Why is neutralization reaction exothermic in nature? Why is heat liberated in such reaction. Isn't it just a simple rearrangement of ions? Hydrogen ion joining hydroxyl ion.

You are right in that the reaction is $\mathrm{H}^+ + \mathrm{OH}^- \rightarrow \mathrm{H_2O}$, but recognise that bonds are formed in this reaction. Bond formation releases energy (and similarly bond breakage is endothermic), and thus neutralisation reactions are exothermic.


Energy of photons emitted by de-excitation of electrons When an electron of an atom returns from an excited state to the ground state, it emits energy in the form of a photon. How does the change in energy level compare to the energy of the emitted photon?

When an electron of an atom returns from an excited state to the ground state, it emits energy in the form of a photon.

That is one of several pathways for deexcitation. Often, internal conversion, the radiationless deeexcitation happens instead.

  How does the change in energy level compare to the energy of the emitted photon?

It depends on the relaxed geometry of the excited state, as compared to the ground state.

Remember that the excitation happens on a femtosecond scale. That is amazingly fast, light travels a distance of just $\mathrm{0.3\,\mu m}$ in $\mathrm{1\,fs}$.

Since it takes longer than some femtoseconds to move the much heavier nuclei, the molecule in its electronically excited state still has the geometry of the ground state. This can be far from the lowest vibrational level (the relaxed geometry) of the excited state.

Before emission of a photon occurs, ectronically excited molecules relax and adopt their lowest vibrational state.

This means that the energy of the emitted photon often is (much) lower than the excitation energy. This effect is known as Stokes shift.


Activated complex theory vs. consecutive reactions Activated complex theory, tells us that due to the collision between the molecules of the reactants, they form a transition specie before the product is formed, which is called active complex. On the other hand we have consecutive reactions on which is also formed a intermediate product before forming the actual product we're interested on. My question is :

Where is the difference between the activated complex and the intermediate product, since they are both formed before  the actual product ?

The key difference is that transition states occur at a maximum of the potential energy curve for the reaction whereas intermediates occur at a local minimum. Take this example of the reaction profile for an $\ce{S_{N}1}$ reaction:

You will see that the products and reactants occur at minima on the curve and the intermediate also occurs at a minimum, albeit a higher energy one. In between the minima are located the maxima where you find the transition states.  Unless the activation energy is zero (see this question for rare examples) there will always be a transition state located between any two minima.

Transition states are usually represented using dashed bonds to show bonds in the process of being broken or formed as opposed to being fully formed as in intermediates or products. Additionally the Hammond postulate says that the transition state will most resemble the stable species closest to it in energy, in this case the carbocation, and this can be used to help predict the structures of transition states.

Transition states are very short lived because they immediately 'roll downhill' on the potenetial energy curve to reach an intermediate or product. By contrast some intermediates are actually quite stable and can be isolated.


How is the specific heat of an alloy obtained? I was wondering:

How is the specific heat of an alloy obtained?

My first attempt is to assume a thermal equilibrium between a mechanical solution and an homogeneous solution of two metals, A and B.

\begin{align*}
    Q_{A+B}&=Q_A+Q_B\\
    n_{A+B}C_{A+B}(T_{eq}-T_{A+B})&=n_{A}C_{A}(T_{eq}-T_A)+n_{B}C_{B}(T_{eq}-T_B)\\
    C_{A+B}(T_{eq}-T_{A+B})&=\frac{n_{A}}{n_{A+B}}C_{A}(T_{eq}-T_A)+\frac{n_{B}}{n_{A+B}}C_{B}(T_{eq}-T_B)\\
    C_{A+B}(T_{eq}-T_{A+B})&=X_AC_{A}(T_{eq}-T_A)+X_BC_{B}(T_{eq}-T_B)\\
    C_{A+B}&=\frac{X_AC_{A}(T_{eq}-T_A)+X_BC_{B}(T_{eq}-T_B)}{(T_{eq}-T_{A+B})}
\end{align*}
Where Q=heat, C=specific heat, n=moles, X=molar fraction and T=temperature.

Is it correct?

Is there another way of doing this?

I have previously searched and this are some relevant things on Chem.SE what I have found:

What is this equation for?
Properties of steel and aluminum alloys

I did not find a solution or an approximation in any of those links.

My aim is to prepare a class and teach my students how to determine the specific heat of any alloy, if possible, without going to a lab. (There is no basic infrastructure for special labs).

The enthalpies and calorific capacities of certain mixtures are tabulated in standard references. If you do not have such data, you can use the following approximation:

  Rule 1: For a mixture of gases or liquids, calculate the total change in enthalpy as the sum of its changes in the pure components of the mixture. In fact, the enthalpy changes associated with component mixing are neglected, which is an excellent approximation for gas mixtures and similar liquid mixtures such as pentane and hexane, but is bad for different liquids, such as nitric acid and Water.

  Rule 2: For highly diluted solutions of solids or gases in liquids, neglect the change in enthalpy of the solute. This approach is better, as the solution is more diluted.

  The calculation of the enthalpy changes for heating or cooling a known composition mixture can often be simplified by estimating the calorific capacity of the mixture as follows:
  \begin{equation}
    C_{p,m}(T) = \sum_i y_i\cdot C_{p,i}(T)
\end{equation}

  Where:

  $C_{p,m}(T)$ Calorific capacity of the mixture

  $y_i$ Mass or mole fraction of the i-th component

  $C_{p,i}(T)$ Calorific capacity of the i-th component

Reference:

Richard Mark Felder and Ronald W. Rousseau. Elementary Principles of Chemical Processes. John Wiley & Sons Inc.; 3rd Revised edition. 702 pages.


Hybridization, MOT and Paramagnetism In what way can hybridization or molecular orbital theory be used to explain paramagnetism?

For instance, when something is hybridized to make enough bonding electrons, do all the electrons end up paired? Or are some left unpaired, explaining paramagnetism?

Paramagnetism is a form of magnetism whereby certain materials are attracted by an externally applied magnetic field. Valence bond theory (VBT) and hybridisation doesn't really do a good job at predicting whether a molecule is paramagnetic or diamagnetic (isn't attracted by an external magnetic field). This is why molecular orbital theory (MOT) is so useful as it is successful at predicting whether a molecule is paramagnetic.

For a molecule to be paramagnetic, it needs to have an overall magnetic moment meaning that it needs an unpaired electron. If all the electrons are paired, then the molecule is diamagnetic. So by seeing whether a molecule has an unpaired electron, we can predict if it is paramagnetic or not.

Now lets consider the example of $\ce{O2}$. Experimentally, $\ce{O2}$ is known to paramagnetic. According to VBT, $\ce{O2}$ should look like this:

]1

As you can see, all the electrons are paired. Therefore VBT predicts that $\ce{O2}$ should be diamagnetic.

Now lets examine how the electrons are arranged according to MOT. In MOT, unlike VBT, it involves the creation of bonding and anti-bonding MOs. MOs are basically the supposition of the wavefunctions of atomic orbitals. In $\ce{O2}$ the 2s AOs of each oxygen atom constructively and destructively overlap with each other while their 2p AOs also constructively and destructively overlap with each other The resulting MOs for $\ce{O2}$ looks like this:

Now that we got the MOs, all we have to do is fill them with electrons using the same method that we use for AOs. By doing that we get:

Note that we have 2 unpaired electrons. Therefore MOT correctly predicts that $\ce{O2}$ should be paramagnetic, unlike VBT which predicts that $\ce{O2}$ is diamagnetic.


Why is ionization energy measured in gaseous phase? When I looked up the definition of ionization energy, it says:-

  it is the energy required to remove one electron from the shell to
  infinity in an atom in gaseous state.

Why is it measured in gaseous phase and not in any other phase?

You want to determine the energy for a single molecule and avoid stabilization effects through intermolecular interactions.


How can I model cellular membranes in terms of topology? I'm reading from Principles of Physical Biochemistry by Kensal E van Holde (1998-01-06) in order to review some biochemistry and work my way toward a more mathematical approach to understanding chemistry.

On page 18, chapter 1 (Biological Macromolecules), I read

  "The structure of an integral membrane protein can be thought of as being inverted relative to the structure of a water soluble-protein, with the hydrophobic groups now exposed to the solvent, while the hydrophilic atoms form the core. An example of this inverted topology is an ionic channel (Figure 1.12). The polar groups that line the internal surface of the channel mimic the polar water solvent, thus alloweding charged ions to pass readily through an otherwise impenetrable bilayer."

I don't have a picture handy to show you Figure 1.12, but essentially it is a channel protein.

What's interesting to me about this passage is their use of the word topology. It reminded me of algrabraic topology's notion of a topology on a set, but I can't seem to state this membrane chemistry in terms of the mathematical formalism.

In algebraic topology, a topology $\tau$ on a set $X$ is a subset of the powerset $\mathcal{P}(X)$ where the elements of $\tau$ are open, meaning they satisfy

Containing the original set and the empty set:$$X \in \tau, \emptyset \in \tau$$
Unions of sets in the topology are closed. $$\{ U_i \}_{i \in I} \subseteq \tau \implies \bigcup_{i \in I} U_i \in \tau$$
Closure under finite union. $$\{ U_i \}_{i=1}^n \subseteq \tau \implies \bigcap_{i=1}^n U_i \in \tau$$

My initial thinking on this is that the membrane could be one set, the extracellular matrix a second set, and the intracellular matrix as a third set. Each of these sets might be thought of as their own topologies, and that their mutual intersections should be empty so long as you cleanly partition the spaces. But then again, membranes have a lot of intermolecular spaces within them; seems structurally complicated.

While knowingly extrapolating beyond the original intended meaning of the authors, how can I model  cellular membranes in terms of topology?

As Buck Thorn and MaxW already said in the comments, there is less to it than the textbook excerpt might suggest. The use of the term topology in this context is non-mathematical, similar to when someone says there has been a quantum leap in product development.

The actual topology of most proteins is the same as that of a sphere, and the topology of the typical cartoon channel is that of a donut. Having hydrophilic residues in the channel or outside of the channel is not an inversion because those parts of the surface are topologically equivalent.

  [OP] I can't seem to state this membrane chemistry in terms of the mathematical formalism.

That's because it is non-mathematical. There is some real topology associated with membranes, and that is the concept of inside and outside. Membrane proteins insert in one direction, and they don't "flip" directions over time, even though they are perfectly mobile to move around in the membrane (or have lipids move around them) while pointing in one direction or the other one. During endocytosis and exocytosis, the relative orientation of membrane proteins is conserved but inverted, so that parts that are inside organelles (such as glycosylation) end up on the outside of the plasma membrane.


What does the slope of a Levich plot represent? I know that the limiting current is proportional to the square root of the rotation speed, but what does the slope of the plot represent?

if you mean the following plot:

(Electrochemical Methods. fundamentals and applications, p. 341).

If a bulk concentration is the same as the concentration at the surface of the electrode (the mass-transfer is very high) than the current will be independent on $\sqrt\omega$ and will be constant at the value $i_K$.

The slope of the bold line represent deviation form Levich line and tend towards the limit $\lim_{\sqrt\omega \to \infty} i = i_K $. The higher the speed of rotation is, the less is the electrode reaction limited by mass-transfer effects.

The Levich line (dashed; $\frac{i}{\sqrt\omega}$ ) is constant only when $i_K$ is much larger.


At what temperature does carbon dioxide decompose into carbon and oxygen? If you were to super-heat a mole of carbon dioxide gas, at what point would it decompose into carbon and oxygen? If it decomposes into carbon monoxide and oxygen, at what point do those products decompose? I'm interested in the process from a hypothetical perspective, so energy is not a concern, nor is industrial feasibility.

I'm looking for the heat required for carbon dioxide to break down into atomic carbon and oxygen, if carbon monoxide is being formed then its not hot enough yet!

Disclaimer: I am too lazy to check whether the reaction $\ce{CO2(g) <=> CO(g) + 1/2 O2(g)}$ is more favourable.

There is no single tipping point where all carbon dioxide will suddenly be converted to carbon and oxygen. As you heat carbon dioxide, the percentage of oxygen present would gradually increase.

You're interested in the reaction

$$\ce{CO2(g) <=> C(graphite) + O2(g)}$$

for which, at $298~\mathrm{K}$, $\Delta_\mathrm{r} H^\circ = +393.51~\mathrm{kJ~mol^{-1}}$ and $\Delta_\mathrm{r} S^\circ = -2.862~\mathrm{J~K^{-1}~mol^{-1}}$ (data from Atkins, Physical Chemistry 10th ed., pp 974-7). The equilibrium constant for this reaction is remarkably simple:

$$K = \exp{\left(\frac{-\Delta_\mathrm{r}G^\circ}{RT}\right)} = \frac{a_{\ce{O2}}}{a_{\ce{CO2}}}$$

and if we assume ideality, then both activities are proportional to the partial pressures, which are in turn proportional to the number of moles of both substances:

$$K \approx \frac{n_{\ce{O2}}}{n_{\ce{CO2}}}$$

Therefore if $K = 100$, then you have a hundred times more oxygen in your system than carbon dioxide. If $K = 1$, then you have equal amounts. The pressure and volume of the system do not matter in this interpretation of the value of the equilibrium constant. Our goal is now to find out the variation of $K$ with temperature.

The variation of $\Delta_\mathrm{r} H^\circ$ with temperature is given by Kirchhoff's law, and to simplify things, let's assume that the heat capacities at constant pressure are independent of temperature over the temperature range we are interested in:

$$\begin{array}{cc}
\hline
\text{Compound} & C_{p,\mathrm{m}}^\circ\text{ / }\mathrm{J~K^{-1}~mol^{-1}} \\
\hline
\ce{C(graphite)} & 8.527 \\
\ce{O2(g)} & 29.355 \\
\ce{CO2(g)} & 37.11 \\ \hline
\end{array}$$

$$\Delta_\mathrm{r}H^\circ(T) = +393.51~\mathrm{kJ~mol^{-1}} + (0.77~\mathrm{J~K^{-1}~mol^{-1}})(T - 298~\mathrm{K})$$

One can derive a similar expression for the entropy change, which is "left as an exercise for the reader":

$$\Delta_\mathrm{r}S^\circ(T) = -2.862~\mathrm{J~K^{-1}~mol^{-1}} + (0.77~\mathrm{J~K^{-1}~mol^{-1}})\left[\ln{\left(\frac{T}{298~\mathrm{K}}\right)}\right]$$

All that remains is to calculate the Gibbs free energy change

$$\Delta_\mathrm{r}G^\circ = \Delta_\mathrm{r}H^\circ - T\Delta_\mathrm{r}S^\circ$$

and the equilibrium constant

$$K = \exp{\left(\frac{-\Delta_\mathrm{r}G^\circ}{RT}\right)}$$

You can do this easily with a spreadsheet. I've thrown together some very messy Mathematica code and I'm sure someone will come along and tell me that there's a much more concise way of doing it, but:

Manipulate[{Power[10,logT],Exp[((393510+(0.77*(Power[10,logT]-298)))-(Power[10,logT]*(-2.862+(0.77*(Log[Power[10,logT]/298])))))/(-8.31446*Power[10,logT])]},{logT,2.47422,12}]

The first number in the curly brackets gives you the value of $T$ in Kelvins, and the second gives you the value of $K$ at that temperature $T$. Here are some of the numbers:

$$\begin{array}{cc}
\hline
T\text{ / K} & K \\
\hline
298 & 7.513 \times 10^{-70} \\
10^3 & 2.073 \times 10^{-21} \\
10^4 & 0.007895 \\
10^5 & 0.6899 \\
10^6 & 1.307 \\
10^9 & 2.598 \\
10^{12} & 4.9257 \\
\hline
\end{array}$$

$K = 1$ when $T = 2.52 \times 10^5~\mathrm{K}$.

This is well above the temperature at which carbon vaporises ($4827~\mathrm{^\circ C}$ according to Wikipedia). If you redo the calculations using instead the equation

$$\ce{CO2(g) <=> C(g) + O2(g)}$$

then you get these new expressions (thermodynamic data again from Atkins):

$$\Delta_\mathrm{r}H^\circ(T) = +1110.19~\mathrm{kJ~mol^{-1}} + (13.083~\mathrm{J~K^{-1}~mol^{-1}})(T - 298~\mathrm{K})$$

$$\Delta_\mathrm{r}S^\circ(T) = +149.498~\mathrm{J~K^{-1}~mol^{-1}} + (13.083~\mathrm{J~K^{-1}~mol^{-1}})\left[\ln{\left(\frac{T}{298~\mathrm{K}}\right)}\right]$$

and much more favourable numbers:

$$\begin{array}{cc}
\hline
T\text{ / K} & K \\
\hline
298 & 1.637 \times 10^{-187} \\
1000 & 1.470 \times 10^{-50} \\
2000 & 3.417 \times 10^{-21} \\
5000 & 0.003130 \\
6000 & 0.351794 \\
7000 & 10.65 \\
10000 & 5594.31 \\
\hline
\end{array}$$

This time, $K = 1$ at $T = 6276~\mathrm{K}$. You can play around with the code here:

Manipulate[{T,Exp[((1110190+(13.083*(T-298)))-(T*(149.498+(13.083*(Log[T/298])))))/(-8.31446*T)]},{T,298,10000}]

Why does $K$ for this reaction increase so drastically with temperature? It is likely to be because of the entropy contribution: because the entropy change is much more positive, increasing the temperature makes the term $-T\Delta_\mathrm{r}S^\circ$ much more negative, lowering $\Delta_\mathrm{r}G^\circ$ and increasing $K$.


Regioselectivity in Friedel–Crafts acylation of thiophene In the above reaction, I  understand that $\ce{AlCl3}$ is acting as a Lewis acid to generate an acylium ion, which is a strong electrophile.

Would this electrophile attack at carbon 2 or 3?

I think it would attack at C-3, as the resulting positive charge at C-2 would be stabilized by resonance using the lone pair on sulfur.

Friedel-Crafts acylation reactions are well documented on thiophenes, and go with high regioselectivity for the 2-position (or 5-position, which is equivalent in the case of unsubstituted thiophene) rather than the 3-position (or 4-position). Joule and Mills' Heterocyclic Chemistry (5th ed., p 327) gives the following two examples:

The regioselectivity can be explained through consideration of the intermediates involved when the thiophene attacks the electrophilic carbenium species.

One is able to see that the intermediate species produced by attack at the 2-position is able to be drawn in three resonance forms, whilst the intermediate produced by attack at the 3-position is only able to be drawn in two resonance forms. As a general rule of thumb, the more resonance forms one can draw, the more 'stable' and intermediate, and hence the energy required to access it is lower, hence in the case of thiophene we get preferential electrophilic substitution at the 2/5-position.

(When looking at the intermediates, we can also consider the type of conjugation involved. Attack at the 2-position yields a linearly conjugated intermediate, which is lower in energy than the cross conjugated intermediate from attack at the 3-position).


Size consistency in computational chemisty methods I am looking for a clear answer that would explain what exactly size consistency is, why configuration interaction (CI) is not size consistent and how this leads to errors when using CI for large systems.

Size consistency is essentially the idea that if you were to determine the energy of two noninteracting systems A and B, you should obtain the same result if you sum the energy of A and B separately and if you calculate the energy of A and B as one supersystem (usually with A and B separated by a large enough distance where they don't interact). Expressed mathematically, we might write: $$E(A+B)=E(A)+E(B)$$ provided that A and B don't interact.
As to the second part of your question, I want to address a finer point. Full CI is size consistent; the issue comes when you have to truncate CI to make it computationally feasible. Truncated CI is only exactly size consistent up to the point of truncation. For example, CISD (CI with single and double excitations) would only be exactly size consistent for systems with $N=2$ electrons.
The issue becomes a problem for large systems because this lack of size consistency leads to the wrong limiting behavior of the correlation energy. Specifically, truncated CI leads to the false conclusion that $\lim_{N\rightarrow \infty}\frac{E_{corr}}{N}=0$, while the correlation energy should vary linearly with $N$.
For more information on this, I would read chapter 4, section 6 of Szabo and Ostlunds Modern Quantum Chemistry. The chapter as a whole gives a thorough primer on CI, with this particular section discussing the size inconsistency problem.


slow down ice melting in water Having to separate freshwater crabs for commercial fisherman while on the boat, we dump the crabs in ice water, as crabs aren't as aggressive and bite less when cold.  There's only so much room on the boat for carrying ice, and the ice melts too quickly in the wooden box filled with water and ice in order to "slush" the crabs.  We've been unable to work full days, running all of our crab traps, as we run out of ice before a full-day's work.  Any ideas of how to slow the melting process of the ice while in the water used for slushing the crabs?  Thanks for your time and knowledge.
"

Unfortunately there isn't much you can do to slow down the melting of the ice-water slush besides insulating your container. If possible, you can enclose the wooden box in styrofoam to prevent heat from entering the system.

Another note you should keep in mind is that the water in the slush speeds up the melting of the ice. A better scenario would be to keep a large (minimized surface area) block of ice in one container, and the slush in another. Use an ice pick to break up the block and add pieces to the slush as needed.


Why anions of weak acids are weak bases? This is what I saw from Chemistry: The Central Science 13th ed. by Theodore brown, page 698. It stated that there are two types of weak bases: one is those with lone pairs of electron and the other is anions of weak acids, which means conjugate bases of weak acids. My question is, aren't conjugate bases of weak acids strong bases? Because weak acids are less readily to give off proton but it's conjugate base can accept proton easily.

Your understanding is correct - on a relative scale, the conjugate base of a weak acid is a strong base, and the conjugate base of a strong acid is a weak base. For example, hydroxide ion (conjugate base of water) is a stronger base than acetate ion (conjugate base of acetic acid) which is a stronger base than chloride ion (conjugate base of hydrochloric acid).


Which of the following isomers of 2,3‐dihydroxy‐4‐methoxy‐4‐oxobutanoic acid are identical? Which of the following are identical?

A and B are identical
A and B are diastereomers
A and C are enantiomers
A and B are enantiomers

What I know is that when we rotate a Fisher projection by 180°, then we get identical compounds (I might be wrong). But how to tell if the compounds are enantiomers of each other or not?
For that how to get the same group, i.e. $\ce{-COOH}$ in this case at the top to compare the compounds?
Also, is there any trick to tell that if they are diastereomers, enantiomers or identical by the hydroxyl group positions in Fisher projections?

On rotating the compound B by 180°, we get the two -OH groups on the left side and the two H atoms on the right. In other words, B would then be an enantiomer of A.

Note: on rotating a Fischer projection by 180°, the compound remains the same; it's just that all the atoms have to be rotated, i.e. the atoms on C3 in this case would go to C2 and vice versa.


Why is the IUPAC name of glycerol expressed in the way that it is? I have come across the name trihydroxy propane for glycerol, which I think should've been Propan-1,2,3-triol.

Why is it named this way? Is there any  other rule of nomenclature working?

Glycerol and trihydroxypropane are common names. The IUPAC name is propane-1,2,3-triol. The common name glycerol comes from the root glyco- which means sweet. Glucose, glycogen, and glycerin (another name for glycerol) have the same etymology.

Trihydroxypropane makes sense as a common name because there are only 3 locations for the hydroxyl groups on the 3 carbon chain, so the name is basically unambiguous - multiple hydroxyl groups on the same carbon are not stable (see geminal diols).


Will bleach combined with any of these chemicals create poisonous gas? I need to know if a combination of bleach and the chemicals below will create poisonous gas. Will it create chloramine, and will there be any visible gas?

1) Sodium laureth sulfate
2) Sodium citrate
3) C12-14 Pareth-3

I need to know if a combination of bleach and the chemicals below will create poisonous gas.
Sodium laureth sulfate

Sodium laureth sulfate is $\ce{CH3(CH2)11(OCH2CH2)_{\bf{3}}OSO3Na}$, where the boldface 3 is an approximation and can actually vary between 1 and 5 or so.  The key point is it contains no nitrogen.

Sodium citrate

Sodium citrate can refer to mono-, di-, or tri-sodium citrate, which have formulae like $\ce{C6H_{7-\it{n}}O7Na_{\it{n}}}$, where $n$ is 1, 2, or 3, respectively.  Again there is no nitrogen.

C12-14 Pareth-3

According to Google, "C12-14 Pareth-3" is a "polyethylene glycol ether of a mixture of synthetic C12-14 alcohols with an average of 3 moles of ethylene oxide."  The most representative compound would presumably be the C13 alcohol with exactly 3 moles of ethylene oxide.  That would have the formula $\ce{CH3(CH2)11CH2O-(CH2CH2-O)3H}$, or, if I did my math right, $\ce{C19H40O4}$.  Again the key point is there is no nitrogen.

Will it create chloramine, and will there be any visible gas?

None of the compounds have nitrogen, so chloramine cannot be created from bleach (chemically known as sodium hypochlorite) and these chemicals.
I think the likely products of bleach treatment would be non-volatile compounds such as inorganic sulfate plus organic compounds like tridecanoic acid, tridecanol, ethylene glycol, as well as oxidized derivatives of all of those organics, possibly including some small amount of $\ce{CO2}$ gas.  But $\ce{CO2}$ gas is not poisonous and not visible.


What is a complete reaction? For a lab assignment, I combined $\pu{10 mL}$ of $\pu{1 M}$ sodium sulfate solution with $\pu{10 mL}$ of $\pu{1 M}$ calcium chloride solution. The ionic equation I came up with is:

$$\begin{multline}
\ce{2Na^+ (aq) + SO4^2- (aq) + Ca^2+ (aq) + 2Cl^- (aq) ->}\\
    \ce{CaSO4(s) + 2Na^+ (aq) + 2Cl^- (aq)}
\end{multline}$$

I was then asked, in a lab question, the following:

  If the reaction called for $\pu{1 M}$ chloride ion with $\pu{1 M}$ calcium ion, would you still use $\pu{10 mL}$ of each solution for a complete reaction? Explain your reasoning.

To my understanding, I initially had $\pu{2 M}$ chloride ion and $\pu{1 M}$ calcium ion. With only $\pu{1 M}$ chloride ion instead, the ionic equation would change to this, or so I'm led to believe:

$$\begin{multline}
\ce{2Na^+ (aq) + SO4^2- (aq) + Ca^2+ (aq) + Cl^- (aq) ->}\\
 \ce{CaSO4(s) + 2Na^+ (aq) + Cl^- (aq)}
\end{multline}$$

As a (balanced) molecular equation, to my understanding, this would be:

$$\begin{multline}
\ce{Na2SO4 (aq) + Ca (aq) + Cl (aq) ->}\\
\ce{CaSO4(s) + NaCl (aq) + Na (aq)}
\end{multline}$$

Is this a complete reaction? I have seen conflicting definitions of a complete reaction online:

answers.com says that a complete reaction means all of the reactants are reacted into products.
However, someone on chemicalforums.com says that a complete reaction means that all of the limiting reactant is used up. The University School of Milwaukee seems to back up what the user on chemicalforums.com said ("all of at least one of the available reactants is used up and converted into products").

My main question is thus: what is a complete reaction? No matter my final equation for this lab question, I still need to know what a complete reaction is so that I can identify them in general.

My intuition says that this is a complete reaction, since all of the calcium ions and the chloride ions are converted to products. Is this correct? The leftover sodium ion makes me suspicious, so I don't know how to answer.

I later spoke to my lab instructor and he clarified that the question is asking for me to come up with two arbitrary aqueous solutions, one which contains $\pu{1M}$ of chloride ion and another which contains $\pu{1M}$ of calcium ion, e.g. $\ce{KCl(aq)}$ and $\ce{Ca(NO3)2 (aq)}$.

So in this case, I would need twice the volume of potassium chloride to get the number of chloride ions I need to balance the equation and prevent explosions, i.e. no extra sodium ions leftover.

I suspect that you would have both these molecular equations:

\begin{align}
\ce{2KCl(aq) + Na2SO4(aq) &-> K2SO4(aq) + 2NaCl(aq)}\\
\ce{Ca(NO3)2(aq) + Na2SO4(aq) &-> 2NaNO3(aq) + CaSO4(aq)}
\end{align}

To answer your question with "What is a complete reaction?":

  A complete reaction is one where all of at least one of the available reactants is used up and converted into products.

So for example $\ce{KCl}$ and $\ce{Na2SO4}$: In this equation you will need $\pu{2 M}$ $\ce{KCl}$ and $\pu{1 M}$ $\ce{Na2SO4}$ in order for all the reactants to react. However, you only have $\pu{1 M}$ $\ce{KCl}$ (so $\pu{1 M}$ chloride ions). This means that, as you mentioned, you will need the double concentration for the $\ce{KCl}$ solution to achieve a complete reaction.

So in this case the only way for all $\ce{Na2SO4}$ molecules to react is to add the double amount of $\ce{KCl}$ solution. Then this is a complete reaction : at least one reactant is used up in the reaction.

But say we have a situation in which we have a $\pu{2 M}$ $\ce{KCl}$ solution, but a $\pu{2 M}$ $\ce{Na2SO4}$ solution, then all the $\ce{KCl}$ would react, but still \pu{1 M} of $\ce{Na2SO4}$ will not have reacted.If in your lab assignment it is the meaning that you react all $\ce{Na2SO4}$, then in this case you don't have a complete reaction. Only when a two times larger volume of the $\ce{KCl}$ is added (to 2M $\ce{Na2SO4}$), then you will once again achieve a complete reaction. Furthermore you would just get a solution of solvated ions.

In the case of $\ce{Ca(NO3)2}$ and $\ce{Na2SO4}$: In this equation you will need $\pu{1 M}$ $\ce{Ca(NO3)2}$ and $\pu{1 M}$ $\ce{Na2SO4}$ in order for all the reactants to react. You have $\pu{10 ml}$ of both compounds, each $\pu{1 M}$ in concentration: so in this case there is no need to use a larger volume of aqueous $\ce{Ca(NO3)2}$.


How are the two formulas for enantiomeric excess/optical purity equal? I’ve been given two formulas to use for enantiomeric excess or optical purity:
$$\frac{\text{observed rotation}}{\text{specific rotation of pure enantiomer}}\:\:\:\:\:\times\:100\%\tag{1}$$
$$\frac{|d-l|}{d+l}\:\times\:100\%\tag{2}$$
However, I cannot figure out how they are equal to each other. You could rewrite equation (1) as
$$\frac{\alpha_d-\alpha_l}{[\alpha]}\:\times\:100\%$$
and because $\alpha=[\alpha]cL$,
$$(c_dL-c_lL)\:\times\:100\%$$
However, that’s as far as I got. I can’t see how you would eliminate L, and even doing so would not bring it to resemble the second equation. I would appreciate it if someone could offer a mathematical proof or—if this isn’t even a matter of mathematics at all but rather a conceptual misunderstanding—clarify things for me.

EDIT

$d$ is the mass of the dextrorotary enantiomer and $l$ is the mass of the levorotary enantiomer. $c$ refers to concentration, $L$ the pathlength travelled by light, [$\alpha$] specific rotatio, and $\alpha$ (without the brackets) means observed rotation.

The equations are trivially equal since each is equal to the enantiomeric excess. To prove this, we need only show each equal to the enantiomeric excess.

Start with the second equation. Because both enantiomers have the same molar mass, we can divide both numerator and denominator by the molar mass of the compound and rewrite the equation as $$\text{EE} = \frac{|\text{difference in moles of enantiomers}|}{\text{total moles of enantiomers}} = |\text{difference in mole fractions of enantiomers}|,$$ which is the definition of enantiomeric excess. This equation is not very useful, since most of the time the masses of each of the enantiomers in solution are not known.

The first equation, as you have it, is incorrect. Following Ben Norris's suggestion, we write $$\text{EE} = \frac{|[\alpha]_\text{obs}|}{[\alpha]_\text{pure}} = \frac{|[\alpha]_\text{obs}|}{[\alpha]_\text{pure}} = \frac{|\chi[\alpha]_\text{pure}-(1-\chi)[\alpha]_\text{pure}|}{[\alpha]_\text{pure}} = |\chi-(1-\chi)|,$$
where $\chi$ represents the mole fraction of one enantiomer, and we once again obtain the same result.

Had your original equation been right, $L$ would have cancelled in your derivation and you would have obtained $(c_d-c_l)/(c_d+c_l)$, which is readily brought to equation $(2)$.


What would the van der Waals parameters look like for gaseous metals? So basically I'm trying to do some calculations on laser based machining, and it involved vaporizing a small mass from a target metal workpiece. I was assuming that a certain m mass of the metal vaporizes nearly instantaneously, and the resulting vapor expands from that initial volume.
I was using the van der Waals corrected gas equation, and wasn't sure how to quantify a and b for gaseous metals (take Al for example). I tried searching online, but couldn't find any values for gaseous metals.
Any help or resources I could refer to would be appreciated.

Aside from the metal vapor oxidizing, the pressure of gas over your weld is unlikely to be much different from one atmosphere whereas the critical pressure of almost everything is 10 atmospheres or (often a lot) more.  Generally gases become non-ideal when the pressure is a large fraction of the critical one; we are well below pressures where that occurs.  So the Ideal Gas Law works -- until the metal vapor oxidizes.


What is the precipitate of this reaction of magnesium? $\ce{MgSO4}$, on reaction with $\ce{NH4OH}$ and $\ce{Na2HPO4}$, forms a white crystalline precipitate.
Identify the precipitate.

My approach: I guess the white crystalline precipitate is $\ce{Mg(OH)2}$, since it is white and insoluble in water.

With $\ce{(NH4)2HPO4}$, $\ce{Mg^2+}$ forms a white, crystalline precipitate of magnesium ammonium phosphate hexahydrate:

$$\ce{Mg^2+ + HPO4^2- + NH4^+ + OH- + 5H2O \rightleftharpoons Mg(NH4)PO4 \cdot 6H2O\downarrow}$$

This is a very sensitive test for $\ce{Mg^2+}$; however, calcium, strontium, barium and other heavy metal ions also form precipitates of phosphates under these conditions. $\ce{Mg(NH4)PO4 \cdot 6H2O}$ can be distinguished by the form of its crystals when they are examined under a microscope. It forms rhombic crystals with prismatic shape, and is less microcrystalline than other earth alkali metal phosphates (sources: 1, 2).


Acidic nature of benzene and other alkenes Is benzene acidic? It has $sp^2$ hybridization so doen it mean that it is acidic in nature? Are alkenes acidic?

A sp2-hybridized carbon is more electronegative than an sp3-hybridized carbon, and you would expect it to be more acidic.

Experiments support this prediction.  Benzene has a pKa of ~ 43, compared to alkanes, which have a pKa of ~ 50.  You can prepare lithioaromatics from bromo/iodoaromatics and butyllithium via metal-halogen exchange.

The situation with alkenes is slightly more complicated; the pKa of vinylic protons is in the same ballpark as of aromatic protons, but allylic protons are slighty more acidic.

Consult any of the usual pKa tables on the web.


"Don't pour oil into lye, pour lye into oil" I was making soap last weekend and I got distracted, and thence I put too much lye into my oils blend. Basically, the whole soap paste is unusable to make soap because the cure will never be enough to neutralize all the lye contained in the soap paste.

I wanted to recalculate the lye quantity and add the quantity of oil needed to salvage my soap paste but I remembered a piece of advice I read on the recipe: never pour oil into lye, always pour lye into the oils.

Why is that?

At first I thought it was because of similar reasons to why one shouldn't pour water into acid (as these two substances react strongly when put together and the hydration reaction may provoke projections). It has been a while since my high school chemistry lessons but I think that a similar thing happens with water and basis. However, oil has pretty much nothing in common with water when it comes to chemistry. Another reason may be that what a user from a soap making community calls the "splash potential":

  You are more likely to splash whatever is being poured into than what is being poured. You do not want lye water splashing out onto you and your surroundings.

This sounds more plausible but it isn't a scientific explanation. So, is there a scientific reason behind the restriction: "do not pour oil into lye"?

Lye is denser and tends to pour through the lighter oil, improving mixing (and possibly cutting down on the splattering of lye).  The same is true of concentrated mineral acids added to water.

In any case the hazards are minimized by also adding slowly and with stirring.


Why is HCl the acid of choice for swimming pools? Why is muriatic acid the go to acid to put into pools? Does it work better than nitric acid, sulfuric acid, or another strong acid? I mean they all will add the same amount of H+ right? Note, we use some cyanuric acid as well, but mainly HCl for adjusting pH.

There are several reasons not to consider these acids to be interchangeable for use in swimming pools. A total answer covers aspects of chemistry, biology, safety, availability, and cost. As hydrochloric, nitric, and sulfuric acids are all strong acids and can provide the desired acidification, they are very different when you consider their respective counter ions. Do you want chloride, nitrate, or sulfate in your pool water?

Safety First:

Hydrochloric, nitric, and sulfuric acid are all strong acids that are capable of causing chemical burns and present inhalation hazard from their fumes.
Additional safety information can be obtained from PubChem.:
Hydrochloric acid
Sulfuric acid
Nitric acid

Chemistry:

Nitric and sulfuric acids are more corrosive to copper and stainless steel. Copper (I) chloride is significantly less soluble in water than copper salts of nitric and sulfuric acids. Swimming pool heaters are likely to contain copper pipe.
As an added benefit for salt pools, which use an electrochemical chlorine generator, the Cl- ions are converted to chlorine for disinfecting the pool water.

Biology:

As pointed out in another response, nitrates can promote algae growth.

Availability and Cost:

Hydrochloric acid is commonly available as muriatic acid (31.5% HCl, ~10 molar) from home centers, building and construction suppliers, and hardware stores, because it is used in many construction-related activities, with cost ranging from \$5 - 9 (USD) per US gallon. A gallon will contain approximately 37 mole equivalents of H+.
Sulfuric acid is available from auto parts retailers as battery acid (37% H2SO4) and costs approximately \$10 (USD) per US quart ($40 per gallon). A gallon of battery acid will contain approximately 36 mole equivalents of H+ (adjusting for normality). Sulfuric acid costs 4-8 times more than hydrochloric acid for equivalent pH reducing capacity.
Nitric acid is less readily available to the consumer market. One on-line supplier offered reagent grade (70%) for \$125 (USD) per liter.

Added Note:

Cyanuric acid is a weak acid and serves as a stabilizer. Quoting from PubChem:

  Cyanuric acid is used in swimming pools to lower the rate of photochemical reduction of chlorine, hypochlorous acid and hypochlorite ion.


Amine addition to nitro group What is the mechanisms of this reaction? How feasible is this reaction? Can it happen under relatively mild conditions if it is intermolecular?

A reasonable mechanism for the transformation is shown below. Since this is under basic conditions, I've written a mechanism that avoids cationic intermediates. Deprotonating the hydrazine gives a very nucleophilic nitrogen that can attack the nitro group. Proton transfer gives a nitrogen anion, which can kick out hydroxide. Two additional proton transfer steps lead to the product, HOBt.

A SciFinder search didn't reveal any examples of the intermolecular variant of this reaction, although I didn't find this exact intramolecular reaction either.


Relationship between the symmetry number of a molecule as used in rotational spectroscopy and point group While doing a problem set I noticed that the symmetry number of a molecule turns out (usually) to be half the number of symmetry elements that the point group the molecule belongs to.

When I say symmetry number I refer to the symmetry number used in rotational spectroscopy — i.e. the number of indistinguishable orientations of the molecule. This accounts for suitable symmetrisation of the rotational wavefunction and the nuclear wavefunction of the molecule.

For example benzene belongs to point group $D_\mathrm{6h}$, which has 24 symmetry elements. It also has symmetry number 12, i.e. 1/2 the number of symmetry elements in it's point group.

Another few examples:

Water. Point group $C_\mathrm{2v}$, with 4 symmetry elements. Has 2 as symmetry number.
Ammonia. Point group $C_\mathrm{3v}$ with 6 symmetry elements. 3 as symmetry number.
Ethene. Point group $D_\mathrm{2h}$ with 8 symmetry elements. 4 as symmetry number.

This seems to be a general trend. It sort of makes sense intuitively that the 2 should have a relationship, but I can't formalise the reason for this.

Obviously for diatomics/linear molecules this doesn't apply. This is because rotational motion around the axis of symmetry is not physical — moment of inertia around this axis is zero, as all mass along axis. Thus there are no rotational levels associated with this axis. But ignoring diatomics/linear molecules, where we have neglected/ignored an axis of rotation, the above relationship seems to hold.

Why is this the case? I'd appreciate both a mathematical derivation and a more intuitive argument.

This is not in general true
Consider molecules a point group not containing inversion symmetry, e.g. $C_2$ hydrogen peroxide

The $C_2$ group has only two elements, $E$ and $C_2$, and the $C_2$ rotation operation maps between two identical arrangements of atoms.  Both the symmetry number and order of the group are 2.
Matrix representations
All rotations in 3D space can be represented by a orthogonal 3x3 matrix with determinant 1, and the composition of rotations about the origin can be represented by matrix multiplication of the transformation.
More formally: the group SO$(3)$ (3D rotations) is isomorphic to the group of orthogonal real 3x3 matrices with determinant 1 under multiplication.
The number of symmetry elements in a point group formed from rotations (and which hence preserve chirality) will be the same as the number which can be represented by such matrices and the same as the number of equivalent orientations of the molecule (the symmetry number) as they are all representation of the same thing - 3D rotations.
Inversions
The group of all transformations that keep the origin fixed, including reflections as well as rotations is O$(3)$ - or the product of SO$(3)$ with the set $\lbrace I,-I\rbrace$ or the inversion operation.  It maps rotations onto improper rotations (which include reflections - improper rotation by 0 degrees) and hence all symmetries in 3D space.
All point groups with inversion symmetry, such as the 3 listed in the question, contains this $-I$ element.  $-I$ will multiple all the elements of the rotation matrix by $-1$ and create an orthogonal 3x3 matrix with determinant -1.  By composition with each of the rotations elements of the point group it creates another element - an inversion, improper rotation or reflection - doubling the size of the group without increasing the symmetry number.
Conclusion: chirality matters
This rule is true for non-chiral groups.  You can invert the molecule to get another symmetric copy for every rotationally equivalent copy, so the order of the group is twice the symmetry number.
For chiral groups that will create the enantiomer which isn't symmetrically equivalent so the order of the group and symmetry number are the same.


Isotropic Fermi Contact Couplings vs. Dipolar Coupling Constant I need to calculate a Dipolar Coupling Constants for some fancy triplet molecule that has been measured by EPR spectroscopy.

My "normal" output from Gaussian gives me the following:

Isotropic Fermi Contact Couplings
Spin Dipole Couplings
Anisotropic Spin Dipole Couplings in Principal Axis System

Am I right that none of these are Dipolar Coupling Constants? And If so ... how can I compute them either directly with the program or by hand through some formula?

The output sections you mention are the major components of the electron-nuclear hyperfine interaction. There is also an orbital contribution, but this tends to be small compared to the Fermi contact and spin dipole terms, unless you go down the periodic table.

These are the terms in the effective spin Hamiltonian that contribute to the hyperfine interaction (from Wikipedia):

$$
\begin{align}
\hat{H}_{\textrm{hyperfine}} = & 2 g_{I} \mu_{N} \mu_{\mathrm{B}} \frac{\mu_{0}}{4\pi} \frac{1}{L_{z}} \sum_{i}^{N_{\mathrm{elec}}} \frac{\hat{l}_{zi}}{r_{i}^{3}} \mathbf{I} \cdot \mathbf{L} \\
&+ g_{I} \mu_{N} g_{e} \mu_{\mathrm{B}} \frac{\mu_{0}}{4\pi} \frac{1}{S_{z}} \sum_{i}^{N_{\mathrm{elec}}} \frac{\hat{s}_{zi}}{r_{i}^{3}} \left\{ 3(\mathbf{I}\cdot\hat{\mathbf{r}})(\mathbf{S}\cdot\hat{\mathbf{r}}) - \mathbf{I} \cdot \mathbf{S} \right\} \\
&+ \frac{2}{3} g_{I} \mu_{N} g_{e} \mu_{\mathrm{B}} \mu_{0} \frac{1}{S_{z}} \sum_{i}^{N_{\mathrm{elec}}} \hat{s}_{zi} \delta^{3}(\hat{\mathbf{r}}_{i}) \mathbf{I} \cdot \mathbf{S}
\end{align}
$$

The first term is the orbital contribution due to spin-orbit coupling, the second term is the spin dipole term due to the electrons and nuclei behaving as dipoles in the presence of each other's magnetic fields, and the third term is the Fermi contact term arising from the non-zero probability of the unpaired electron(s) being inside the nucleus.

Isotropic Fermi Contact Couplings are the eigenvalues of the Fermi contact part of the effective spin Hamiltonian. Because of the presence of $\delta^{3}(\hat{\mathbf{r}}_{i})$, there is only one value, hence the "isotropic" descriptor.

Spin Dipole Couplings and Anisotropic Spin Dipole Couplings in Principal Axis System are two different presentations of the spin dipole term. Spin Dipole Couplings shows the eigenvalues of the spin dipole part of the effective spin Hamiltonian (in a.u.). Because this is a symmetric 3x3 tensor, only the 6 unique elements are shown. Diagonalizing the tensor gives the 3 principal values Baa, Bbb, Bcc as the eigenvalues and the vectors from the origin for each principal value as the eigenvectors. These are the values you're interested in, most commonly reported in MHz by the computational/theoretical chemistry community and in Gauss by the spectroscopy community.


Why is there a lone pair in thionyl fluoride? Why is there a lone pair in $\ce{SOF2}$? I drew its structure, which according to me should look like this:

Why is there a lone pair on sulfur? Isn't its octet complete? If yes, why should it expand its octet and gain more electrons?

When you want to write a Lewis structure, I suggest you start from considering how many valence electrons each atom has. In your case, you would have:

S = 6
F = 7; total = 14
O = 6

The final total would be 14 + 6 + 6 = 26. The structure you guessed for is correct, sulfur is the central atom. You start by drawing a single bond per each atom, getting something like this:

F
|
S–O
|
F

Then, you can draw a double bond for oxygen:

F
|
S=O
|
F

If you do the math, by counting the electrons that you put in the previous structure, you would have 2 + 2 + 4 (two S-F bonds and a S=O) = 8.
Then we will add in the lone pairs: three on each halogen, two for oxygen, and one for sulfur. Therefore we add 3 × 2 × 2 = 12 electrons (three lone pairs on each fluorine) plus 4 electrons on O (the two lone pairs), and we have a total of 24. The final 2 , adding up to 26, come from the lone pair on sulfur.

This makes sense because sulfur is in period 3, so it is possible for it to have more than 8 electrons: in other words, the octet rule applies not only for 8 electrons, if that makes sense.


What is the difference between a bond, force, and interaction? I am trying to explain why some solids are more soluble in water than others:

So the energy released from the formation of bonds between the solid solute molecules and water molecules compensate the energy needed to break the intermolecular bonds between water molecules and the intermolecular bonds between the solute molecules.

The problem is that I'm not sure if I have used the correct terminology, for example I have never used intermolecular 'bonds', I've only used intermolecular forces before.

Here is how I think of the situation.

The term "bond" is generally reserved to describe an attractive, stabilizing interaction between atoms or groups in the same molecule or in different molecules, that has a magnitude of at least several kcal/mol.  Everything from hydrogen bonds at the low end up to full chemical bonds at the high magnitude end are included.
The terms "force" and "interaction" can be used to describe any attractive or repulsive situation, of any magnitude, between atoms or molecules. So these terms would be used to describe the van der Waals "interaction", where the "forces" involved are much less than those involved in bonding.  However, the terms "force" and "bonding" can also be used to describe situations involving larger magnitude interactions such as, "bringing the atoms together led to a bonding interaction."

"Force" and "interaction" can be used to describe any situation - attractive, repulsive, high energy, low energy; whereas "bonding" is typically reserved to describe stabilizing, higher energy interactions.


Why is the entropy of deuterium greater than that of hydrogen gas Why is the entropy of deuterium more than that of hydrogen gas?

I thought about it but couldn't figure out a reason for why that would happen. Why would the number of neutrons in the nucleus affect the entropy of the gas?

The entropy can be split into three parts arising from translational, rotational and vibrational energies and using statistical mechanics each of there can be calculated from measure properties. The gases $\ce{H2}$ and $\ce{D2}$ differ in each of these three energies due to their differing mass and nuclear spin. The experimental values are hydrogen $130.7$ and deuterium $144.9 ~\pu{J mol^{-1}K^{-1}}$

The translational entropy is given by the Sakur-Tetrode equation,
$$S_{trans} = R\left(\frac{5}{2}+\ln \left(\frac{V}{N} \right)+ \frac{3}{2}\ln\left( \frac{2\pi mkT}{h^2} \right )\right)$$
where m is the mass, V volume, N Avogadro’s number, h the Planck constant and k the Boltzmann constant. Assuming that the gases behave ideally the difference in entropy is small as it depends only on the mass difference all other terms being the same, so $S_{trans}$ is greater for $\ce{D2}$ than $\ce{H2}$ by $(3R/2)\ln(2) \approx 8.64~ \pu{ J mol^{-1}K^{-1}}$.

The vibrational entropy can be ignored at low temperature, for example at room temperature $kT \approx 210~ \pu{cm^{-1}}$ whereas the vibrational quantum are $4395$ and $3118~ \pu{cm^{-1}}$ for $\ce{H2}$ and $\ce{D2}$ respectively.

The contribution from the rotational motion depends both on the mass, as the moment of inertia determines the rotational constant, and on the nuclear spin which via the symmetry properties of Fermions and Bosons affects the weighting of the rotational energy levels.

To calculate the entropy the partition function $Z$ is required. Once this is obtained the entropy is calculated as
$$ S = k\ln(Z)+\frac{U}{T}$$
where the internal energy $U$ is
$$U= kT^2\left ( \frac{\partial \ln(Z)}{\partial T} \right)_V$$

The partition function for odd mass number, assuming a rigid rotor model, is (see this answer for details Why do spin isomers of hydrogen (ortho and para hydrogen) change their nuclear spin with temperature variance? )
$$Z_{odd}= \frac{g_n(g_n-1)}{2}\sum _{J=0,2,..} (2J+1)\exp(-B_eJ(J+1)/(k_BT)) +\\ \frac{g_n(g_n+1)}{2}\sum_{J=1,3,..} (2J+1)\exp(-B_eJ(J+1)/(k_BT)) $$

where $g_n = 2s_n+1$ where $s_n$ is the nuclear spin, $1/2$ for hydrogen and $1$ for deuterium. The rotational quantum numbers are J and $B_e$ the rotational constant assumed to be the same for all J but which differs between hydrogen ($60.85 \pu{cm^{-1}}$) and deuterium ($30.44 \pu{cm^{-1}}$) by a factor of two. (These rotational constants are not typical, values for most molecules which are heavier and have longer bonds are far smaller, for example, $\ce{F2}, B_e =1 \pu{cm^{-1}}$)

The partition function for even mass number
$$Z_{even}= \frac{g_n(g_n+1)}{2}\sum _{J=0,2,..} (2J+1)\exp(-B_eJ(J+1)/(k_BT)) +\\ \frac{g_n(g_n-1)}{2}\sum_{J=1,3,..} (2J+1)\exp(-B_eJ(J+1)/(k_BT)) $$
where the only difference is that due to the weighting (those terms in $g_n$ ) in from of the summations which run over even and odd values of J. Hydrogen has weightings $1$ and $3$, deuterium $6$  and $2$ respectively in the equations above.
Thus Z for hydrogen with $g_n=2$ is
$$Z=1\cdot\Sigma (even~J) + 3\cdot\Sigma(odd~J)$$
and for deuterium, $g_n=3$
$$Z=6\cdot\Sigma (even~J) + 2\cdot\Sigma(odd~J)$$

At low temperatures the argument in the exponential is large meaning that the exponential is small. e.g. $exp(-large ~number) \rightarrow 0$  for all J terms except $J=0$ where the exponential is $1$. Thus in hydrogen the limiting partition function at low temperatures is $Z_{T\rightarrow 0}= 1 + 3(small~number) \approx 1$ and the limiting entropy is $S_{T\rightarrow 0 } R\ln(1) \rightarrow 0$; (and $U(0)=0$)

In deuterium the limit is $Z_{T\rightarrow 0 } = 6 + 2(small~ number) \rightarrow 6 $ and thus the limiting entropy is $S_{T\rightarrow 0 } R\ln(6) \rightarrow 14.9~ \pu{J mol^{-1}K^{-1}}$.
The change of rotational entropy with temperature is shown in the figure.


Monoepoxidation of 1,2-dimethylcyclohexa-1,4-diene using m-CPBA What is the product formed when 1,2-dimethylcyclohexa-1,4-diene undergoes epoxidation with one equivalent of m-CPBA (meta-chloroperbenzoic acid)?

In the epoxidation mechanism, the π-electrons of the double bond attack m-CPBA:

So, methyl groups attached to the double bond should increase its nucleophilicity. On the other hand, since it is a nucleophilic attack, the methyl groups should also cause considerable steric hindrance. So what should the answer be?

The interaction of the alkene with m-CPBA is primarily nucleophilic/electrophilic in nature, i.e. the π MO of the alkene attacks the O–O σ* orbital. Therefore, the more electron-rich alkene (i.e. more highly substituted alkene) preferentially reacts.

With methyl groups there probably isn't too much steric hindrance. m-CPBA approaches above/below the alkene and doesn't directly attack over the methyl groups.

Indeed, the monoepoxidation of 1,2-dimethylcyclohexa-1,4-diene is reported to proceed selectively at the more electron-rich tetrasubstituted double bond (J. Org. Chem. 2006, 71 (4), 1653–1657):


Is KF the most ionic compound? I saw somewhere (can't recall where) that KF is the most ionic compound.  I expected CsF.  Does the greater polarizability of Cs allow it to more easily form covalent bonds compared to K?  Does this overcome the fact that K being in n = 4 should bond better with F in n = 2?

Yes, this is a subtle thing.

Using the Pauling electronegativities, one would expect CsF to have the larger electronegativity difference (3.2). So in principal, it should be "more ionic."

Unfortunately, an ionic bond requires separating charge, so $\ce{Cs+F-}$. The problem is that $\ce{Cs+}$ is much larger than $\ce{K+}$ and so the dipole moment for a fully ionic $\ce{Cs+F-}$ would be much larger.

We find that the charge isn't fully separated, likely because of the need to stabilize the large dipole moment.

So some people plot a curve of "% ionic character" as determined by the actual dipole moment vs. the expected dipole moment (i.e, full charge separation). We find that this curve reaches a limit ~75-85% ionic character.

IIRC, I think on that basis LiF is more Ionic than KF.

For more, I really like Bruce Robinson's lecture notes:
http://courses.washington.edu/bhrchem/c152/Lec23.pdf via the Internet Archive


Acid catalysed synthesis of ethyl ethoxymethylenemalonate One synthesis of quinolones begins with the formation of an ethyl ethoxymethylenemalonate, as seen in this Organic Syntheses paper.

I've been asked if the malonate derivative would be formed if methylmalonate was treated with trimethyl orthoformate with a catalytic amount of $\mathrm{H^+}$, without heating. The nature of the solvent isn't mentioned, but I guess the reaction is carried out in a non-aqueous medium with an ion-exchange resin, since I think trimethyl orthoformate acts as a water trap.

My hunch is that there shouldn't be a significant quantity of product formed. We're dealing with an activated methylene, why not simply use a base and then carry on with our addition? I can't imagine how, under normal circumstances, the elimination product ethyl ethoxymethylenemalonate will be formed.

How can this reaction happen in a mechanistically reasonable way?

Ok, I will try my hand on the mechanism. In my opinion the reaction should work with a catalytic amount of $\ce{H+}$ since the key step for the reaction is the hydrolysis of the orthoester. This is an entropically very favourable reaction since from one reactant molecule you get two product molecules.

The resulting molecule is extremely reactive. So it can easily react with the enolized malonate. The enolization might be facilitated by the $\ce{ZnCl2}$ and/or the cat. $\ce{H+}$.
(I'm not entirely sure whether the $\ce{ZnCl2}$ really plays the role I've given it below in this reaction. So, if anyone knows good arguments against it, please let me know.)

The protonated product can then take part in an $\mathrm{E}1$ elimination reaction to yield ethyl ethoxymethylenemalonate. The carbocation which results when $\ce{EtOH}$ leaves the molecule is stabilized by the neighboring ethoxy group. The hydrogen on the central carbon atom is very acidic, since it is placed between three carboxylate groups, so it can even be removed by rather non-basic molecules, e.g. by an $\ce{EtOH}$, a malonate or an othoester molecule.

So much for the acid catalysed mechanism. Now for the question, why one doesn't simply use a base to facilitate the addition of the activated methylene compound. Well, this is due to the fact, that orthoesters are quite stable against basic hydrolysis. So, under basic conditions you don't get the carbonyl component of the addition reaction.


Do two Miller planes need to have similar arrangement of atoms? For two Miller planes, do I need to have the same arrangement of atoms in both of them?

Does it need to look like the same plane translated a little?

I have this question because if this is not the case, then how do you do the calculations for interference, etc, using Bragg's law, where we assume the atom of the bottom plane is directly below the atom of the top plane?

In answer to you first question, no if hkl differ, yes if they do not. I try to explain below.

In a crystal the unit cell defines the repeating unit. Inside the unit cell the atoms are arranged as they are in the molecular structure with the molecule being at the same angle and position within each of the unit cells. Each atom scatters the x-ray radiation depending on how many electrons the atom has and on the position the atom has in the unit cell. The many repeating identical unit cell structures cause constructive interference and so spots are observed on the x-ray detector. Thus the structure is determined from the intensity of the spots on the detector not their position.

The hkl planes are a construction we devise to allow determination of the x-ray amplitudes because the constructive interference that causes scattering has to occur from atoms in the repeated positions as determined by the unit cell. (There are thousands to millions of unit cells in a crystal). Thus we define planes starting at  hkl = (0 0 0) and continue to larger values such as  (1 2 0 ) etc. etc. (100 50 1) etc. (Negative values have a line above the number $\bar 3$ for example.)

Suppose we choose the plane (1 2 3), this is  parallel to (2 4 6) to (4 8 12) and so on and these planes are separated by a distance we will call $d_{hkl}$. We can reduce all these  planes to (1 2 3) and the reason for this is that the phase of the x-ray wave from plane (2 4 6) is $2\pi$ from that of (1 2 3) and that of (4 8 12) is $4\pi$ shifted. But as the phase repeats itself over $2\pi$ all these larger phases are simply the same as $2\pi$. A phase change of $2\pi$ is equivalent to a path difference of $\lambda$ or a whole wavelength, which might be easier to envisage as then waves add together exactly.

Now, if an atom is found bang on the plane (1 2 3) it will have a phase of $2\pi$ (or path-length change of $\lambda$) with respect to the origin so has the maximum scattering intensity possible for that plane. However, consider that the atom in not on the (1 2 3) plane, but at a position $x$ away (and so between two (1 2 3) planes) then the phase it has is now $\displaystyle \frac{x}{d_{hkl}}2\pi $ and so the scattered x-ray waves are out of phase by this amount and so the scattered amplitude is reduced. Now, of course, every the atom in the molecule will be at some distance from one of the parallel (1 2 3) planes and so will have its own particular scattering amplitude to add to the total for that plane.

The same reasoning applies to each of the unique hkl planes and to each atom. It is the un-ravelling of these amplitudes that produces the x-ray diffraction structure and it is made more complicated because the x-ray detectors measure the intensity of the scattered light (which is  amplitude squared (as $I^*I$)) which means that the phase information is lost. This causes great difficulties in solving the x-ray structure.


Simulation of a particle and chemical molecule I am currently studying the basics of simulation and want to try out some experiments. It is a known fact that gold nanoparticles can bind to chemical groups which contains thiol (R–SH group). I want to simulate the binding of a thiol group to a gold nanoparticle. How can I do this? Where do I have to start? Which software can be used for this purpose?

Density Functional Theory is worth investigating if you're looking for a fairly efficient way of investigating these interactions at an electronic level. The term 'nanoparticle' is wonderfully vague as it encompasses two to three orders of size magnitude - the size of the nanoparticle will inform the specific choice of method:

Small (as in, only a couple of atoms) nanoparticles are amenable to computation with atom-centered basis sets in programs such as Gaussian, ADF, ORCA, GAMESS and so on. The latter two may be acquired free for academic use. If your particle/ligand complex is highly symmetric and/or you have supercomputer time you might be able to work with a moderately large particle. I recall a paper abstract where researchers did TDDFT on a 120 atom silver nanoparticle in ADF (here it is!), aided by high symmetry.

Large systems, where it's a) computationally unreasonable to model the whole particle and b) to a first-order approximation the ligand-particle interaction can be treated as occurring on an infinite plane or edge, may be more amenable to periodic DFT using plane-wave basis sets (PW-DFT). There are many codes available, ranging from free to very expensive. Off the top of my head, VASP, SIESTA, Quantum Espresso, Abinit and CASTEP are fairly popular.

There's also an atom-centered periodic franken-version of ADF called ADF-BAND which can model 1D to 3D periodic systems, which is worth mentioning. It is not particularly well documented, however.


Does CN4 with the following structure exist? I am a mathematician, not a chemist, so my questions are rather combinatorial. I am intrigued by the possibility of the following binary compound:
$$\ce{\overset{-}{N}=\overset{+}{N}=C=\overset{+}{N}=\overset{-}{N}}$$
which ought to be isoelectronic with carbon suboxide, and thus possibly stable. Does this compound exist, even in interstellar space?

All kinds of weird molecules exist in space. The molecule is formally known as bis(diazo)methane according to chemdraw. By looking at the structure it contains a Diazomethane moiety. Resonance would also take place to introduce the charges at both nitrogen atoms.  The structure doesn't seem to exist on any chemical database nor any official papers have been released relating to the exact structure. It may exist considering the fact that all atoms have a total of 8 surrounding electrons.
I hope this provides some clarity


How to use Lewis theory to differentiate between an acid and a base? According to Lewis theory a material is an acid if it accepts an electron pair, and a material is base if it donates an electron pair. But if we compare the structures of $\ce{HNO3}$ and $\ce{NaOH}$, i.e.

$$
\ce{H-\underset{.\!.}{\overset{.\!.}{O}}-\underset{\underset{\huge :O:}{||}}{\overset{+}{N}}-\underset{.\!.}{\overset{.\!.}{O}^}\!:^-}
\quad
\text{and}
\quad
\ce{[Na]+ [:\!\underset{.\!.}{\overset{.\!.}{O}}H]-}
$$

there are so many lone pair of electrons which can be shared, and in both structures there is a positive atom, so they can accept a lone pair. How can I say that one of them is acid and the other one is a base?

First of all, within the Lewis framework, you may have substances that can act as both acid and base (though most of the time as one or the other, not both at the same time).$^{1}$ The adjective for such substances is "amphoteric".

It comes down to a bit of chemical intuition to decide which is which. It is true that almost anything that has a lone pair can act as a base - though it may take a strong acid to form an actual Lewis pair. For acids, one should look for an obvious gap in the electron cloud ($\ce{AlCl3}$) or candidates for complex formation ($\ce{FeCl3}$).$^2$

For nitric acid, the situation is pretty clear: the hindrance provided by three oxygen atoms around the central nitrogen prevents access, so it will hardly act as a Lewis acid. It can act as a base, however, weak Brønsted bases are usually also weak Lewis bases because they are content on their own.

For sodium hydroxide, one must realize that in an aqueous medium, the two ions will fully dissociate. The hydroxide anion will act as a Lewis base, but the sodium cation with its full outer electron shell will hardly act as an acid.$^3$

$^1$ Adducts of e.g. $\ce{CO2}$ in frustrated Lewis pairs, wherein the $\ce{CO2}$ acts as a sort-of bridge between the two Lewis sites, are an example of when "both at the same time" can occur.

$^2$ I am aware that these compounds will form dimers or more complex structures in reality. If you are able to spot this behavior and understand why, you are already able to spot Lewis acids.

$^3$ Note that the sodium cation in water will be solvated, i.e. some water molecules will orient their lone pairs towards the ion. The border between solvation and Lewis adduct formation can be blurry.


Safe coating to make a 3D printed respirator mask airtight What is a good example of something safe that could be applied as a coating layer to the walls of a 3D printed respirator mask to make them truly airtight (seal pores and fissures) while (ideally) satisfying the following criteria?:

Not air permeable at the pressures created when a person inhales
No (or low) VOC out-gassing from the coating
Does not impose health risks to the respirator wearer
Can be applied by dipping (or maybe spraying)
Forms a thin layer that is flexible enough not to crack during normal use of the mask

Here are some materials I have been considering for the job:

White glue (school PVA glue): Seems safe. Is is airtight?
Water based liquid rubber: some SDS suggest it's safe. It's marketed as watertight (but is it AIRtight?)
Some sort of common paint?

If someone here could suggest some potential coating materials or knowledge bases where I could check their non-permeability/safety, I would be really grateful.

As I'm not a native speaker, I apologize for any broken grammar.

The problem: It is well known that 3D printed objects are usually not exactly airtight, due to micro-fissures. To make things worse, I ordered a DIY 3D printed respirator mask (composed of ABS plastic) that is visibly permeable to air (one can literally see the tiny gaps all across it's walls' surfaces through which air and viruses could come in when the wearer inhales).

P.S.: Silicone caulk might not be an option, since it's too viscous and would be a mess to apply to the outside of the respirator, rendering an unprofessional finishing. I should note that I could possibly apply it only to the inside of the walls, but I intend to coat all surfaces, just to be extra safe.

You need something flexible and airtight. Could neoprene fabric, like it's used for diving suits do the trick, esp. for prototyping? Otherwise, I thought of injection moulded silicone, as it's used in diving goggles.


Calculating internal energy of methane I've recently found out my calculated value of internal energy of methane largely deviates from the ab initio output (at $\pu{1000 K}$ $0.058~\text{Hartree} = \pu{152.3 kJ mol-1}$). I used HF method, I know it's not highly-accurate, but I just need what internal energy should roughly be.

The way how I calculate:

$$H = U + PV = U + Nk_\mathrm{B}T$$

where $U$ is the internal energy, $H$ is the enthalpy.
In this case, $\Delta_\mathrm{f}H = \pu{-74.9 kJ mol-1}$ (I found the enthalpy value of methane from heat of formation table on Wikipedia).

So, I simply substitute the temperature $T = \pu{1000 K}$ into the equation and I got $\pu{-83.2 kJ mol-1}$. First, the internal energy of a system cannot be negative. Second, even it's positive, why it deviates that much from ab initio calculations? Can someone give me a hint?

I believe the answer to your question is summarized well by the documentation of Psi4, an open source quantum chemistry package which can do similar calculations. I am fairly certain this is the problem you're running into but I could be wrong.

  It is important to know that PSI4, like any other quantum chemistry program, does not compute the usual enthalpies, entropies, or Gibbs free energies of formation provided by most reference books. Instead, quantum chemistry programs compute “absolute” thermodynamic properties relative to infinitely separated nuclei and electrons, not “formation” values relative to elements in their standard states. If you are computing thermodynamic differences, like a reaction enthalpy computed as the enthalpy of the products minus the enthalpy of the reactants, then these “absolute” enthalpies are perfectly valid and usable. However, they cannot be mixed and matched with enthalpies of formation from reference books, since the zero of energy is not the same. Additionally, the “thermal energies” reported in kcal/mol are the finite-temperature corrections to the electronic total energy, and not the overall thermal energies themselves. If in doubt, use the reported Total Energies in Hartree/particle.

Link to documentation

In other words, you're value may be perfectly reasonable once you use it to calculate a change in energy, but as an absolute energy both values are fairly meaningless because they both pick a reasonable, but arbitrary, zero of energy.

This is also why you're internal energy can be negative. This simply means it has a smaller internal energy than whatever the reference internal energy is. I didn't check your work, but it is possible to have a negative energy depending on the reference value.

Also, here is the very thorough documentation (with examples!) from Gaussian. They also note,

  This is generally not the same as the output from Gaussian for a calculation on an isolated gas-phase atom. These values are referenced to the standard states of the elements. For example, the value for hydrogen atom is 1.01 kcal/mol. This is ($H_{H_2}^{\circ}(298K)- H_{H_2}^{\circ}(0K))/2$, and not ($H_{H}^{\circ}(298K)- H_{H}^{\circ}(0K))/2$ , which is what Gaussian calculates.

So you're going to have the same problem. Read through there and I think it tells you how to make the correction. I think you just need to inclue some zero-point energies.


Retention time and dead time Why does ts start just after tm ?
Does the analyte starts spending time in the stationary phase only after the mobile phase has passed and reached the detector ?
Wouldn't there be an overlap between the two ?

The column is filled with the carrier (liquid or gas) before the sample is injected. Thus if there is no interaction between the sample and the column, then the fastest that the sample can get to the detector is the dead time denoted by $t_M$ in the diagram. If the sample does interact with the column, then it is retarded by a time $t_S$  shown in the diagram. Thus the total retention time from injection to detection is $t_M + t_S$.


Why do Ketones Have Lower Wavenumbers than Esters? The average wavenumber for a ketone is about $\pu{1720 cm-1}$ and the average wavenumber for an ester is about $\pu{1740 cm-1}$. This, however, does not make sense, as the carbonyl group of an ester should have a greater single bond character than the ketone due to resonance from the adjacent oxygen atom. This greater single bond character should thus result in a lower wavenumber for the ester, but it does not. Is there an explanation for this?

For an undergraduate starts to learn IR spectroscopy, the stretching frequency of any $A-B$ chemical bond, $\over\nu$ (in $\pu{cm-1}$) can be calculated by using following equation:
$${\bar\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}= \frac{1}{2\pi c}\sqrt{\frac{k(m_A+m_B)}{(m_Am_B}}$$
where $k=\text{the force constant of the bond}=\text{bond strength}$, $m_A= \text{mass of atom  }A$, $m_B= \text{mass of atom  }B$, and $c= \text{speed of light}$. The equation shows $\bar\nu$ is at least depends on two factors, reduced mass $\mu$ and the force constant of the bond $k$.

The dependency on $\mu$ would be explained by the difference in stretching frequency of $\ce{C-H}$ ($\pu{\approx 3000 cm-1}$) and $\ce{C-D}$ ($\pu{\approx 2200 cm-1}$). The force constants of $\ce{C-H}$ and $\ce{C-D}$ are approximately equal.

The most important fact when compared to different carbonyl bond stretching frequencies is force constants of the bonds of interest. For example, ring strain in a cyclic ketone usually increases the $\ce{C=O}$ stretching frequency. That of cycloheptanone is $\pu{\approx 1702 cm-1}$, cyclohexanone is $\pu{\approx 1714 cm-1}$, cyclopentanone is $\pu{\approx 1747 cm-1}$, and cyclobutanone is $\pu{\approx 1783 cm-1}$. Therefore, we can generally conclude that the stretching frequency of the bond increase with the increase of the reactivity of carbonyl bond. Likewise, reactivity of carbonyl compounds in increasing order is: acid chlorides ($\pu{1780-1820 cm-1}$) > acid anhydrides ($\pu{\approx 1760 cm-1}$ and $\pu{\approx 1810 cm-1}$) > esters ($\pu{1730-1750 cm-1}$) > aldehydes ($\pu{1720-1740 cm-1}$) > ketones ($\pu{1705-1725 cm-1}$) > carboxylic acids ($\pu{1700-17205 cm-1}$) > acid amides ($\pu{1630-1680 cm-1}$). The reduction of double bond character in ester (average bond length of methyl acetate is $\pu{\approx 123.2 pm}$) is minimal but has a better leaving group than in ketones (average bond length of acetaldehyde is $\pu{\approx 123.1 pm}$) when subjected to react.


How to determine the mole fraction of the liquid phase from a given mole fraction of the vapour phase for a benzene/toluene mixture? A quantity of benzene, $\ce{C6H6}$, and toluene, $\ce{C6H5CH3}$, is placed in a $\pu{1 L}$ evacuated vessel at $\pu{25 ^\circ C}$. At equilibrium, a small volume of liquid is visible at the bottom of the container. A sample of the vapour phase is analysed and found to contain $\pu{53 mol-\%}$ benzene.
  What is the mole fraction of benzene in the liquid phase?

  Vapour pressures at $\pu{25 ^\circ C}$: $P_\mathrm{vap}(\text{benzene})= \pu{0.125 atm}$, $P_\mathrm{vap}(\text{toluene}) = \pu{0.037 atm}$

First of all, I am not even certain if this a Raoult's law problem. I just assumed so, since Henry's constant was not given in the problem.

First I'm getting the total liquid pressure using Raoult's law.
\begin{align}
P(\text{benzene}) &= X(\text{benzene}) \cdot P_\mathrm{vap}(\text{benzene}) &
                  &= 0.53 \times 0.125 &&= 0.06625~\mathrm{atm}\\
P(\text{toluene}) &= X(\text{toluene}) \cdot P_\mathrm{vap}(\text{toluene}) &
                  &= 0.47 \times 0.037 &&= 0.01739~\mathrm{atm}\\
P_\mathrm{total}  &&&= 0.06625 + 0.01739 &&= 0.08364~\mathrm{atm}\\
\end{align}

Then using Dalton's law, I am getting the mole fraction of Benzene in the liquid phase.
\begin{align}
P(\text{benzene}) &= Y(\text{benzene}) \cdot P_\mathrm{total}\\
Y(\text{benzene}) &= \frac{P(\text{benzene})}{P_\mathrm{total}}\\
                  &= \frac{0.06625}{0.08364} = 0.792.
\end{align}

However, my answer is incorrect according to the answer key that has $0.25$ as the answer. I have a feeling my whole approach may be wrong since my answer is off by a lot. What am I doing wrong?

The question gives you the mol percentage of benzene ($53 \%$) in the vapour phase, and not in the liquid phase.

It's always good to write down what you have first. (note: I use $y$ to denote mole fractions in the vapour phase and $x$ to denote mole fractions in the liquid state)

First of all, we have $y_\text{benzene} = 0.53$. Additionally, we have been given the vapour pressure of pure benzene and toluene (denoted with a superscript $*$):

$$p^*_\text{benzene} = 0.125\ \mathrm{atm}  \qquad p^*_\text{toluene} = 0.037\ \mathrm{atm} $$

and we require $x_\text{benzene}$

We also, know $p_\text{benzene} = x_\text{benzene}\ p^*_\text{benzene}$, and the total vapor pressure is given by $$P = p_\text{benzene} + p_\text{toluene} = x_\text{benzene}\ p^*_\text{benzene} + x_\text{toluene}\ p^*_\text{toluene} = x_\text{benzene}\ p^*_\text{benzene}+ (1- x_\text{benzene})\ p^*_\text{toluene}$$

Finally,

$$y_\text{benzene} = \frac{p_\text{benzene}}{P} = \frac{x_\text{benzene}\ p^*_\text{benzene}}{x_\text{benzene}\ p^*_\text{benzene}+ (1- x_\text{benzene})\ p^*_\text{toluene}}$$

It is now, trivial to solve this equation. Since I am lazy, and was not in the mood to write down two lines of algebra and reach for a calculator, I chucked it into Mathematica, and the result is:

Thus, $x_\text{benzene} = 0.25$, which is indeed the result given in your answer key.


How big is the influence of basis set and functional on ZPE and thermal corrections? Most quantum chemistry packages are capable of computing ZPE and thermal corrections for a given temperature. But how big is the influence of the basis set chosen and the functional used?

Is it sufficient to calculate this at a lower level than the corresponding final high-level single point calculation? What about differences between the energies - if there is an error, is it systematic so that energy differences are not affected?

This is more of a comment, apologies.

Consider this paper (Ruth L. Jacobsen, Russell D. Johnson III, Karl K. Irikura, Raghu N. Kacker: Anharmonic Vibrational Frequency Calculations Are Not Worthwhile for Small Basis Sets, J. Chem. Theory Comput., 2013, 9 (2), pp 951-954). The magnitude of the computational-level-dependent scaling factors for the frequencies reported therein should give you a sense of the magnitude (I would also think that at least some other studies of the topic are cited).


Why is OTs a good leaving group and how does it work? I've been told that $\ce{OTs}$ is a good leaving group for substitution reactions, but I'm not sure how it works or why it is a good leaving group. Can someone please explain?

$\ce{Ts}$ comes from Toluenesulfonyl chloride, and when the $\ce{Ts-}$ ion replaces and H in an alcohol, it becomes an $\ce{OTs}$ group.

Tosylate groups ($\ce{CH3C6H4SO3-}$) are good leaving groups because their conjugate base forms are quite stabilized by resonance:


How do acids help 'solubilise' compounds? I've heard that root hair cells in plants secrete acids to help 'solubilise minerals' so that they dissociate and are in ion form... this supposedly allows them to be absorbed.

How so can acids solubilise certain compounds? It must be noted: I am a student and my solubility knowledge is fairly basic. I understand that ionic compounds dissolve in water because of the water molecule and the compound's polarity, but it goes no further than that.

Thank you in advance.

There are multiple ways in which plant root and root hair cells increase the availability of nutrients and positively affect the plant’s ability to absorb nutrients from the soil.

One of these ways is by excretion of $\ce{H+}$ ions from malaic acid. This promotes a cation exchange process that results in cationic nutrients being brought into the cell in order to maintain charge balance.  This is the process to which you are referring in your question.

Additionally, enzymes, for example phosphotases, can convert chemically unavailable nutrients, like organic phosphorus, into bioavailable inorganic phosphates.  Phosphate uptake is a case in which acidic soils actually inhibit the availability of this nutrient to plants. (Annals of Botany, 112: 317 – 330, 2013)

There are also proteins excreted by root hairs that can complex metal ion nutrients like iron, facilitating the uptake of such nutrients.


Conformational analysis determining forces So the determining forces in the conformational analysis (what will be the more stable state and what will be the more excited state) according to my textbook are:

Steric interaction
Electron repulsion
Interaction of bonding and anti bonding molecular orbitals (for example $\sigma$ and $\sigma^{*}$ in ethane)

Electron repulsion is quite clear. Filled (molecular) orbitals repulse each other. I have no clue however what is meant with steric interaction, my textbook makes a small remark about it saying it is a repulsive force when atoms come to close to each other. It notes that in the conformation of ethane this hardly plays a role because the hydrogen atoms are to small to interact. But isn't this the same as electron repulsion? What else makes (neutral) atoms want to avoid each other?

Then the last part of bonding and anti bonding orbital interaction (overlap) stabilizing the staggered conformation of ethane is also not so clear to me. How does partially putting electrons in a higher energy molecular orbital (anti bonding) make a molecule's conformation lower in energy?

what is meant with steric interaction
isn't this the same as electron repulsion?

I agree with you Jori, the basis of steric interactions is electron repulsion.  When anyone stops to think about "why are adjacent bulky groups destabilizing" they'll soon arrive at the destabilizing interpenetration of electron clouds.  However, the use of the term "steric" to denote interactions due to size and bulk has been around so long that it has simply become common usage - the way things are said. The following figure shows trans-1,2-di-t-butylethylene to be more stable than its cis counterpart.  It's easier to say and understand that the bulky t-butyl groups produce steric strain in the cis isomer, rather than to say electronic clouds interpenetrate in the cis isomer.  The word "steric" immediately creates a picture of sorts in a chemist's mind.  Tetra-t-butylethylene has yet to be synthesized; attributing this to steric bulk around the double bond provides a clear image of the problem.

Then the last part of bonding and anti bonding orbital interaction
(overlap) stabilizing the staggered conformation of ethane is also not
so clear to me. How does partially putting electrons in a higher
energy molecular orbital (anti bonding) make a molecule's conformation
lower in energy?

First off, let's recognize that the orbital interaction you are referring to is a way to understand the effect of hyperconjugation in various conformations.  Also, we are not putting electrons in a higher energy molecular orbital.
Here are some diagrams that may help explain this.  The first diagram on the left doesn't relate to hyperconjugation, but since it is part of the picture, let's explain it.  The Pauli Exclusion principle says that we can't have electrons with the same quantum numbers in the same orbital. In ethane, each pair of electrons in one C-H bond have the same quantum numbers as the pair electrons in any other C-H bond.  Therefor, the closer we bring 2 C-H bonds, the closer we are to violating the Pauli Exclusion principle.  This argument tells us that the eclipsed conformation should therefor, be higher in energy than the staggered conformation.
The figures in the middle and the right show the effects of hyperconjugation upon the eclipsed and staggered conformations of ethane, respectively.  According to Frontier Molecular Orbital theory we examine the interaction between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO) to see if stabilization or destabilization results. In the hyperconjugated resonance structure for ethane we have a proton and a pair of electrons in the carbon $\ce{sp^3}$ orbital, that $\ce{sp^3}$ orbital with two electrons is our HOMO.  The only LUMO for it to interact with is the LUMO for the C-H (the $\sigma^{*}$ orbital) bond on the adjacent carbon atom.  We can mix the $\ce{sp^3}$ and $\sigma^{*}$ orbitals to produce 2 new MOs, one bonding (no nodes between these 2 orbitals) and one antibonding (1 node between these two orbitals).  What is pictured below is the bonding (in-phase, no nodes) combination of these two orbitals.  We see that in the eclipsed (middle) conformation the interaction in this orbital is destabilizing, there is no constructive overlap.  However, in the staggered (right) conformation the overlap within this orbital is now constructive - there is positive overlap.  The hyperconjugation argument also favors the staggered conformation being at lower energy.


What fraction of hydrogen in air results in an explosive mixture? I would like to know if there is documented evidence of the quantity of hydrogen to air that must be present in the mixture to risk explosion or flame.

Hydrogen is explosive in mixtures in air at room temperature and ambient pressure when it is between 4% and 75% by volume.  These limits are respectively called the "lower explosive limit" (LEL) and the upper explosive limit (UEL).

Values for hydrogen (and many other gases) are provided in many locations on the internet.  Engineering toolbox is one such site.

More info on LEL and UEL is available from Wikipedia.


Miscibility of Molten Metals While reading about the size of atoms and ions of the Group 1 elements in the textbook "Concise Inorganic Chemistry" by JD Lee, I came across this line:

  The Li+ is much smaller than the other ions. For this reason, Li only mixes with Na and above 380 Celsius, and it is immiscible with the metals K, Rb and Cs, even when molten; nor will Li form sunstitutional alloys with them. In contrast the other metals Na, K, Rb, Cs are miscible with each other in all proportions.

I am unable to understand why the elements cannot form alloys with Li and what makes the other elements miscible with each other in all proportions.

I observed that the metallic radius of Li is 1.52 $\unicode{xC5}$ and that the difference between its radius and that of Na is about 0.34 $\unicode{xC5}$. But the difference between the radii of the other successive elements in this group is also about the same, so it doesn't help explain miscibility.

I would really appreciate some insight and clarification.

Questions like this are not prone to easy answers. But, lets look at what is known about the liquids, then at the various binary diagrams.

Lithium - Melting point 453K, density at melt is 0.512 g/cm$^{3}$, or 0.0737 mol/cm$^{3}$

Sodium - Melting point 371K, density at melt is 0.926 g/cm$^{3}$, or 0.0403 mol/cm$^{3}$

Potassium - Melting point 336K, density at melt is 0.828 g/cm$^{3}$, or 0.0211 mol/cm$^{3}$

Calcium - Melting point 1112K, density at melt is 1.376 g/cm$^{3}$, or 0.0343 mol/cm$^{3}$

Clearly Li has by far the smallest atomic volume in the melt.

The Li-Na binary, as you point out, shows a miscibility gap, tilted towards the Li side, but peaking around 300C. The K-Li system is less well studied, but it also shows a miscibility gap which peaks at perhaps a similar value of 300C.

In sharp contrast, the Ca-Li system exhibits an intermediate compound (Li$_{2}$Ca) and can be described with a 'virtually ideal' liquid phase (C.W. Bale and A.D. Pelton, Bulletin of Alloys Phase Diagrams 8(2) 125-127 (1987)).

Given that the molar density of liquid Ca lies between those of Na and K it is clear that you cannot ascribe the observed miscibility gaps solely to atomic radii.

To further complicate matters, lets keep looking at more binary diagrams. The K-Na diagram shows an intermediate compound, KNa$_{2}$ and no miscibility gap.  KNa$_{2}$ has the same crystal structure as Li$_{2}$Ca (both are MgZn$_{2}$ prototype). The Ca-Na binary shows -- surprise! -- a miscibility gap towards the Na side, peaking at 1200K. So, Ca and Na, even with fairly similar atomic volumes, do not want to mix in the liquid, while Ca and Li are perfectly happy to.

Bottom line - atomic radii do not really mean that much when predicting binary phase diagrams.  Even with comparable radii, you would be surprised at the weirdness than can be found on phase diagrams. As one example, Ag-Cu (both fcc metals) do not form a continuous solid solution, and have a slightly positive enthalpy of mixing in the liquid (not enough to form a miscibility gap).  Yet, Au-Cu (Ag and Au have similar atomic radii) form a fcc continuous solid solution (with intermediate ordered fcc-based phases at lower temperatures), and has a strongly negative enthalpy of mixing in the liquid.

Have fund exploring binary phase diagrams (and ternary, ...), but don't rely on simple heuristics.


Why is weight of 1 mole of substance equal to atomic/molecular mass in grams? Why is weight of 1 mole of substance equal to atomic/molecular mass in grams?

According to me, it happens because mole has been defined in such a way. It is defined as the numbers of particles in $\pu{12 g}$ of $\ce{^{12}C}$. If it were $\pu{24 g}$ instead of $\pu{12 g}$, then the weight of 1 mole of substance would equal 2 times the atomic/molecular mass in grams.

Please correct me if I am wrong somewhere (or maybe everywhere).

Why is weight of 1 mole of substance equal to atomic/molecular mass in grams?
According to me, it happens because mole has been defined in such a way.

Yes! That is correct.

It is defined as the numbers of particles in 12 g of C12. If it were 24 g, instead of 12 g, then the weight of 1 mole of substance would equal 2 times the atomic/molecular mass in grams.

Also correct, assuming that the definition of unified atomic mass units (amu) remained the same. @Martin's answer is correct, but we can also arrive at the same conclusion using a simple dimensional analysis approach.
First we need the definition of an amu:
$$\pu{1 atom}~ \ce{^12C} = \pu{12 amu}$$
Now take the real definition of a mole:
$$\pu{1 mol}~ \ce{^12C} = \pu{12 g}$$
Now, divide the first equation by the second:
$$\frac{\pu{1 atom}~ \ce{^12C}}{\pu{1 mol}~ \ce{^12C}} = \frac{\pu{12 amu}}{\pu{12 g}}$$
Cross-multiply and reduce:
$$1~\frac{\pu{g}}{\pu{mol}~ \ce{^12C}} = 1~\frac{\pu{amu}}{\pu{atom}~ \ce{^12C}}$$
What this tells us is that the ratio of g/mol to amu/atom is exactly one - and we made sure it would work out that way by carefully choosing how we defined moles and amus.
Since the masses of all elements are defined in terms of amu, which are ultimately based on the mass relative to carbon-12, this ratio holds for all any atom or molecule.
Let's take it further and put in your hypothetical doubling of the number of grams of carbon-12 per mole:
$$\frac{\pu{1 atom}~ \ce{^12C}}{\pu{1 mol}~ \ce{^12C}} = \frac{\pu{12 amu}}{\pu{24 g}}$$
Now when we cross-multiply and reduce, we get:
$$2~\frac{\pu{g}}{\pu{mol}~ \ce{^12C}} = 1~\frac{\pu{amu}}{\pu{atom}~ \ce{^12C}}$$
Since we kept the definition of an amu the same, but changed that of the mole, what we see is that 1 mole of carbon-12 using these units would have to weigh 24 g (the molar mass), even though its atomic mass would still be 12.
That's a lot harder to keep track of mentally, and so it is a good thing for us that the definitions of amus and moles were chosen so carefully!


Planes of Symmetry and Chirality I am trying to identify whether the following objects possess planes of symmetry or not, but my answers seem different, from the textbook solutions (Klein Organic Chemistry 2e, Page 220, Question 23). Some of them I am unsure of and would like to make sure I know the proper reasoning for. Can someone please clarify?

(b) The correct response is yes. Is the plane of symmetry I have drawn acceptable?
(c) The correct response is no, but I thought it is possible to have a plane of symmetry like the one I have drawn.
(e) The correct response is yes, which I think I understand. I think this has three planes of symmetry.
(f) The correct response is no, which I agree with. However, it doesn't seem to concur with the response for the follow-up question.

Follow-up: Which image has three planes of symmetry? The correct answer is apparently, (f). My response would have been (e).

Let us address the elephant in the room first and foremost.  All of these objects do not have any symmetry if you wanted to get real technical about it.  However, let us just assume certain things for this useful exercise to be relevant.

b.) The teacup has a vertical plane of symmetry splitting the teacup in half along the handle axis and intersecting with the centroid of the cup and saucer.  I would say that your guess is correct though it would be useful to have drawn in a plane rather than simply a line (for all examples).

c.) At first glance the pliers have a couple planes of symmetry BUT upon further thought it may not due to the way the rotating JOINT is constructed.  See the example below:

While it is difficult to see in your picture, below it is made very clear.  Therefore I will say that this object does not contain a plane of symmetry.  It would be POSSIBLE that it could have a plane of symmetry, one that lies in the plane of your monitor screen, cutting the object in half from tip to handle end... but it would be impossible to tell without seeing the other side. Many times pliers come with an inscription on one side of the joint, thereby breaking this symmetry.

e.) The brick does indeed have multiple planes of symmetry.  One vertical plane bisects the brick along its width and one bisects the brick along its length.  In addition, one plane of symmetry lies in the plane of the brick (the plane the brick is lying on), cutting the brick in half (think of it as coming in through the side of the brick).

f.) The hand has no plane of symmetry that I can see at all.  The back of the hand doesn't look like its front... the left and right side of the hand are clearly not mirror images of each other...

The brick has three planes of symmetry.