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What is the difference between ∆G and ∆G°?
In Brady's Molecular Nature of Matter, I read that $\Delta_\mathrm{r} G^{\circ}$ is $\Delta_\mathrm{r} G$ at $25~^\circ\mathrm{C}$. But later, it gives a value for $\Delta_\mathrm{r} G^\circ$ at another temperature.
What is the meaning of $\Delta_\mathrm{r} G^\circ$?  Specifically:

Does it only refer to $25~^\circ\mathrm{C}$?
Does the $\Delta_\mathrm{r} G^\circ = \Delta_\mathrm{r} H^\circ - T \Delta_\mathrm{r} S^\circ$ equation only refer to $T=298~\mathrm{K}$?
If at $25~^\circ\mathrm{C}$, a system that has reached equilibrium ($\Delta_\mathrm{r} G=0$), and thus

$Q=K$
$\Delta_\mathrm{r} G^\circ=0$
$\Delta_\mathrm{r} G^\circ=0=-RT\ln K$
$RT\ln K=0$
$\ln K=0$
$K=1$

But doesn't $K$ vary from reaction to reaction? It shouldn't always equal 1, should it?

Related: Is the Gibbs standard free energy always constant?

Short answer

Does it need to be at $25~^\circ\mathrm{C}$?

No. $\Delta_\mathrm{r} G^\circ$ can be defined at any temperature you wish to define it at, since the standard state does not prescribe a particular temperature. If you change the temperature, $\Delta_\mathrm{r} G^\circ$ will change.

Does $\Delta_\mathrm{r} G^\circ = \Delta_\mathrm{r} H^\circ - T\Delta_\mathrm{r} S^\circ$ always use $T = 298~\mathrm{K}$?

No. You use whatever temperature you are running your reaction at.

(...maths...)

Yes, at equilibrium, $\Delta_\mathrm{r} G = 0$ and $Q = K$.
However, everything after the first bullet point is wrong. You cannot conclude that $\Delta_\mathrm{r} G^\circ = 0$, nor can you conclude that $K = 1$. The equation $\Delta_\mathrm{r} G^\circ = -RT \ln K$ does not analogously translate into $\Delta_\mathrm{r} G = -RT \ln Q$! The accurate relation is:
$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + RT\ln Q$$
Setting $Q = K$ and $\Delta_\mathrm{r} G = 0$ in this equation does not tell you anything about the value of $K$. In fact, if you try doing it, all you will find out is that $\Delta_\mathrm{r} G^\circ = -RT \ln K$ - no surprises there!

Long answer
Any book that writes that $\Delta_\mathrm{r} G^\circ$ is the "special case" of $\Delta_\mathrm{r} G$ at $T = 298~\mathrm{K}$ is wrong.
The Gibbs free energy of a system is defined as follows:
$$G = H - TS$$
Under constant temperature and pressure (from now on, I will just assume constant $T$ and $p$ without stating it), all systems will seek to minimise their Gibbs free energy. Equilibrium is reached when $G$ is minimised. When $G$ is at a minimum, any infinitesimal change in $G$, i.e. $\mathrm{d}G$, will be $0$. Therefore, this is equivalent to saying that the condition for chemical equilibrium is $\mathrm{d}G = 0$.
Clearly, we need a way to relate this quantity $\mathrm{d}G$ to the actual reactants and products that are in the system. This can be done by using the Maxwell relation (see any physical chemistry text for details):
$$\mathrm{d}G = V\,\mathrm{d}p - S\,\mathrm{d}T + \sum_i \mu_i\,\mathrm{d}n_i$$
Under constant $T$ and $p$, $\mathrm{d}p = \mathrm{d}T = 0$ and therefore
$$\mathrm{d}G = \sum_i \mu_i\,\mathrm{d}n_i$$
where $\mu_i$ is the chemical potential of species $i$, defined as a partial derivative:
$$\mu_i = \left(\frac{\partial G}{\partial n_i}\right)_{n_{j\neq i}}$$
So, we now have a refined condition for equilibrium:
$$\mathrm{d}G = \sum_i \mu_i\,\mathrm{d}n_i = 0$$
We can go further by noting that the values of $\mathrm{d}n_i$ for different components $i$, $j$, etc. are not unrelated. For example, if we have a reaction $i + j \longrightarrow k$, then for each mole of $i$ that is consumed, we must also use up one mole of $j$; this means that $\mathrm{d}n_i = \mathrm{d}n_j$.
This can be formalised using the idea of a stoichiometric coefficient $\nu_i$, which is defined to be positive for products and negative for reactants. For example, in the reaction
$$\ce{3H2 + N2 -> 2NH3}$$
we have $\nu_{\ce{H2}} = -3$, $\nu_{\ce{N2}} = -1$, and $\nu_{\ce{NH3}} = 2$.
By stoichiometry, if $1.5~\mathrm{mol}$ of $\ce{H2}$ is consumed, then $1~\mathrm{mol}$ of $\ce{NH3}$ has to be produced. We could write $\Delta n_{\ce{H2}} = -1.5~\mathrm{mol}$ and $\Delta n_{\ce{NH3}} = 1~\mathrm{mol}$. These quantities are proportional to their stoichiometric coefficients:
$$\frac{\Delta n_{\ce{H2}}}{\nu_{\ce{H2}}} = \frac{-1.5~\mathrm{mol}}{-3} = 0.5~\mathrm{mol} = \frac{1~\mathrm{mol}}{2} = \frac{\Delta n_{\ce{NH3}}}{\nu_{\ce{NH3}}}$$
The quantity $0.5~\mathrm{mol}$ is a constant for all chemical species $\ce{J}$ that participate in the reaction, and it is called the "extent of reaction" and denoted $\Delta \xi$ (that is the Greek letter xi). If the reaction is going forward, then $\Delta \xi$ is positive, and if the reaction is going backwards, then $\Delta \xi$ is negative. If we generalise the above result, we can write
$$\Delta \xi = \frac{\Delta n_i}{\nu_i}$$
and if we make $\Delta n_i$ smaller and smaller until it becomes an infinitesimal, then:
$$\begin{align}
\mathrm{d}\xi &= \frac{\mathrm{d}n_i}{\nu_i} \\
\mathrm{d}n_i &= \nu_i\,\mathrm{d}\xi
\end{align}$$
If we go back to our condition for equilibrium, we can substitute in the above to get:
$$\mathrm{d}G = \sum_i \mu_i\nu_i\,\mathrm{d}\xi = 0$$
Now, $\mathrm{d}\xi$ is no longer dependent on $i$, since we have established already that $\Delta \xi$ (and by extension $\mathrm{d}\xi$) is a constant for all chemical species. So, we can "divide through" by it to get:
$$\Delta_\mathrm{r} G \equiv \frac{\mathrm{d}G}{\mathrm{d}\xi} = \sum_i \mu_i\nu_i = 0$$
where $\Delta_\mathrm{r} G$ is defined to be $\mathrm{d}G/\mathrm{d}\xi$.
Note that $\Delta_\mathrm{r} G$ is an intensive property and has units of $\mathrm{kJ~mol^{-1}}$, since $\mathrm{d}\xi$ has units of $\mathrm{mol}$. This ensures that the units we use are consistent: since we know that $\Delta_\mathrm{r}G = \Delta_\mathrm{r} G^\circ + RT\ln Q$, $\Delta_\mathrm{r}G$ must have the same units as $RT$.
How do we interpret the physical significance of $\Delta_\mathrm{r} G$, or in other words, what does it even mean? There are two ways, each based on a different mathematical expression.

We have $\Delta_\mathrm{r}G = \sum \nu_i \mu_i$. This means that $\Delta_\mathrm{r}G$ is simply the difference between the chemical potentials of the products and the reactants, weighted by their stoichiometric coefficients. For the reaction $\ce{3H2 + N2 -> 2NH3}$, we have:

$$\Delta_\mathrm{r} G = \sum_i \mu_i\nu_i = 2\mu_{\ce{NH3}} - 3\mu_{\ce{H2}} -\mu_{\ce{N2}}$$

We have $\Delta_\mathrm{r}G = \mathrm{d}G/\mathrm{d}\xi$. This means that it is the slope of a curve of $G$ against $\xi$:

Note that up to this point, we have not stipulated any particular temperature, pressure, amounts of species present, or any conditions whatsoever. We have only said that the temperature and pressure must be constant.
It is important to realise that $\Delta_\mathrm{r}G$ is a well-defined quantity at all $T$, all $p$, and all possible values of $n_i, n_j, \cdots$! The shape of the curve will change when you vary the conditions. However, no matter what the curve looks like, it is always possible to find its gradient ($= \Delta_\mathrm{r}G$) at a particular point.

What exactly, then, is $\Delta G^\circ$? It is just a special case of $\Delta G$, where all the reactants and products are prepared in a standard state. According to IUPAC, the standard state is defined as:

For a gas: pure ideal gas when the pressure $p$ is equal to the standard pressure $p^\circ$.
For a liquid or solid: pure liquid or solid at $p = p^\circ$
For a solution: ideal solution when the concentration $c$ is equal to the standard concentration $c^\circ$.

$p^\circ$ is most commonly taken to be $\pu{1 bar}$, although older texts may use the value $\pu{1 atm} = \pu{1.01325 bar}$. Since 1982, IUPAC has recommended the value $\pu{1 bar}$ for the standard pressure (Pure Appl. Chem. 1982, 54 (6), 1239–1250; DOI: 10.1351/pac198254061239). However, depending on the context, a different value of $p^\circ$ may prove to be more convenient. Likewise, $c^\circ$ is most commonly – but not necessarily – taken to be $\pu{1 mol dm-3}$.
Note that in the above definitions, no temperature is specified. Therefore, by defining the standard Gibbs free energy, we are fixing a particular value of $p$, as well as particular values of $n_i, n_j, \cdots$. However, the value of $T$ is not fixed. Therefore, when stating a value of $\Delta_\mathrm rG^\circ$, it is also necessary to state the temperature which that value applies to.
When a reaction vessel is prepared with all its substances in the standard state, all the components of the system will have an activity of exactly $1$ by definition. Therefore, the reaction quotient $Q$ (which is a ratio of activities) will also be exactly equal to $1$. So, we could also say that $\Delta_\mathrm{r}G^\circ$ is the value of $\Delta_\mathrm{r}G$ when $Q = 1$.
Returning to the graph of $G$ against $\xi$ above, we note that at the left-most point, $Q = 0$ since there are only reactants; at the right-most point, $Q \to \infty$ as there are only products. As we move from left to right, $Q$ increases continuously, so there must be a point where $Q = 1$. (In general, the point where $Q = 1$ will not be the same as the equilibrium point.) Since $\Delta_\mathrm{r}G$ is the gradient of the graph, $\Delta_\mathrm{r}G^\circ$ is simply the gradient of the graph at that particular point where $Q = 1$:

The gradient of the graph, i.e. $\Delta_\mathrm{r}G$, will vary as you traverse the graph from left to right. At equilibrium, the gradient is zero, i.e. $\Delta_\mathrm{r}G = 0$. However, $\Delta_\mathrm{r}G^\circ$ refers to the gradient at that one specific point where $Q = 1$. In the example illustrated above, that specific gradient is negative, i.e. $\Delta_\mathrm{r}G^\circ < 0$.
Again, I reiterate that the temperature has nothing to do with this. If you were to change the temperature, you would get an entirely different graph of $G$ versus $\xi$. You can still find the point on that graph where $Q = 1$, and the gradient of that graph at the point where $Q = 1$ is simply $\Delta_\mathrm{r} G^\circ$ at that temperature.

We have established the qualitative relationship between $\Delta_\mathrm{r} G$ and $\Delta_\mathrm{r} G^\circ$, but it is often useful to have an exact mathematical relation.
$\Delta_\mathrm{r} G^\circ$ is exactly the same as $\Delta_\mathrm{r} G$ except for the imposition of the standard state. It follows that if we take the equation
$$\Delta_\mathrm{r} G = \sum_i \mu_i \nu_i$$
and impose the standard state, we get
$$\Delta_\mathrm{r} G^\circ = \sum_i \mu_i^\circ \nu_i$$
Thermodynamics tells us that
$$\mu_i = \mu_i^\circ + RT\ln{a_i}$$
where $a_i$ is the thermodynamic activity of species $i$. Substituting this into the expressions for $\Delta G$ and $\Delta G^\circ$ above, we obtain the result:
$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + RT\ln Q$$
where the reaction quotient $Q$ is defined as
$$Q = \prod_i a_i^{\nu_i}$$
When equilibrium is reached, we necessarily have $\Delta_\mathrm{r} G = 0$ (see the discussion above). The equilibrium constant $K$ is defined to be the value of $Q$ at equilibrium. Therefore, at equilibrium, $Q = K$. Plugging this into the equation above gives us the famous equation:
$$\Delta_\mathrm{r} G^\circ = -RT\ln K$$
Again, no temperature is specified! In general, $K$ depends on the temperature as well; the relationship is given by the van 't Hoff equation.

Comments are not for extended discussion; this conversation has been moved to chat.

I think the answer to these questions can be simplified considerably.  My answer would start out with the part of orthocresol's answer which says:
What exactly, then, is $\Delta G^\circ$? The truth is that is is simply a special case of $\Delta G$, where all the reactants and products are prepared in a standard state. According to IUPAC, the standard state is defined as:

For a gas: pure ideal gas at $p = 1~\mathrm{bar}$
For a liquid or solid: pure liquid or solid at $p = 1~\mathrm{bar}$
For a solution: ideal solution at $c = 1~\mathrm{mol~dm^{-3}}$

I would then add the following:
For the cases of pure gases, liquids, and solids (not solutions), I like to think of the pure reactants and pure products as actually being in separate containers, each at 1 bar pressure.  So, in these cases, the 0 superscript refers specifically to reactants and products all at 1 bar pressure.  It can be at any specified temperature.  So $\Delta G^0$ refers to the change in free energy in transitioning from thermodynamic State 1 to thermodynamic State 2, where
State 1 = Pure reactants (in stoichiometric proportions) in separate containers at 1 bar and temperature T
State 2 = Pure products (in corresponding stoichiometric proportions) in separate containers at 1 bar and temperature T
Then, for $\Delta G$, it is basically the same thing, except that here the pure reactants and pure products are at arbitrary specified pressures in their separate containers, not necessarily at 1 bar.  If these pressures happen to coincide with the partial pressures in an equilibrium mixture of reactants and products (assuming ideal gas behavior), then the $\Delta G$ is zero.
The trick in all this is to figure out a reversible path between pure reactants and pure products at 1 bar (such that the changes in G can be evaluated).  That's where the van't Hoff equilibrium box comes in.

Comments are not for extended discussion; this conversation has been moved to chat.

I see it in different way. You are right in most textbooks $\Delta G^0$ means free energy change at $25~\mathrm{^\circ C}$ but in this case you have to think in a different way. I see it as segregation of pressure and temperature dependence.
$$\Delta G(T,P)=\Delta G^0 (T)+RT\ln(K_\mathrm{eq})$$
At equilibrium you must have $\Delta G(T,P)=0$
So, it helps me to find the trade-off between free energy contribution due to temperature and due to activity of the reactant-product system at equilibrium condition.
In microkinetic modeling, I often need to include lateral interaction and in that case my $\Delta G^0 (T) $ is not constant at a particular temperature but a function of coverage, that's why I have a dynamic relation between $\Delta G^0$ and $K_\mathrm{eq}$. But in most of the practical system it's just a static relation and only means one particular point on thermodynamic landscape.

If the value of $\Delta_\mathrm{r} G^\circ$ depend on temperature then for a chemical reaction does this mean that the equilibrium constant will have different value if we evaluate $\Delta_\mathrm{r} G^\circ$ at different temperature?


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Synthesis of phosphonate salt not working
I have tried four times in the past few weeks to synthesise 2-ethoxyvinyl phosphonyl dichloride by reacting ethyl vinyl ether with phosphorus pentachloride in toluene.
I have been following the literature procedure (see below) successfully up to the point where it says that I need to bubble sulfur dioxide through the reaction mixture - at which point it will turn clear. The reaction turns a pale yellow color; the next step is where I'm at my wit's end - the literature says to obtain the 2-ethoxyvinyl phosphonyl dichloride by direct distillation and every time I've tried this, it's resulted in a black, opaque solution. The literature does not say what color the final product is although my supervisor has said that it should be a yellow oil. I'm thinking that it could have decomposed upon heating, although the boiling point according to the literature is 142 °C and the bp of toluene is 110. Previously I tried the same reaction in $\ce{CCl4}$, which distilled off at 76 °C and also turned black.
I would really appreciate help on this, since it's part of my final year project and I only have a handful of lab sessions left to finish this. Thank you in advance! (Also, I've attached an image of the literature procedure below if it might be helpful).

$%edit$

Make sure to ask this on researchgate.net.

What is the source of the phosphorus-bound O atom in the final product?  It must be the $\cf{SO2}$ I suppose but it isn't clear how it gets there.  What is the intermediate that you believe forms after mixing the ether and the pentachloride?  How long should forming that intermediate take?

can you add a link to the literature?

@CurtF. - the SO2 is the oxidising agent so I believe that's how the phosphorus-bound O gets there. We think the intermediate formed after mixing ether with pentachloride could be CH3-CH2-O-CH=CH-PCl4+ and PCl6-.

@A.K. - The method isn't exactly from the literature, it's part of a US patent file which contains many other methods which aren't relevant. The image I attached in my first post is the only information I have to go on, unfortunately.

Looking at the procedure (and considering the structure of the desired product) it appears that its boiling point is 142C at 27 mm (not at atmospheric pressure). So you should try a vacuum distillation to purify your product. Heating it to try to distil it at 1 atm is probably causing decomposition. Also, be sure to protect the reaction mixture and the final product from atmospheric moisture.

Thanks for this! Vacuum distillation seems to be the best way forward at this point. I'll take your advice and post the results here when I have them.

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Introduction to ball tumbler chemistry?
Chemistry that takes place in a ball tumbler is distinctly different from aqueous chemistry. Is there a good introduction to this method?

Are you referring to grinding in a "ball mill"?

@MaxW How did you guess?

Grinding isn't "chemistry"...

@MaxW But it sure is sexy: http://www.ch.cam.ac.uk/files/posters/SandersPoster_Shanghai.pdf

:-( I should have known that there would be exceptions. I really think this sort of reaction is more of a curiosity than something which would be generally applicable.

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Why are protons in the same position in 1-chloro-4-nitrobenzene magnetically different?
Protons in ortho positions towards the nitro group are said to be magnetically different, even though they're chemically equivalent. Same goes for protons ortho towards the chlorine. Why is that so?

The hunch lies in the definition of magnetic equivalence:

For two protons to be magnetically equivalent they not only have to have the same chemical shift, but they must also each have the same $J$ coupling to other magnetic nuclei in the molecule.

(Quoted from Hans Reich’s NMR course online syllabus; emphasis present in the original.)
What does that mean for our specific example? Compare the molecule we have in figure 1 below.

Figure 1: 1-chloro-4-nitrobenzene with labelled protons.
Needless to say as you already figured that $\ce{H^1}$ and $\ce{H^3}$ are chemically equivalent (one is transformed into the other by $C_2$‡) meaning that they will display the same chemical shift; likewise for $\ce{H^2}$ and $\ce{H^4}$.
However, for magnetic equivalence we need to consider their coupling constants to each other. If $\ce{H^1}$ and $\ce{H^3}$ are magnetically equivalent, then $J_{\ce{H^1 H^2}} \overset{!}{=} J_{\ce{H^3 H^2}}$. But, $J_{\ce{H^1 H^2}}$ is an ortho coupling with a value of $J \approx 7~\mathrm{Hz}$ while $J_{\ce{H^3 H^2}}$ is a para coupling with $J \approx 0~\mathrm{Hz}$.
Thus we can easily see that $J_{\ce{H^1 H^2}} \ne J_{\ce{H^3 H^2}}$ and therefore the nuclei $\ce{H^1}$ and $\ce{H^3}$ are magnetically non-equivalent.
The same argument can be made for $\ce{H^2}$ and $\ce{H^4}$.

‡: They are also transformed into each other by the $\sigma_{\mathrm{v}1}$ plane. However, planes of symmetry create enantiotopic protons while rotations create isotopic ones. In this case — a non-chiral environment and a non-chiral solvent — enantiotopic protons cannot be distinguished and give a common signal, and thus are sometimes incorrectly called chemically equivalent. The argument is not of relevance here though, since the $C_2$ axis renders the protons isotopic thus equivalent.

Nice answer! One small point, I agree that $\ce{H^1}$ and $\ce{H^3}$ are chemically equivalent due to the presence of the $C_2$ axis; if only the $\sigma_{\mathrm{v}1}$ plane existed wouldn't the two protons be enantiotopic instead of equivalent?

It isn't a "hunch" but a "twist" or "quirk" in the definition. You need to also explain that if the two chemically equivalent H atoms were also magnetically equivalent then there would be just two peaks in the H NMR where each peak was for two protons. Rather the observed spectra is for two sets of peaks where each set has a value of two and each set is split into multiple lines. $ce{H^1}$ and $ce{H^3}$ are each contribute half in each of the multiple peaks in one of the sets.

@MaxW About that first word, feel free to edit in a better one. About the rest: I disagree. It’s not like we could say one half of the peak is $\ce{H^1}$ and the other half $\ce{H^3}$. Those two signals are overlaid together into a single peak much like the three protons of a methyl group give a single peak.

@Jan  Not to belabor the point, but two items are either equivalent, enantiotopic or distereotopic.  In your 1-iodobutane example the proton pairs on the same carbon are enatiotopic, they are not equivalent.  In a chiral solvent they could display different chemical shifts and different reactivities.

@ron Thinking about it again, you are correct. I hope I have re-edited my post to reflect that.

I'm confused at this point. A NMR can be found here. http://www.hanhonggroup.com/nmr/nmr_en/RB03060011.html The multiplet at 8.180 is for two H atoms and the multiplet at 7.528 is for two H atoms. Within each peak of the multiplet at 7.528 each "B" hydrogen contributes half to each peak. (I don't understand why the multiplet is split into 9 peaks, or why the 8.180 multiplet is split into 11 peaks.)

@MaxW If you take a macroscopic look at that spectrum, I see a pair of doublets. Each doublet is then further (and complexly) split as all aromatic signals are. At least two minor peaks around each main peak should be $\ce{^{13}C}$ satellites, too.


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How would you make/produce sulfuric acid with tools only the romans had?
I have been reading about sulfuric acid and its other compounds the romans had, yet I don't understand how they where getting sulfur rich compounds and how they convert it to sulfuric acid. Was it from the volcanoes around them, or was it from sulfur rich rocks/mineral deposits?

There are several naturally occurring sulfate minerals such as copper(II) sulfate (pentahydrate mineral name: chalcanthite) that will decompose into $\ce{SO3}$ when heated to a few hundred degrees, which produces sulfuric acid when dissolved in water.
These compounds were known as vitriols due to their glassy appearance and sulfuric acid was known as oil or spirit of vitriol since it was derived from these minerals.
Chalcanthite seems to be a natural oxidation product of copper deposits.
Copper(II) sulfate in particular was known as Roman vitriol, but I don't know if this means that that was the primary source for the Romans.
In any case, ancient civilizations did not have any means of mass producing sulfuric acid and industrialized processes involving sulfur dioxide from the combustion of elemental sulfur arose much later.

authors: عبسي الطرابلسي


text:
Increase in Yield of products in an exothermic reaction
Given an exothermic reaction $\ce{N2 + 3H2 -> 2NH3}$ (which was initially in equilibrium). The temperature is then increasing with time. I am supposed to predict the yield of ammonia with time.
According to an exothermic reaction, the equilibrium constant K is supposed to decrease with increase in temperature. But it is found that the yield increases initially before decreasing continuously as expected. What causes the initial increase in the yield?
I suppose that the heat overcoming the activation energy could provide an explanation, but is there any other concept which is not fully reliant on the kinetics of the reaction?

Can you provide the reaction?

Reaction: N2 + H2 -> NH3

@A.K Can you answer now? I'm still not clear with this question.

No your question is on hold you should rephrase it and elaborate to show more effort to understand the material as the hold states.

@A.K. Is this fine now?

I'm sorry again about the phrasing. It may sound like that of a homework question,

Look, you say: "which was initially in equilibrium". Then the temperature are stable. In the reaction you have four mols in the reactants and 2 mols in the products. When you increase the temperature the system pressure increase too.
So to increase the stability the system "walk" for the reactants, because this decrease the pressure (2 mols in products and 4 mols in reactants).
If you remember the gas ideal law, the pressure is directly proportional to the mol numbers. So this is the reason because the temperature favor the products.

Thank you for commenting on this post. How can you conclude that the pressure increases? No constraint is given to the volume of the system. And I was asking why the yield increases initially, which I suppose Le Chatelier's principle cannot explain.

The reactants and products are gases, if the system stay open they will travel across the Universe. :)
So, the system must be closed. If the system is closed a increase in the temperature increase the pressure.
So the system will choose a condition where the gas mol number be smaller.

What about variance of K with temperature? How can you say that yield increases? The yield is supposed to decrease for a reaction where heat is released. I am confused. Thanks again.

For example, let's think that K = 1 (1 atm) when you increase the temperature the reaction goes to a new equilibrium point.
Using PV = nRT  (R and V are constant). If the temperature increase, P increase and n decrease. If the n decrease that means the reaction will go to products (because in the products the mol number of gas is smaller in relationship to the reactants).
Now, at the new temperature we have a new K, the equilibium was displaced. The must be  K' > 1.

Oh, and the Le Chatelier's principle can explain this situation. The "system response" to temperature increase is: displace the equilibrium to te products for decrease the pressure.

There also exists a relation between K1 and K2 at 2 different temperatures, that is the Vant Hoff equation. What about the enthalpy of the reaction? It's also a factor, which does not associate with Le Chatelier's principle. I think simply using the ideal gas law will not work in this situation.

To answer your question in short, according to Le Chatelier's Principle the endothermic(reverse) reaction will be favored. So from the equation, it is clear that the reverse reaction produces a greater number of moles as compared to the forward one. Therefore because the amount of moles of a compound is directly proportional to the mass of that compound  then the reverse  reaction will produce more yield that the forward reaction which will produce less yield.Thus less ammonia will be produced. Hope this helps.


text:
Synthesis of 2-benzylcyclohexan-1-ol from cyclohexanone
There are no conditions given and a mechanism is required.
I first thought it could be along the lines of Claisen rearrangement but that needs an aromatic ring. Since the Claisen rearrangement can't be used I considered a ($\mathrm{S_N2}$) reaction of benzyl bromide by the alpha carbon followed by reduction of the carbonyl.

Ok so I found an answer:

Deprotonate at the alpha carbon using LDA which gives a negative charge.
This then attacks $\ce{BrCH2Ph}$ at the $\ce{CH2}$ and kicks off the bromine in the process ($\mathrm{S_N2}$).
Then to form the alcohol the ketone is reduced using $\ce{LiAlH4}$ using THF as a solvent.

Alkylation of an enolate is not a reliable reaction. You have to be concerned about over-alkylation and O-alkylation. There are better approaches to this target.

How would one prevent over alkylation? Should I have introduced a protecting group first like a ketal for the carbonyl? How would I protected the other alpha carbon? @jerepierre

You can treat cyclohexanone with dimethyl carbonate (THF, NaH) to give the anion of a beta-ketoester. Now alkylate with benzyl bromide followed by aqueous acid hydrolysis and decarboxylation. Extra step but more selective than direct alkylation. One might consider an enamine alkylation.

Making a carbon-carbon bond at the alpha position of a carbonyl suggests using some kind of enolate chemistry. In contrast to an SN2 type reaction with an alkyl halide (which can give over-alkylation and competitive O-alkylation), an aldol condensation is a reliable way to construct the requisite C-C bond. Here cyclohexanone reacted with benzaldehyde under basic conditions gives an enone. Two subsequent reduction steps, hydrogenation followed by hydride addition to the ketone, give the target compound.

However you would probably have to be concerned about dialdol here.

@Jan Yes, I would... SciFinder had lots of examples of mono-aldol, and there's probably enough of a reactivity difference between cyclohexanone and the mono-aldol product to get good enough selectivity here. Way better than alkylation at least.

Hmm, wouldn't NaBH4 probably produce the saturated alcohol directly from the α,β-unsaturated carbonyl?

@orthocresol You might be right... I thought NaBH4 was pretty selective for the ketone, but quick Google searches support what you're suggesting. Looking thru SciFinder for similar transformations, it looks like more commonly people reduce the ketone first with hydride (NaBH4 or LAH) and then follow up with hydrogenation. I'm guessing that hydride will give a mixture of the fully reduced and the allylic alcohol. In practice, you're right. Try NaBH4 first, and if it goes all the way, great. If not, hydrogenation is easy and will finish it off.

...or hydrogenate enone over PtO2 or Ni(R) to get double reduction.


text:
What does A485 mean? And what does the values on the y-axis expresses?
This question is from past exam paper. I don't want an answer to the multiple choice question below, I just want to understand what A485 means in the context of the question.

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

I did try the question but I am not getting what is A485 and 485nm, how this is related to this question and what are the steps to be taken to get the answer.

Then do as instructed; put in precise details of everything you did. Then I'll vote to reopen your question so it'd get answered.

Am I okay now, sir?

What did you do that didn't get answers? Enlighten us by [edit]ing your question.

I got it. I think, I don't need to edit the question. May be, if I ask next time any question I will try to ask according to the instruction.

@Ahmad For further reference, please also keep the following in mind: A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors. Further I invite you to take the [tour] to learn about the site and review the homework policy.

$A_{485}$ means the same thing as the text of the multiple choice question "absorbance of the solution at 485 nm".
"nm" is an abbreviation for nanometers.
In other words, the absorbance of light having a wavelength of 485 nanometers is plotted on the y-axis of the graphs.


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Successful predictions of MO software?
There have been many successful predictions of MO software. Is there a website or journal dedicated to recording them all?

No. There are far too many and it wouldn't really be very useful.

I'd expand bon's answer a bit. MO models are used to predict some particular behavior of a molecule rather that how the molecule would behave under all circumstances. So for example it is just too complicated as of yet to "seamlessly" model NMR, IR, UV-Vis, mass spec, boiling point, melting point, and chemical reactivity of a molecule in the solid, liquid and gas phases in a single MO model.

@MaxW is there a table associating which models describe certain phenomena - ideally under stated ranges of conditions?


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How to tell which species has the highest ionization energy?
I am preparing for my final exam, and I am very confused about ionization energy. An example question would be:

Between the species $\ce{Ne, Na+, Mg^2+, Ar, K+, $\&$~Ca^2+}$, which one has the highest ionization energy?

I thought that ionization energy increased from left to right in a period and from down to up in a group in the periodic table, so I thought that $\ce{Ne}$ would be the one with the highest ionization energy. But the right answer is $\ce{Mg^2+}$. I assume that this has something to do with the electrons, but I don't know what.

Take a neutral atom. Remove one electron. That took some energy. Now remove a second electron. Does that take more energy, or less energy than to remove the first electron?

Those aren't compounds.

Like most chemical conundrums you need to simplify the problem using some knowledge of chemistry.
First there are two electron configurations. $\ce{Ne, Na+}$ and $\ce{Mg^{2+}}$ all have the configuration ${1s^2 2s^22p^6}$. $\ce{Ar, K+}$ and $\ce{Ca^{2+}}$ all have the configuration ${1s^2 2s^22p^6 3s^2 3p^6}$.
For $\ce{Ne, Na+}$ and $\ce{Mg^{2+}}$, $\ce{Mg}$ has the highest atomic number so of these three, it would be most difficult to ionize $\ce{Mg^{2+}}$.
For $\ce{Ar, K+}$ and $\ce{Ca^{2+}}$, $\ce{Ca}$ has the highest atomic number so of these three, it would be most difficult to ionize $\ce{Ca^{2+}}$.
Now it is just necessary to compare the ionization of $\ce{Mg^{2+}}$ to $\ce{Ca^{2+}}$. Since the $2p$ orbitals are closer to the nucleus than the $3p$ orbitals, it will be harder to ionize $\ce{Mg^{2+}}$ than $\ce{Ca^{2+}}$.

Axiom 1: Ionisation energy increases from left to right along a period.
Axiom 2: Ionisation energy decreases from top to bottom along a group.

Both of these are often or mostly correct. However, there is more. Remember that the electron you are extracting carries a negative charge ($\ce{e-}$). So:

Axiom 3: Ionisation energy increases dramatically for every positive charge introduced.

And also, to state the concept of core versus valence electrons:

Axiom 4: The first ionisation energy of a core electron will be dramatically more than that of a valence electron.

In your case, you have six atoms to compare which all have core electrons only (if you count the fully populated shell of the noble gases as core electrons which isn’t strictly correct but close enough). Thus we do not need to consider axiom 4. Axiom 3 tells us, that $\ce{Mg^2+}$ and $\ce{Ca^2+}$ are the only contestants for maximum ionisation energy since they both have the highest charge. And comparing the two according to axiom 2, we arrive at $\ce{Mg^2+}$ having the highest ionisation energy of these atoms.

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Why does treatment of a geminal dihalide with sodium amide lead to an alkyne and not an allene?
I've come across the following problem in Klein's Organic Chemistry, 2nd edition:

However, I'm a little bit confused about why the terminal alkyne is produced. I believe that an allene should be formed via stepwise elimination of HBr, using the strong base NaNH2. The first elimination should give the more substituted, more thermodynamically stable alkene (Zaitsev's rule):

Why does the former occur instead of the latter?

monosubstituted acetilenes are deprotonated by sodium amide, allenes are not. Since alkenes and allenes can converse to each other in amides via deprotonation, typically even if you put a nonterminal allene or alkyne into sodium amide solution, it transforms into terminal alkyne.

I think the first proton is abstracted from the primary carbon due to the following reasons: Primary is less sterically hindered and hence is easily accessible by $\ce{NH2-}$ ion. Also, even a partial negative charge on carbon in the transition state of the reaction is better off on primary carbon.

Sodamide is bloody dangerous.  If your bottle is too old, it likely has begun to oxidize some and you have potential azides or other polynitrogen compounds that may lead to an explosion.
Even with good stuff, generally the reagent is not heated to avoid similar complications.  This means normally this reagent works at low temperatures (often times dry ice/acetone).  Under these conditions kinetic products are favored over thermodynamic ones.
The exomethyl group is much more accessible for deprotonation, and so kinetically, I would expect the amide ion to abstract the terminal hydrogen.

After that, the resulting bromoalkene still has the possibility of forming the allene product, but the transition state resulting in elimination to the alkyne is favoured as the vinylic hydrogen and the bromine are always locked in an anti conformation due to alkene geometry, as indicated in the above diagram. On the other hand, the allylic hydrogen may not necessarily be anti to the bromine due to C–C bond rotation.

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If chemical bonding is quantum mechanic, is saying they bend and vibrate like springs wrong?
I don't know much about quantum mechanics, but I do know that it's quite far removed from the easily observable and that anything "quantum" involves things that are discreet. So, how close is the analogy between molecular bending and vibration with that of springs? Is there maybe a better way to think, maybe a better analogy, to describe molecular bonds?

The "nasty" fact is that there isn't any one model that creates an adequate analogy for chemical bonding. What model to pick depends on what phenomena you're modeling. So the spring model is useful in thinking about infrared spectra, but useless in pondering NMR spectra. Just adding "springs" to atoms though doesn't describe spring lengths or coil tension either. So "physical models" must be used with care.

I am reminded of this (unfortunately true) xkcd comic.

@NicolauSakerNeto Those moments when you realise that a flaw of a visualisation was directly on the tip of your nose for years but you didn’t see it … Oh the pain ^^'

That's a good question, but remember, an analogy is just that: comparing dissimilar things on the basis of structure or function. Though there may not be a physical similarity, the analogy is effective in describing what happens, and can even be used to calculate outcomes, within limits.
In a similar way, the concept of a mass and a spring are used to describe an electrical R-L-C tuned circuit. Both have resonant frequencies and other properties that can be calculated from the physical properties of the components.
One of the difficulties in describing quantum phenomena is that until the physical properties are measured, they may be indeterminate. An electron is not in a specific place, but has a probability of being detected in a specific place. So though the analogy is not physically comparable, it helps understanding of the chemical bond, and can even produce useful results, such as IR spectra, and its use is valid.
This begs the question, though; perhaps there is another analogy that you might find more useful... Any suggestions from others are welcome!

I think there are two important points I would add to the good answer presented already. First is that this idea of modeling molecular vibrations as springs has been used to describe quite complex things in a fairly simple way. So, to some extent it is a good analogy. One example of this is an idea that Einstein actually came up with to find the heat capacity of solids. He treated every atom in a solid as a spring system which, as you point out, had to vibrate at discrete frequencies... and that's pretty much it. Using that idea alone he came up with an equation which predicts the heat capacity of solids quite accurately (gets worse at low temperatures). You can read about Einstein solids at that link if you want. It's kind of dense though...
One other thing is that molecular vibrations are clearly more complex than just a simple harmonic oscillator, but, as it happens, one can make corrections to the simple harmonic oscillator and end up with quite a good description of reality.
Specifically, the energy of a vibration is described by:$$E(v)=(v+\frac12)\omega-\omega_{xe}(v+\frac12)^2$$where $v$ is the vibrational energy level (things are discrete as you point out) $\omega$ is the vibrational frequency in wavenumbers, and $\omega_{xe}$ is the anharmonicity of the vibration. The interesting thing is this correction to the harmonic oscillator which the second term makes can be further expanded so that one has an infinite series which converges to the exact energy value of the vibration. So, in the sense that this can give the correct answer and is built on the idea of a simple harmonic oscillator, a spring is very much how atoms vibrate. Hope that is helpful.
That above equation coms from the Morse potential
Here is a picture comparing the two


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How are the boiling points of Tungsten and other metals determined and proved?
The boiling point of Tungsten is 10,030 degrees Fahrenheit. How was this determined and proved? And more generally how are the boiling points of metals determined and proved?
Is it really so simple as "heat it up as high as you can and see if (and at what temperature) it evaporates"?
I feel like there is a more sophisticated process, but I can't really guess at what that might be, especially if the boiling point was determined years and years (100+?) ago.
The Internet is really good at telling me the boiling point of a lot of (or all, I haven't exhaustively checked) elements and alloys, unfortunately it seems it's not very good at telling me how these seemingly arbitrary temperatures were determined.

The melting (and boiling) points of elements are certainly not arbitrary - they are well defined thermodynamically. To find them, yes, you heat things up until they melt, measuring the temperature along the way. Ther is nothing particularly hard about it. In fact, the temperature gets easier to measure as things get hotter.

I realize they're not completely arbitrary. I'm sure there are number of factors that affect them so many in fact that perhaps they may seem arbitrary to someone who hasn't studied chemistry, or physics in such depth.

But that seems too easy. I'd like to know details of the process as well.

You have to understand that there are dozens of ways to measure temperature. The particular technique used depends on many factors, including how accurate a determination is needed. You could easily earn a PhD refining an element and measuring the melting point more accurately.

According to the 1913 paper THE VAPOR PRESSURE OF METALLIC TUNGSTEN, The Physical Review, vol. II, pp. 329-342:

The vapor pressure (p) of a pure substance is related to its evaporation rate (m) in a vacuum by the following relationship:
$$ m= \sqrt{\frac{M}{2\pi RT}}p$$

Evaporation rates were measured at temperatures in the range from 2400K to 3100K, heating by electric resistance of a filament, and determining the evaporation rate by weighing the filament before and after the evaporation.
The temperature was determined by observing the color and intensity of the light emitted by the filament. (Paper says they will publish a separate paper about the details of the temperature determination technique).
Vapor pressure was extrapolated to a boiling point of 5110 K using the Clausius-Clapeyron relationship.

That's the answer I was looking for! Thanks Dave.

You certainly cannot measure >6000 K (Kelvin I am using SI units here) using a thermocouple because your probe would instantly evaporate itself. However, you can use the electromagnetic radiation spectrum to measure the temperature of the element. The radiation spectrum of a black body radiator is very well understood and can be measured using a spectrophotometer. Once you have the radiation specrum, you can use Planck's law of black-body radiation to calculate the temperature.
What you then need to take care of is that you melt/evaporate the element of interest and not the container and that you do not burn the element instead of vaporizing it. In case of tungsten, this can be achieved in a setup similar to a xenon lamp. A xenon lamp is nothing else than a glass container filled with inert xenon gas in which a tungsten wire is suspended and electrically connected. By carefully measuring the electromagnetic spectrum while increasing the voltage until the tungsten wire melts you should be able to measure the melting temperature. I am not sure if this also works to measure the evaporation temperature because the wire might disconnect from your power source once molten (commonly referred to the lamp being burnt out) It may be possible to shape the wire such that it does not disconnect from your power source while melting and that the heat dissipates fast enough at the electrical connections to avoid damaging those. Alternatively (with todays technical means) one could use a laser to heat the tungsten until it evaporates.


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opperating a distillation column far from flooding
I know that flooding will ruin the performance of a distillation column. There are many correlations (Eckert, kister, billet, etc.) that will tell you when a column will flood based on your fluid conditions. Most design guides suggest choosing a diameter that will lead 70-80% of the flooding condition. This makes sense as the size of the column (IE, the cost) is as small as possible while still giving you a 20-30% margin of error.
However, what if due to other constraints you want the column to be large so that it is only operating at 10% of the flooding conditions (IE, a much larger diameter than normal). Would this column still operate? Is this type of operation just avoided because it is not economically practical and a smaller column would do the same thing operating closer to flooding?

Using a much too large column means you loose a lot more heat through its walls, meaning it will take longer for the equilibrium temperature/concentration gradient in it to form, and the destillate comes not only later but also considerably slower.
And you'll have to put more power into heating the still, with following danger of decomposition, unwanted side-products, etc.
If you use an expensive column with silvered vacuum insulation, this effect might not be too large even at only 10% of the maximum capacity. I wouldn't want to judge without trying, though.
Education and experience taught me to always use the proper type and size of equipment. It's faster, cheaper, and gives higher yields. I you have a large column that costs thousands, you can sure afford an additional smaller one at a tenth of the price, can't you?

Good answer. The issue is I am using a diabatic column so the vapor flow rate changes significantly over the height of the column. The average flooding condition is 30%, the max is 90%, and the min is 10%. Its pretty linear over the entire height of the column.

So you can imagine the issue. Either half the column is overdesigned or we have a shrinking diameter over the column. Our cost is pretty low in either case so it seems simpler to just have it be overdesigned for half. But I want to make sure it will still be functioning in that realm.

Help me out, what is a "diabatic" column?

diabatic just means that the heat is added at each tray instead of all at the bottom. Here is a nice link: http://www.sci.sdsu.edu/~salamon/OptDiabDist.pdf

Ah. Well, the bad disadvantages mentioned above don't apply then, but i guess a sort of "conical" column would still deliver your destillate initially faster , and at generally lower energy consumption. Should be easy to calculate the flux here, with constant temperatures everywhere. If you custom build it anyway, why no go for the optimum?

Not a bad idea. Just harder to make. I guess we will see how much more it would cost and then go from there. Thanks for the help!


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Concentration of Iodide Ions in water

You add 10.50 g of solid lead(II) iodide to 525 mL of water. Assuming that the lead(II) iodide does not add to the volume of the solution, what is the concentration of iodide ions in solution?

Here’s what I have done:
$10.50\ \mathrm g\ \ce{PbI2}\cdot \frac{1\ \mathrm{mol}\ \ce{PbI2}}{461.00\ \mathrm g\ \ce{PbI2}}\cdot \frac{2\ \mathrm{mol}\ \ce{I-}}{1\ \mathrm{mol}\ \ce{PbI2}}= 0.04555\ \mathrm{mol}\ \ce{I-}$
$\frac{0.04555\ \mathrm{mol}\ \ce{I-}}{0.525\ \mathrm L\ \text{solution}} = 8.68\cdot 10^{-2}\ \text{Molar}\ \ce{I-}$
What went wrong?

I think you have to consider the solubility rules and determine how much of the lead (II) iodide will actually dissolve in that given amount of water rather than assuming all of it dissolves.

Yep. I think it's insoluble, so the concentration would be zero?

No, it's not completely insoluble, it's just after a certain point, it cannot be saturated any further. Try searching solubility problems regarding Ksp to better understand the problem, try this video: link.


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How to identify cis-trans alkene isomers?
Is the alkene with the representation $\ce{CH3CH=CHCH2CH3}$ a cis-trans isomer? If so, why? I mean, it is not like the other cis-trans alkenes, because all of them were symmetrical somehow?

First of all, let's talk about geometrical isomerism from the beginning, as I think that you don't understand it properly.
Geometrical isomerism is different spacial arrangements of atoms or groups across any bond under the following conditions:

Restricted rotation about a bond in the molecule.
Both, substituted atom or group on each C-atom, about which rotation is frozen or restricted should be different.
In case of cumulenes, odd and even number of double bond and the distance between the atom or group should be taken into consideration.

Now coming to your question,
$\ce{CH3-CH=CH-CH2-CH3}$ can show geometrical isomerism because it follows conditions 1 and 2 given above. It can therefore form cis and trans isomers.
Which of them will be the major product depends upon the stability of the compound.
As you can see, in the given conditions there was no mentioning that the compound has to be symmetrical to form cis-trans isomerism. You don't have to look for symmetry, but for the above conditions.
I mentioned earlier that geometrical isomerism is a different spacial arrangement of atoms or groups across any bond, se let me give some examples for that:

Compounds containing $\ce{C=C}$, $\ce{C=N}$, $\ce{N=N}$
Compounds containing cyclic structures, homocyclic compounds (benzene type), heterocyclic (eg. pyridine), fused rings (eg. naphthalene), etc.
Compounds  containing restricted rotation about single bonds due to steric hindrance. Attached groups are so bulky, such that on rotation repulsion increases thus, decreasing the stability.

Yes, the formula could be for either:
cis-2-pentene

trans-2-pentene

Sorry but I don't get it. I can't see the difference could you please clarify more ?

The trans-2-Pentene isn't drawn quite the same way as the cis version. Redraw it with the two H atoms and you should get it. see: https://en.wikipedia.org/wiki/Cis%E2%80%93trans_isomerism

I'll add that the $\ce{-CH3}$ group can spin about the single $\ce{C-C}$ bond, but the $\ce{C=C}$ is "locked." So the hydrogen atoms are either on the same side (cis isomer), or opposite sides (trans isomer) of the $\ce{C=C}$ bond.


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What would the name and formula be for Mo+O+Se?
I do cross-stitch and want to spell the word "moose" (for a project focusing on the animal) with squares from the periodic table and present a compound formula and name that encapsulates all three elements, Mo+O+Se, in that order. I'm not looking for strictly realistic chemistry, just a formula and a name that would work IF the compound had actually existed. The number of atoms of each element is up to you. Can you help me with a formula and a name for Molybdenum + Oxygen + Selenium?

I’m voting to close this question because this is not a Chemistry question and pertains to wordplay which doesn't fit in this site.

Well, molybdenum oxysulfide is apparently a real compound with the formula $\ce{MoOS2}$ or $\ce{MoO2S}$.  Since selenium is a chalcogen and has (some) chemistry that is analogous to sulfur, I think molybdenum oxyselenide would be an appropriate choice.
More formally:

Molybdenum(IV) oxide is a compound with the formula $\ce{MoO2}$.
Molybdenum(IV) selenide, also called molybdenum diselenide, has a formula of $\ce{MoSe2}$
Molybdenum sulfide selenide exists and has the approximate formula $\ce{MoSSe}$.

Thus, perhaps molybdenum(IV) oxide selenide would be another choice.

Wonderful, Curt, thank you! And would that simply make the formula MoOSe? (Please forgive my cluelessness!)

So, Curt F., with the edits to the original answer, can I assume that I can go with at least one of the following three combos? Molybdenum oxyselenide (MoOSe), or Molybdenum dioxyselenide (MoO2Se), or Molybdenum oxydiselenide (MoOSe2)? (And again, please forgive my cluelessness!)

Material Scientists and Engineers would likely use the latter  molybdenum(IV) oxide selenide in publication so I would go with that.

And what would the formula for that be?

The formula would be MoOSe.  Isn't that what you were asking for?


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Products of radical bromination and mechanism
My organic chemistry final is next Tuesday, and I have worked through the practice final. The last question is the following: Give and name all the compounds resulting from the radical monobromination 2,3-dimethylpentane. Indicate the major and minor products.
I drew:

My prof says that I am missing my R/S. I can clearly see my chiral centres, but how do I determine the orientation of the H? Red is the major product, blue is the minor product, and magenta is the very minor product.
The second question is: write the mechanism of this reaction identifying the initiation, propagation, and termination steps for the formation of one of the products. This is what I drew:

My prof says that I am missing a propagation step with another bromine-bromine molecule.
EDIT:

My prof says that I am missing my R/S. I can clearly see my chiral centres, but how do I determine the orientation of the H? Red is the major product, blue is the minor product, and magenta is the very minor product.

You don't determine the orientation of the H you just acknowledge the formation of both the R and S stereo isomers in your products in 50/50 proportions.

My prof says that I am missing a propagation step with another bromine-bromine molecule. I would greatly appreciate your input!

All you need to show is the organic radical reacting with diatomic bromine to form you brominated molecule and another bromine radical.

Thank you very much! That's what I thought. So I simply mention the 50/50 ratio where appropriate (i.e. where there is a chiral carbon), right?

That's all.  Your starting material is chiral, and all of the products will chiral.

Thanks! I have edited my question: I can see it now. The molecules have either 1 or 2 chiral centres.

Not quite.  do 2-bromo-2,3-dmethylpentane and 2-bromo-3,4-dmethylpentane have 2 chiral centers or 1?

A.K. Oups! 2-bromo-2,3-dimethylpentane seems to have 1 chiral center(2), ,but  2-bromo-3,4-dimethylpentane seems to have 2 centers... Am I forgetting something? :)

No, you're good now.  I just forgot that a chiral center was formed on 2-bromo-3,4-dimethylpentane.

Let us continue this discussion in chat.


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Effect on the partition function upon the mixing of two ideal gases

Consider a box that is separated into two compartments by a thin wall. Each
  compartment has a volume V and temperature T. The first compartment contains N
  atoms of ideal monatomic gas A and the second compartment contains N atoms of
  ideal monatomic gas B. Assume that the electronic partition functions of both gases
  are equal to 1. The molecular partition function for each component is given by
  $$q_i = \frac{V}{Λ_i^3}$$

Firstly I am asked to write the total initial canonical partition function which is given in the question as $Q_{initial}=Q_AQ_B$. The algebra is fine, but why would the two separate gases have a total canonical function?
I was then asked to show that after the gases mix (and do not react) that the following is true: $$\frac{Q_{mixed}}{Q_{initial}}=4^N$$.
I got that $Q_{initial}=\frac{q_A^Nq_b^N}{(N!)^2}$. Therefore, the only reasonable expression for the mixed function is:
$$Q_{mixed}=\frac{(2q_A)^N(2q_b)^N}{(N!)^2}$$
However, I cannot see why the individual molecular partition functions will have doubled. Especially since A and B are distinct gases.

The fact that $Q_{\mathrm{initial}} = Q_{A}Q_{B}$ implies that the two subsystems, respectively consisting of gas A and gas B, are independent. Because $\log Q_{\mathrm{initial}} = \log Q_{A} + \log Q_{B}$, it follows that various extensive thermodynamic quantities (e.g., internal energy, free energies, entropy) of the total system are simply given by the sum of the corresponding quantities of the subsystems.
The partition functions are doubled because each gas occupies double the initial volume after mixing, i.e., $V \rightarrow 2V$.


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How does pH affect the degradation of ascorbic acid (vitamin C)?
I know that vitamin C degrades over time (as I understand through oxidation). However, I am having trouble understanding how changing the pH of the solution in which the vitamin C is affects the rate at which the vitamin C degrades.
Among others, Roig et al. [1] demonstrated that there is indeed a change in degradation based on pH.
References

Roig, M. G.; Rivera, Z. S.; Kennedy, J. F. A Model Study on Rate of Degradation of L-Ascorbic Acid during Processing Using Home-Produced Juice Concentrates. International Journal of Food Sciences and Nutrition 1995, 46 (2), 107–115. https://doi.org/10/fgwc5c.

see http://pubs.acs.org/doi/abs/10.1021/jf9805404

How does pH affect the electron densities of various reactive centres of the molecule?

If you draw the structures of the ascorbic acid and the ascorbate, and if you analyze, what is easier to oxidate, and what is the oxidation product, it should be obvious, why it is pH dependent, considering AA is a weak acid.

Ascorbic Acid is a known antioxidant molecule and its stability has always been an issue especially for pharma industry which often makes use of AA in formulation. Now as much as pH is a culprit in AA degradation, its only one among other factors like solvent, exposure to light/uv, and the nature of solution (viscous/non-viscous). There are quite a lot of studies on stability of Ascorbic Acid like this one
I normally prepare Ascorbic Acid freshly for each experiment, but it would be an interesting experiment to check the stability of AA for an extended time period at different pH values.
Here is an interesting simulation demonstrating degradation of Ascorbic Acid.

Thank you for your answer, and welcome to the site! Are you sure your answer addresses the question's fundamental inquiry (How is AA degradation affected by pH)?

A comment above cites a publication claiming a role for the hydrogen ion in the decomposition of Vitamin C.
I would agree especially if a low pH fosters a Fenton/Fenton-like chemistry. The latter could proceed in the presence of transition metals impurities, which can engage in Fenton chemistry liberating hydroxyl radicals, directly attacking Vitamin C and consuming H+. The chemistry is further promoted in the presence of light/UV (Photo-Fenton).
Also, a role for oxygen leading to reactive oxygen species (like H2O2 which can feed a Fenton reaction).
Dust particles can be rich in metal oxides and air exposure can also be a source of water vapor. So just frequent opening of a container may accelerate decomposition.
Also, elevated temperatures, in general, could impact redox reaction rates, so cold/dry storage may assist in limiting the impact of a reduction of pH on Vitamin C in the presence of metal impurities and exposure to air/water/dust and possible elevated temperatures.


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Is there a way to experimentally measure entropy?
I've been doing thermodynamic chemistry, and recently focusing on Gibbs Free Energy. Whilst doing calculations using, $$\Delta G = \Delta H - T \Delta S$$ I have been given a table of values for $\Delta H$ and $\Delta S$, and a temperature to work with. I was wondering , as the unit for entropy is $\mathrm{J/K}$ or $\mathrm{J\ K^{-1}}$ how exactly one would experimentally measure entropy/change in entropy, or can you?

The most common way of measuring $\Delta S^\circ$ for a chemical reaction is probably by making a van't Hoff plot. You measure the equilibrium constant $K$ at different temperatures and plot $\ln K$ vs $T^{-1}$.  The $y$-intercept = $R\Delta S^\circ$ and the slope = $-R\Delta H^\circ$.
Another option is to measure $\Delta H^\circ$ by calorimetry and measure $K$ by some other means. Then compute $\Delta G^\circ$ from $K$ and solve for $\Delta S^\circ$

Thank you. Very helpful answer, I've done calorimetry before so I know how to go about this method.

This might be a bit unconventional. M.A.Ramezani found an user with your name on it: Jan Jensen. If you recall creating this unregistered version of yourself, you can find out more about merging these accounts.

The entropy change between two thermodynamic equilibrium states of a system can definitely be directly measured experimentally.  To do so, one needs to devise (dream up) a reversible path between the initial and final states.  Any convenient reversible path will do, since the integral of dq/T is the same for all reversible paths.  So you have to identify a path that is easy to implement, and for which the heat flow can be measured easily (say by phase change in a reservoir).  The experiment has to be carried in as close to reversible conditions as possible, since, in the real world, there is no perfectly reversible path.  The hard part would be measuring the amount of heat flow.  In some specific cases it could be done indirectly, such as in the isothermal quasistatic expansion of a gas (say in contact with a heat bath), where the measured amount of work (determined say by gradually removing small weights from a piston) would be equal to the amount of heat added.  Of course, in many cases, it is much easier to calculate the entropy change (based on identifiable reversible paths) than it is to measure.  In my judgement, devising a method to cause the heat flow to occur gradually and quantitatively measuring the heat flow directly would be the hard part.

That's very interesting! Do you have a link that explains how to have a chemical reaction done reversibly and with heat measurement?

@JuanPerez Are you familiar with the vast Hoff equilibrium box concept?  Or, are you familiar with how to carry out an electrolytic reaction by controlling the imposed potential difference so that the reaction occurs reversibly?

Right! An electrochemical reaction can be controlled, makes sense. Never heard of vant hoff box however.

A van't Hoff equilibrium box is a clever conceptual device for carrying out a gas phase reaction reversibly, starting with pure reactants and ending with pure products.  Google it, or see Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics.

Entropy is a measure of randomness.
The definition itself is hard to comprehend at times. And when you go to think, about how one would measure randomness, you get no answer. The term Entropy was introduced as a State Function to check whether the reaction is possible or not.
Well there are formulae to determine the change in entropy. However as far as I have read, all books say that calculating the absolute value of entropy is quite difficult.
According to me if you know the values of Entropy at Standard States you can calculate the Change in entropy for a process from STP to the temperature at which you want its absolute value and then just subtract to get the absolute value. The Standard State entropies are mentioned in quite a few books.
Hope the answer helped.:-)


text:
What is this glassware?
I found this in a second hand store, and I think it has to do with freeze drying. Is there any validity to this assumption? What is this called? Also, the little tuberculina at the bottom are stuffed with fiberglass I believe.

It does look like glass wool plugs in the tubes to trap and liquid droplets which might be formed, so the present purpose use for freeze drying seems reasonable.  But the marks on the tubes would have no purpose in  freeze drying.  In the good old days, every chemistry lab had a glassblower on call who could repair glassware or build specialty glassware. I'd guess that this configuration was built out of some other manifold. A glass blower was expensive, so having something factory made as a starting point would probably be cheaper than building the whole thing by hand.

It looks to me like tubes used to distribute gas flow.
If you work with substances that would react with the oxygen in air, you prevent that by working under an inert gas such as nitrogen, i.e. you fill all the flasks containing your substance with nitrogen gas, keeping up a steady flow of the nitrogen in order to prevent any from air leaking in.
The nitrogen is usually provided through a valve in the fume hood, and then distributed through the tubes you have there. The left end in your picture is connected to the valve, and on the ends on the right side you can attach tubes leading to your reaction flasks. The fibreglass is there to prevent air from getting into the nitrogen distributing system while there's nothing attached to the tube.
The ends of the glass tubes look weird to me though, so I might be entirely wrong.

This was my thought as well.  A google image search for gassing station, a term I've heard microbiologists use to describe such a setup, shows several informative pictures of setups that have components similar to this.

This is a very useful answer, although I have to wonder if it is correct. Every inert gas/vacuum manifold I can find (glass) have valves, and have an input on the side of the manifold.

@iDomo this may be a more 'fix/maintainance-friendly' version with separate individual valves.  Also, if the rubber hoses have very thick walls, it was used to distribute vacuum, not gas flow.

I doubt it has much to do with freeze drying. That would be a glass manifold with rubber hoses leading out with syringe barrels stuck in the ends.  The fiber glass is likely meant to either resist and equalize gas flow or to prevent liquids from boiling into the manifold.


text:
Why is the minus sign (-) not allowed in reaction equations?
Recently in my chemistry class, the teacher told us that while writing "half equations" (such as a sodium atom losing one electron to form a sodium ion) we were not to use a simple minus sign. We were only allowed to use a plus sign. For example, this would not be allowed:
$$\ce{Na $-$ e- -> Na+}$$
The "correct" way of writing it would be:
$$\ce{Na -> Na+ + e-}$$
My question is: Is this conventional, or somehow only related to my school? Could any example of a chemical equation utilising the minus sign please be given for clarification?

Don't get too caught up in notational rigidity. You're "allowed" to use any notation you want, if it gets you to the correct answer. If you really want, go ahead and use $\ce{Na\ - \  e^{-} -> Na+}$, or go all out and use $\ce{-Na^{+}\ - \  e^{-} -> - Na}$. However, as Richard Feynman discovered while learning trigonometry, if you start using too much obscure/uncommon/personal notation, you get into trouble showing others your work, which makes doing science hard. Standardizing notation is just a way to save time and headaches later on.
The most basic chemical equations actually carry very little chemical information, to the point that it's easy to write reactions which are chemically outrageous. One of their most fundamental purposes is simply to establish atom conservation (which is why so much time is spent on teaching how to "balance" reactions). Thus, chemical equations are always equivalent to a system of linear equations, which is why you can treat them like such and add/subtract molecules to both sides, multiply by a number, add two different reactions together, etc. Using negative stoichiometric coefficients would thus be a very trivial extension of this behaviour, and the equation would remain completely valid.
Let's look at an example. Take the intermolecular dehydration of ethanol to produce diethyl ether:
$$\ce{2 C2H5OH -> C2H5OC2H5 + H2O}$$
The equation is just as valid if you subtract a water molecule from each side:
$$\ce{2 C2H5OH\ -\ H2O -> C2H5OC2H5}$$
Not many people would write the above equation. However, we do have another type of standardized notation which is often used, and which involves a negative stoichiometric coefficient. Using this notation, the equation is written as:
$$\ce{2 C2H5OH ->[\ce{- H2O}] C2H5OC2H5}$$
The water molecule has a legitimate negative stoichiometric coefficient of -1, and again the equation is exactly as valid as all of the ones above. Why is it that negative stoichiometric coefficients are rarely written in the reactants or products, but common when written atop the reaction arrow? I have no idea, but ultimately it makes no difference. That's just how notation happened to become standardized, and if everyone understands it, that's fine.

Thanks ! Just one more question: is it standard (or at least common) to use the last form of equation with other things as well? For the sake of example, (using the Sodium equation once again) can I write" Na--> Na+ "(with minus electron written on top of the arrow)? Will that be accepted?

Putting things other than water on top of the arrow is very common. It is done occasionally for electrons, e.g. see this and this. However, if your teacher said not to use negative stoichiometric coefficients, then you probably shouldn't test their resolve. It's good that you questioned the requirement and have learned more in the process, but there's no need to give your teacher a hard time about it.

You can talk about negative stoichiometric coefficients with your teacher if you want, but don't try to be a rebel for the sake of it and go brazenly filling out your tests with them!

Well, that bounty went exceptionally well. May I encourage you to include your comments into the answer, that would be really neat. I'll assign the bounty when I have time to pick for a new target.

@Martin Sorry for the extremely delayed response, but thank you for the exposure. To be honest I feel like my answer wasn't worthy of so much positive attention! I'd rather not bump the question to the top again just to add the comments (hopefully SE visitors always check the comments), but I'll think about it.

We put things over the arrow generally for space saving, or to emphasise products and reactions, or ultimately to write down cycles which would look funny if closed via the common notation. In such a sense is clear why sometimes is better to write - CO2 or something above the arrow. This commented, I like the answer.

It's not a matter of what's conventional and what isn't. In real life, nothing can be subtracted or erased as is done in arithmetic. The law of conservation of mass, energy, etc applies. Thus when a Sodium Atom is Reduced to a Na+ Ion You cannot Right Na - e- = Na+ because the electron isn't disappearing, it's being ionically shared with another element via electrostatic forces of attraction.

This is a bit treating chemical equation as they would be mathematical ones. It is somehow true (what you say) but an equation as Na - e —> Na+ is perfectly readable and sensible. This is because chemical equation aren't mathematical expressions. See also my comment here https://chemistry.stackexchange.com/questions/147401/what-is-the-mathematical-meaning-of-the-plus-sign-in-chemical-reaction-equat

And see answers as well. I suggest that by Farooq.

One thing is a stoichometric equation, another is a reaction mechanism. For stoichometry, we can write an equal symbol as in mathematics
$\ce{Na = Na+ + e-}$
and this is equivalent to
$\ce{Na - e- = Na+}$
and to
$\ce{Na+ + e- - Na = 0}$
This last expression is how we define stoichometrically a reaction. By convention, products have positive stoichometric coefficients (+1 for $\ce{Na+}$ and $\ce{e-}$) and reactants have negative coefficients (-1 for $\ce{Na}$).
For the reaction mechanism, we write
$\ce{Na -> Na+ + e-}$
and the arrow denotes the process that occurs in time. $\ce{Na}$ before the arrow is the reactant (matter before the process), and $\ce{Na+}$ and $\ce{e-}$ after the arrow denotes the matter after the process has completed: the products of the reaction.


text:
Comparing the basicities of guanidine, acetamidine, benzamine, and dimethylamine
The following compounds were given, and I have to predict their order of basicity:

Compound
Name
Chemical formula

A
Guanidine
$\ce{(H2N)2C(NH)}$

B
Acetamidine
$\ce{(H3C)(H2N)C(NH)}$

C
Benzamidine
$\ce{(H5C6)(H2N)C(NH)}$

D
Dimethylamine
$\ce{(H3C)2NH}$

In terms of the resonance structures available to the conjugate acids:

AH+ has 2 resonance structures;
BH+ has 1 resonance structures;
CH+ has 4 resonance structures;
DH+ has 0 resonance structures.

Judging by this, C should be the most basic, as its conjugate acid is most stabilised by resonance. However, it turns out that guanidine A is more basic, despite its conjugate acid having fewer resonance structures. Why is this the case?

Related: Protonation of Guanidine

The actual number of resonance structures is inconsequential: what matters is whether they are significant contributors to the resonance hybrid. This issue has been discussed at length, particularly in the case of carboxylic acids vs phenols.
Along this line of thought, the resonance structures that involve cationic carbon with six valence electrons are not really worth considering. Octet-complete resonance structures are always much more major contributors than octet-deficient ones.

If we only count octet-complete resonance structures of the conjugate acids, then guanidine (A) wins hands down with three; the two amidines (B) and (C) both have two; and dimethylamine (D) only has one.
Between (B) and (C) the choice is tougher. You might be inclined to think that this is where the minor contributors come in: surely benzamidine (C) is more basic than acetamidine (B), since its conjugate base has some of these minor resonance contributors. However, experiment shows that the order of acidity is the other way round: benzamidine (pKa 11.6 in water at 20 °C) is actually a weaker base than acetamidine (pKa 12.52),[1] making for an overall basicity order of (A) > (B) > (C) > (D).
The reason for this is not a resonance effect, but rather an inductive effect. It is true that the benzamidinium ion enjoys marginally more stabilisation by resonance. However, in general, sp2 carbons (and aryl groups) are more electron-withdrawing than sp3 carbons (and alkyl groups). So, the positive charge on nitrogen is also destabilised by the inductively electron-withdrawing phenyl group.
A rare case of inductive effects outweighing resonance effects, but there you are.

Reference

Albert, A.; Goldacre, R.; Phillips, J. The strength of heterocyclic bases. J. Chem. Soc. 1948, 2240–2249. DOI: 10.1039/JR9480002240.

In the case of acetamidine, can't hyperconjugation be a factor for stabilization of the conjugate acid?

@orthocresol: The difference between basicity of acetamidine and benzamidine can be compared to the  difference between acidity of acetic ($\mathrm{p}K_\mathrm{a}$ 4.75) acid and benzoic acid ($\mathrm{p}K_\mathrm{a}$ 4.20) to show the electron donating dominance of acetic group vs phenyl group.


text:
Electrolysis of sodium chloride solution
I am trying to understand electrolysis. My question  is: when electrolysis is done with a NaCl solution I have learnt that sodium metal is not formed. Why is this?

Let's look at these two redox couples:
$$\ce{Na+ + e- <=> Na}  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,E^0(\mathrm{red})=-2.71 V $$
$$\ce{2H2O + 2e- <=> H2 +2OH- }\,\,\,\,  E^0(\mathrm{red})=-0.83 V $$
Because it is much easier to reduce water than $\ce{Na+}$ ions, the only product formed at the cathode is hydrogen gas.

The electrolysis of molten NaCl indeed gives elemental sodium and chlorine gas:
$\ce {2NaCl (l) -> 2Na (l) + Cl2 (g)}$
However, in solution (namely aqueous solution such that you are doing the electrolysis on a solution of NaCl in water), the highly reactive Na produced reacts instantly with water to give sodium hydroxide, hence why no solid Na is ever observed.


text:
Can an amide nitrogen be a hydrogen bond acceptor?
Can $\ce{N}$ (i.e. main chain $\ce{NH}$s of recognised residues in a peptide or protein) be a hydrogen bond acceptor? It is a well known fact that the main chain $\ce{NH}$s can be a hydrogen bond donor (as observed in the α-helix), but can they act as a hydrogen bond acceptor? If yes, help me with an example or even better a reference. And, if no, why can't they act as an acceptor, after all it's an electronegative atom?

see if this reference is help full :http://pubs.rsc.org/en/content/articlehtml/1989/p2/p29890001355 "The extremes of both scales are charted. Alkyl thiols and amines are negligible as proton donors; correspondingly, π-donor hetero-atoms as e.g. in esters and amides are negligible acceptors"

@DavePhD It’s very hard to actually extract data from that article that rules out acceptors from amide nitrogens — since they measure thermodynamic data, any H bonds will probably be accepted by the amide carbonyl which gladly accepts them and there is no way to distinguish.

All the electrons in an amide need to do is induce any dipole in a H- bond donor. If you want an example look at the macromolecule Kevlar as the sheer volume and number of these interactions give it its properties and not just in one plane. When heated it is a liquid crystal. I would look for colligative effects as sources. You've already indicated amide bonds as donors, look at other polymers.

According to the references below, amide nitrogen can be a weak hydrogen bond acceptor: https://pubs.acs.org/doi/abs/10.1021/acs.jpca.7b11013 https://www.ncbi.nlm.nih.gov/pubmed/27166805

If I understood you correctly, you are talking about the peptide bond nitrogens ($\ce{R-C(=O)-\color{red}{N}H-R}$). This is, reduced to its significant chemical functional group an amide, more precisely a carboxylic amide.
The amide nitrogen technically has a lone pair and thus technically could function as a hydrogen bond acceptor when viewed alone. However, this lone pair is actually significantly delocalised towards the carbonyl bond. If we ignore the two $\ce{R}$ residues, we could write this the following way:
$$\ce{O=C-N-H <-> ^{-}O-C=N^{+}-H}$$
The lone pair is not really available as a lone pair on the nitrogen, but forms a π-bond to the neighbouring carbonyl carbon to a significant extent. Thus, these nitrogens cannot be hydrogen bond acceptors.
Further, the fact that a Grignard reagent cannot react with an amide gives great support for the fact that nitrogen’s electrons are delocalized onto the carboxylate.

Finally, thank you Orthocresol for supplying me with the paper given at the end of this post. The authors performed some quantum chemical calculations to extract the energy difference between free formamide and water molecules and various conceivable hydrogen bonds between them. The strongest stabilisation can be found in an amide–amide hydrogen bond of the following type:
$$\ce{H2N-CH=O\bond{...}H-NH-CO-H}$$
At around $-40~\mathrm{kJ \cdot mol^{-1}}$ for a $\ce{H\bond{...}O}$ distance of $285~\mathrm{pm}$ and a $\ce{C=O\bond{...}H}$ angle of $60^\circ$.
$$\ce{H2N-CH=O\bond{...}H-OH}$$
For the interaction of a water molecule donor and the carbonyl oxygen as depicted just above, the stabilisation is given as up to $-27~\mathrm{kJ \cdot mol^{-1}}$ for a distance of $280~\mathrm{pm}$ and a $60^\circ$ $\ce{C=O\bond{...}H}$ angle.
$$\ce{OHC-HN-H\bond{...}OH2}$$
For this interaction where the amide nitrogen is donor and water is acceptor, the stabilisation was calculated to $\Delta E \approx - 30~\mathrm{kJ \cdot mol^{-1}}$ — achieved twice. Once for the geometry $d(\ce{H\bond{...}O}) = 280~\mathrm{pm}$ and $\angle(\ce{H\bond{...}O-H2}) = 30^\circ$ (the angle is measured between the $\ce{H-O-H}$ plane and the amide plane), and once for $d = 280~\mathrm{pm}$ and $\angle = 15^\circ$.
$$\ce{OCH-H2N\bond{...}H-OH}$$
This is the one you are looking for. These energy differences are all given as positive. The lowest is $\Delta E \approx +0.8~\mathrm{kJ \cdot mol^{-1}}$ for a distance $d(\ce{H\bond{...}N}) = 360~\mathrm{pm}$ — $1.3$ times the previous distances. (The angle is not given but from the picture we can assume a $90^\circ$ angle between the amide’s plane and the $\ce{N\bond{...}H}$ axis.) For the $280~\mathrm{pm}$ which brought the most stabilisation in the other cases and which thus seems to be the ideal hydrogen bond distance, they report $\Delta E \approx + 30~\mathrm{kJ \cdot mol^{-1}}$ — this is about the dimension of the stabilisation of the previous hydrogen bond only in a destabilising manner!
These studies let us conclude that it is safe to say the amide nitrogen will never be a hydrogen bond acceptor.
Reference:
A. Johansson, P. Kollman, S. Rothenberg, J. McKelvey J. Am. Chem. Soc. 1974, 96, 3794. DOI: 10.1021/ja00819a013.

Hi, RE: "The lone pair is not really available as a lone pair on the nitrogen, but forms a π-bond to the neighbouring carbonyl carbon to a significant extent. Thus, these nitrogens cannot be hydrogen bond acceptors." A silly question probably, but does this reasoning extend to aniline also?

@GaurangTandon Somewhat, yes, but aniline’s lone pair is not involved in resonance as much.

I understand you mean to say that it cannot be predicted theoretically accurately?

@GaurangTandon It can be predicted. The six-membered benzene ring of aniline is pretty stable in itself so the additional nitrogen lone pair doesn’t add much. The $\ce{C=O}$ double bond benefits much more because it is fully localised by itself.

Sorry for disturbing, but by the same reasoning, can we say that the solubility of benzyl alcohol in water is more than that of phenol? (because the latter has weaker H-bonds, due to a delocalised lone pair, as compared to the former)

Concerning that edit that was approved just earlier: it's a hydrogen bond acceptor (Lewis base), not a hydrogen acceptor (Bronsted base). The edit introduced incorrect terminology.


text:
why do electron pairs not take axial positions in the VSEPR structure of ClF3?
when we make a Valence shell electron pair repulsion theory based structure for an molecule after calculating its hybridisation ; when we have to decide the positions of the ions we put them in such positions so that repulsion is minimized .It follows the basic rule of lone pair-lone pair repulsion> lone pair-bond pair repulsion >bond pair -bond pair repulsion . So when it comes to deciding the position of lone pairs of electrons in ClF3 why dont we put them on the axial positions so that they have an angle of 180 deg between them as compared to 120 deg that they currently have ?

I think your diagram is a bit wrong, isn't it supposed to be a Bent-T shape? Also if the lone pairs occupy axial positions then there will be very high 6(LP-BP)*  repulsion's as compared to earlier very high 2(LP-BP)* repulsion's.

Dupe of this question. Also, Bent's rule...again.

Let's take a look at the electronic configuration of $\ce{ClF3}$.

For $\ce{ClF3}$ actually three isomers are possible.

It is highly unfavorable for the lone pairs to be 90° apart which rules one of the three. In order to rationalize the obervation that the isomer with both lone pairs in equatorial positions is the observed form, it is necessary to count the number of 90° bond pair - lone pair interactions. The observed isomer  has four such angles and the isomer with the lone pairs in axial positions has six which is (presumably) less favorable. The molecule might be called T-shaped based on the atomic positions only.

Here are the links to the other two possible structures http://i.stack.imgur.com/YpQef.gif
http://i.stack.imgur.com/quW3e.gif

Good answer, though poor picture clarity, btw why aren't you adding the other two structure to your answer post?

@SujithSizon i'm unable to add more than two images on my answer.

@VMiranda It's also important to consider the bonding that the various geometries imply. In the top molecule, is that a p orbital holding 4 electrons? If so, that's not possible - you can put at most 2 electrons in a single orbital. In your middle drawing, how many $\ce{sp^2}$ orbitals are there in that molecule? You can't have more than 3. In your bottom isomer, what kind of bonds are those involving the axial fluorines? The correct answer is as much about bonding (specifically hypercoordinate bonding, see my comment above for a relevant link) as electron repulsion.

Invoking d-orbitals in main group elements gives you $-1$ from me.

Thanks @ron, i would've edited my answer but it seems to have been closed.

@Jan whats wrong with d-orbitals ?

Re editing: I think you can edit answers to closed questions. At least I can. Re d-orbitals: They are too far removed energetically from the others (higher than the next-level s-orbital in ground state!) to reasonably take part in bonding for main-group elements.

authors: 13 จารุวรรณ ระบายศรี


text:
Why radioactive elements emit alpha beta and gamma rays
I am confused about this that why radioactive elements emits alpha beta and gamma rays WHILE other elements can't do so.

Too all close voters: I don't think this is necessarily too broad, but it will probably get a more detailed answer on physics.SE. I suggest that you ask it there instead.

The stability of nuclei is really a sophisticated topic in theoretical quantum mechanics. But there is a simple way to think about what is happening that doesn't get too intense with the quantum mechanical theory.
Nuclei are made from two particles: protons and neutrons. But protons are positively charged and repel each other. The electromagnetic force is very strong and therefore this force is very large. So the first mystery is why all nuclei don't just fly apart.
The reason they don't is that there are two very short-range but very strong forces that bind the nucleus together: the strong and weak nuclear forces. Without getting into mind-bending topics in theoretical physics we can understand something about their net effect like this.
The interaction of the electromagnetic force and the two nuclear forces has some structure (it's quantum stuff, just accept it). Some combinations of protons and neutrons are more stable than others. Each combination has an energy level and some combinations have lower energy than others. Nuclei with even numbers of protons and neutrons are more stable than odd-odd combinations and nuclei with wildly unbalanced neutron to proton ratios are less stable. Neutrons act a little like a glue, helping protons stick together (this is an oversimplification as too many neutrons is also a source of instability: this is a consequence of a complicated interplay of several forces). But bigger nuclei are less stable and need a higher ratio of neutrons. And some large nuclei are just too large for the forces to keep them together so beyond a certain point all nuclei are unstable.
Some nuclei can be transformed into a more stable (lower energy) nucleus by various forms of radioactive decay. Nuclei with too many neutrons can emit a beta particle (this decay mode converts a neutron into a proton); elements with too many protons can emit a positron (converting a proton into a neutron). Bigger nuclei can become more stable by kicking out an alpha particle (which makes the nucleus significantly smaller, moving it towards the stable zone). Gamma radiation is associated with some of these modes: the high energy photons "mop up" the excess energy (I'm simplifying a lot).
This picture (more detail on wikipedia) sums up what we know about nuclear stability:

To sum up: nuclei are held together by the interaction of several forces but the forces not equally effective for all possible combinations of protons and neutrons. Some nuclei have so much energy that they are unstable and can decay into more stable nuclei in various ways depending on the mix on neutrons and protons they have. In some cases the forces hiding the nuclei together are strong enough that they never decay.


text:
What is the most accepted mechanism for the reaction of ketones with m-dinitrobenzene?
This is a commonly used confirmatory technique in organic functional group analysis. The test is performed by addition of meta dinitrobenzene to the given organic sample followed by addition of a base such as  sodium hydroxide. The indication is given by a violet colouration.
Is the probable mechanism based on carbanion formation? Also it would be helpful if exceptions to this reaction and other variants in colour produced by different ketones. My main requirement is to know why aldehydes do not respond to this test.

see also: Foster, R.; Mackie,  R. K. Interaction of electron acceptors with bases—V: The Janovsky and Zimmermann reactions. Tetrahedron 1962, 18 (10), 1131–1135.

Thanks for the link. But I am not a registered user and hence cannot access the full document. I would be thankful if you could explain necessary details (of Janovsky reaction) in your answer.

You are on the right track, a strongly colored enolate is formed: https://en.wikipedia.org/wiki/Meisenheimer_complex#Zimmermann_reaction

@Jannis Andreska... Thanks a lot for the link...seriously useful. If you have anything more to add on, could you please expand it into an answer. Especially the link posted by Loong (if you have access to it or have read it). Thanks again.

@Varun Unfortunately, the linked paper is behind a paywall for me. Nevertheless, I have written an answer based on the information that I could gather.

In the Janovsky reaction, an aldehyde or ketone which contains at least one $\alpha$ methylen group reacts with 1,3-dinitrobenzene and a strong base (like $\ce{NaOH}$) to form a colored Meisenheimer complex (source). The enolate acts as a nucleophile which attacks the aromatic ring that is electron-deficient due to the presence of multiple electron-withdrawing nitro substituents.

image source
As the reaction mechanism relies on the formation of an enolate, non-enolizable aldehydes like, for example, formaldehyde or benzaldehyde, do not resond to this test. Likewise, non-enolizable ketones (like benzophenone) cannot react in this way. The color of the formed complex varies with the number of nitro substituents of the benzene ring, with 1,3-dinitro compounds giving blue complexes, while trinitro compounds yield a red coloration (source).
This paper contains a list of the colors produced by various aromatic nitro compounds when reacted with acetone and $\ce{NaOH}$. It is reported that most mononitro compounds do not produce any color at all, with the exeption of 3-nitro-4-aminotoluene (orange) and 3-nitro-4-aminotoluene benzoate (orange-red). Some examples for the colors obtained from the reaction of dinitro compounds are: m-dinitrobenzene (purplish blue), 2,4-dinitrotoluene (blue), 2,4-dinitrophenol (yellow-orange), 2,4-dinitroresorcinol (brownish-green) and p-dinitrobenzene (green-yellow).

It would be great if you could append a list of the colour of various complexes formed in the Janvosky (and Zimmerman reaction). Also, most of the info I have found regarding Janvoski reaction has been confined to Wikipedia (including one that you posted) and a couple of papers (most behind a paywall), so incase you do happen to find more, could you please append to this? Anyways, I have accepted this answer. Thanks a lot.

@Varun I have added some examples for the colors of various complexes.

It would be great if you could append a list of the colour of various complexes formed in the Janvosky (and Zimmerman reaction) I used the Janovsky reaction (e.g. Zimmermann too) for a project about the synthesis of acridines in the seventhies.
I can send you a list as wished published in Microchimica Acta (Wien) 1968, 739-747 and the paper too. The author is L. Legradi (Hungary).
For this purpose, I need your email address. The paper is in German but you can find a summary in English for all the papers of L. Legradi. All these infos are free of charge in Google or Google Scholar.

@JannisAndreska  Can you plz mention that why even enolisable aldehyde doesnt react with  this reagent  because this  is one of the confirmatory test for ketones.


text:
How does a metal atom's size affect the electrical conductivity of it's giant metallic structure?
In giant metallic structures, how does a metal atom's size affect the structure's electrical conductivity? (i.e. what happens if you go down a group of metal elements?)

Why do you think sole size would be important?

@Mithoron: I don't. That's why I'm asking about it's effects on conductivity.


text:
Miscibility of pairwise miscible liquids
If you have a mixture of 3 (or more) liquids which are "pairwise miscible", but which don't react with each other, does that guarantee that the mixture will only have one layer?
By "pairwise miscible", I mean that for every combination of two substances in the mixture, they are miscible with each other, i.e. soluble in all proportions.
For example, ethanol/water/acetone.

I'll say very probably, but I'd bet that there is a black swan somewhere.

So you want to know whether there is a ternary system with an "island of immiscibility". Well, it turns out that people have thought of this before; such systems have a name ("type 0 systems"), and this paper mentions water/DMSO/THF as an example.