any quantum obfuscation to see why Lewis theory doesn't work well and why it is better not to pretend that bonds consist of pairs of electrons. Take benzene as an example.  You could claim that the bonds in benzene contain 18 electrons so make up 9 bonds. But this means alternating double and single bonds. But that can't be right as all the bonds in benzene are the same length. It is better to say than benzene consists of 6 bonds each of which have somewhere between the character of a single and double bond.
This inability of the Lewis model of bonding to explain real world structures is very common. Nitrate ions are one example where all the bonds have characteristics of more than a "single" bond but less than a "double" bond. Many boranes have bonds between 3 atoms but only two electrons to share among them.
So equating bonds to shared pairs of electrons is not helpful. Nor is using terminology that defines a bond that way. So real chemists should be reluctant to count bonds that way. Whether this is the answer your question is looking for is less clear as you might still be at the stage where they have taught you Lewis structures and nothing else.


Does s stand for sigma bond and p for pi bond? When we are talking about hybridization, in $\mathrm{sp/sp^2/sp^3}$ hybridization, does s stand for sigma bond and p for pi bond ?

You got it backwards.

The $\mathrm{s}$, $\mathrm{p}$, $\mathrm{d}$, $\mathrm{f}$ orbitals stand for sharp, principal, diffuse, and fundamental. Wikipedia: Electron Configuration § Notation:

  The choice of letters originates from a now-obsolete system of categorizing spectral lines as "sharp", "principal", "diffuse" and "fundamental" (or "fine"), based on their observed fine structure: their modern usage indicates orbitals with an azimuthal quantum number, l, of $0$, $1$, $2$ or $3$ respectively.

Then, $\sigma$, $\pi$, $\delta$, $\phi$, being the Greek equivalents of $\mathrm{s}$, $\mathrm{p}$, $\mathrm{d}$, $\mathrm{f}$ respectively, came to designate bonds or molecular orbital symmetry, because $\mathrm{s}$ is the first orbital to form $\sigma$ bonds, $\mathrm{p}$ the first orbital to form $\pi$ bonds, $\mathrm{d}$ the first orbital to form $\delta$ bonds, and $\mathrm{f}$ the first orbital to form $\phi$ bonds.

The $\mathrm{s}$ and $\mathrm{p}$ in $\mathrm{sp^3}$ are precisely the $\mathrm{s}$ orbital and the $\mathrm{p}$ orbital. $\mathrm{sp^3}$ means that $1$ $\mathrm{s}$ orbital mixes with $3$ $\mathrm{p}$ orbitals to create $4$ hybrid orbitals known as $\mathrm{sp^3}$ orbitals.


Ionization of hydrogens in hypophosphorous acid Does anyone know why the hydrogen atoms bonded with phosphorus in $\ce{H3PO2}$ are not ionizable?

I tried looking in high school level books, but they only describe the process, not the real reason. I would appreciate if someone has the answer.

To begin with, removing another proton from anionic species is not very favorable. For example, the monohydrogen phosphate dianion is not very acidic $(\mathrm{p}K_\mathrm{a}\sim 12)$ despite it having an $\ce{OH}$ group.

In the case of $\ce{H3PO2},$ the second and third protons are bonded directly to $\ce{P},$ which is much less electronegative than $\ce{O},$ and consequently these P-H bonds are much less polarized than an $\ce{O-H}$ bond. So these protons are unable to ionize appreciably in aqueous solution.


How does conductivity work for non-redoxed ions? Related (very similar, but here I want a mechanism) https://physics.stackexchange.com/q/21827/7433

By the Kohlrausch law, all ions contribute to the conductivity of an electrolyte.

Now, as I understand it, the mechanism of conduction in an electrolyte is thus:

Ions migrate in solution
These ions get reduced or oxidized at the electrodes and converted to electrons
These electrons continue down the wire, leading to an increased/maintained conductivity/current

But, this mechanism doesn't work for ions which do not get redoxed--movement of ions in the solution cannot be translated to movement of electrons in the wire and thus it seems (to me) that conductivity should not increase.

But ions like $\ce{NO3-}$ have a comparable $\lambda$ (ionic molar conductivity)--so they experimentally do increase conductivity.

What is the mechanism for conduction via these ions?

This can arise due to ion-ion interactions brought on by Coulombic fields.

A proper explanation is rooted in transport phenomena. Unfortunately chemists are largely not taught this, but chemical engineers do get the opportunity.

Under standard conditions, ions in a solution will take a random walk during diffusion. This results in no net movement of our ions over some span of time. Such a system perturbed by an applied electric field will cause ions to preferentially move in one direction, we say the ions will drift. We say they have a drift velocity. These velocities are affected by a number of parameters, as there is still continual collision, but we can say that the mobilities will be unequal for ions of unequal sizes.

Different mobilities mean concentration gradients form and charge separation too. Typically we use an electroneutrality field as a reasonable approximation to allow analytic solutions for various equations. This field depends on all the ionic fluxes present and as such will affect the total measurable current.

The result is simple: the introduction of any ion will cause a perturbation in the field that all the ions encounter in solution.


Functionalization of nanoparticles I want to functionalize silver nanoparticles with thiol group.
My question is "how can I find that the thiolated compound is attached to the surface of nanoparticles or not?"

If the thiolate bounds to the Au nanoparticle, both the electronic, as well as the mechanical properties of the thiolate and the Au nanoparticle change.  So you are going to monitor these changes with a spectroscopy susceptible to this, comparing the spectra of the pure starting materials with the one(s) of the decorated NP.

One of the easier available techniques is to rely on FT-IR spectroscopy, as by the established bond between the two partners, some absorption bands newly appear, and others change their position.  The former are the ones more easily taken as a quantitative reference, too (because yes, IR spectroscopy allows both qualitative, as well as quantitative characterisations, too).

In the instance of Au NP, to check for a bonding of the NPs to the mercapto decoration is possible both by IR spectroscopy, as well as by (surface enhanced) Raman spectroscopy.  In a work deploying $\omega$-mercaptoundecanoic acid (MUA) by Tripathy and  Yu, the later was the preferred check to monitor the S-Ag interaction:

(source, doi 10.1016/j.saa.2008.12.004, Spectrochimica A, 72, 2009, 841-844).

Similar as in the case of the Au NP (vide infra), the key feature is the disappearance of the S-H vibration.  To quote from the section about their characterisation by Raman spectroscopy:

  The most interesting feature was the disappearance of S–H stretching
  ($\pu{2547 cm^{−1}}$) peak. The disappearance of S–H stretching indicated the covalent interaction of sulphur and silver, which is in good agreement with the FT-IR data [16]. However some of the expected peaks such as C–Sgauche stretching could not be identified which may be due to their low intensity values.

the earlier answer, erronusly about Au NP

As an example, were gold nanoparticles were decorated by dodecanthiol (DT) chains:

(source, Sharma et. al. in Indian Journal of Pure & Applied Physics
52, 2014, 93-100.)

Other than the absorption band at $694$ and $\pu{728 cm^{-1}}$, the authors equally draw attention that

  the "S-H vibration band at 2576 cm−1 for pure DT is observed while it is almost absent for AuNPs. This suggests that as the thiol gets bound onto the  surface of gold, the S-H bond of thiol breaks in order to form the S-Au bond." (loc. cit., p. 97).

It may be helpful to mention that short wavelengths of light ($\pu{< 400 nm}$) may break the $\ce{Au-S}$ bond between nanoparticle, and thiol, hence if you doubt the presence of thiolate decoration on the NP, select an absorption band and monitor its attenuation while shining the short wavelength light on the sample.

And complementary to IR spectroscopy, is Raman spectroscopy.


Acid-base chemistry - determining stability I am learning about acid-base reactions and how to identify whether an acid is strong or not by looking at the conjugate base. The more stable a conjugate base the stronger the acid.

However, two of the factors that determine conjugate base stability seems contradictory to me.

I understand the bigger the atom, the more stable the base so $\ce{HI}$ is more stable than $\ce{HF}$.
The smaller the orbital the more stable the base. So $\ce{sp}$ is more stable than $\ce{sp^3}$.

Now why is a bigger atom more stable than a smaller one, but a smaller orbital  more stable than a bigger orbital?

When we say that 'bigger the atom, more stable is the anion', we mean to say that the negative charge would be spread over a larger area. So, the interatomic repulsive forces would be spread over a large area, lowering the energy and stabilising the anion. Going down the group, the jump in size is large enough to compensate the decrease in electronegativity, so we use the 'size of atoms' argument to differenciate between different atoms in the same group.

Conversely, we use the 'size of orbital' argument to determine the basicity of the same atom but in different hybridisations. Hence, the size of the atom would be same and the smaller size of the sp orbital would mean that it would be better attracted by the nucleus making the anion more stable.


How to control the purity and quality of fatty acids or their CoA thioesters? Background: We are looking into getting a series of cis-mono-unsaturated very-long-chain (C20-C32) fatty acids (with different, specific double-bond positions) — and/or the corresponding acyl-Coenzyme A thioesters — synthesised by a company. These compounds are to be used in primary research for a biochemical enzyme activity assay.

Question: Would MS or 1H-NMR be preferable for quality control/identification of such (custom-synthesised) compounds? Or would it not matter for these compounds at hand? What would be the advantages/disadvantages of these two options we are being offered?

Short answer: Both methods are (technically) suitable but it depends on what impurities you want to detect. But in your case, I would probably suggest mass spectroscopy in most cases.

Long answer:

It depends on how many different compounds you’re getting, how much you trust the company, which impurities you are expecting and how important it is to have exactly the one compound you need.

Mass spectroscopy, ideally coupled with HPLC beforehand, will give a specific retention time for a specific compound, and of course you will get the compound’s mass. That alone is not sufficient for a priori identification! When one detects purity by mass spectroscopy, one would usually have a standard run of one’s desired compounds and compare it to the unknown sample. To identify, compare the retention time (ideally standardised, if the project goes on for more than, say, a year) and the compound masses.

The different fatty acids will have similar, but not identical retention times. The cis-double bond should allow for easy detection via UV absorption (good for quantitising known impurities) and the mass will, of course, differ only by the number of carbon atoms. Impurities that are not fatty acids will have completely different retention times, different UV spectra and different masses which should allow an easy attribution of ‘good peak, bad peak’. If it is not of paramount importance to have only exactly one fatty acid, one can even rest happily with quantitising the amounts of the different ones, too.

Note that it will take some fine-tuning of the HPLC gradient, since most gradients are designed to fit a wide range of polarities with average accuracies, while what you want is a high accuracy in a low polarity range.

Everything I said about HPLC-MS of fatty acids also holds true for the CoA-thioesters, although it is possible that the HPLC solvent hydrolyses the thioester. Work-arounds would be buffered solvents and similar.

$\ce{^1H}$-NMR is best used on pure substances, but can also be used to detect impurities. It should allow for an easy identification on which bond is a double bond by the characteristic shifts and/or multiplets of the two vinylic protons (shift range $6-7~\mathrm{ppm}$). However, you will most likely get unseperable signals at around $1-2~\mathrm{ppm}$ — the aliphatic region where all the saturated carbons’ protons will resonate. Thus it will not be easy to detect saturated impurities, say for example the corresponding unsaturated fatty acid. However, it will be easy to detect other impurities stemming from the synthesis such as remaining reactands (which will depend on the synthetic route).

If you go to the CoA-thioesters, it will be a lot harder to estimate purity just by an NMR spectrum, because CoA has a lot more signals in the $2-5~\mathrm{ppm}$ range. But it is still possible.

NMR would allow the quantification of impurities if you can identify them just by integrating the proton signals and dividing them by the amount of protons giving identical signals.

On a time scale, a typical HPLC-MS run will be around $40~\mathrm{min}$ while a typical proton NMR can be recorded within $5~\mathrm{min}$.

I would suggest NMR in situations where you have a really high number of samples that need to be analysed very timely, where it does not matter if the saturated acid is present or where identifying the stereochemistry and regiochemistry of the double bond is of prime importance. In any case where saturated fatty acids would impede your results, I would strongly suggest HPLC-MS. In either case, I would suggest acquiring pure samples (at least of the fatty acids) first to run a series of standards.