CuSO4 in SO2/SO3 gas sampling in Sulphuric Acid Production What role does $\ce{CuSO4}$ has in the gas sampling of an $\ce{SO2/SO3}$ gas mix in a catalytic converter for sulphuric acid production?

To give you a little context: We take gas samples from the beds of the catalytic converter to determine the efficiency of the catalyst. We use "cartridges" that we make with glass tubes filled with $\ce{CuSO4}$. The $\ce{CuSO4}$ (lab grade) comes in crystal version as a pentahydrate. We heat it for around $\pu{30 minutes}$ at $\pu{105 ^\circ C}$ and then we use it to make the cartridges. The cartridges are placed between the sampling point and the sampling bag (made of Teflon). This is a process that has been made since many  years but nobody here is able to tell me why. The answer they give me is that the cartridges are used to "remove" the $\ce{SO3}$ from the sample but I can't see how that could happen. My theory is that because anhydrous $\ce{CuSO4}$ is hygroscopic (I've read it's been used as a desiccant) it may be used to avoid the formation of $\ce{H2SO4}$ that could damage the gas analyzer (a SICK S700). I've really googled it but can´t find the answer. Hope someone could enlighten me in the subject.

Heating hydrate copper sulfate at 105 degrees does nothing. The chemically bound water takes a higher temperature to remove all water molecules (water is lost in steps as a function of temperature). This low temperature removes only two water molecules out of five. I think the major role of copper sulfate, besides being a gentle dessicant, is to oxidize sulfur dioxide to trioxide. Cu(II) ion is a mild oxidizing agent, and mild enough that it can oxidize sulfur dioxide to the trioxide.


Is it possible to make broad spectrum light absorbing paint? Is there such a paint / pigment that absorbs most wavelengths of light (beyond the visible light range) that can be used for solar heat generating applications?  If not, would it be possible to make it?  Ideally, this paint would absorb light from the sun in an efficient manner and become heated quickly by solar light.  It should probably at least absorb infrared, and any other wavelengths of light that produce significant heat when absorbed.  This should be far more effective than normal black paint as black paint only "for sure" absorbs light in the visible spectrum, and this desired paint would absorb other types of light from the sun.

Carbon nanotube black is highly absorbent in and outside the visible spectrum, see http://www.extremetech.com/extreme/186229-its-like-staring-into-a-black-hole-worlds-darkest-material-will-be-used-to-make-very-stealthy-aircraft-better-telescopes.

Of course, that does not make it ideal for energy conversion, as it would also need at least the following qualities to be practical.

A good conductor of heat to the heat-exchanger surface below
An enduring finish, resistant to abrasion, heat and environmental degradation
Economical, both in material cost and application
Safety in use and in manufacture


What is a material more easily cut than shattered? I'm currently developing a sci-fi concept for animation which partially involves a typical "the earth has been overrun by XYZ" scenario.

One of the ideas for XYZ we're thinking of is a species of intelligent crystalline beings made of diamond -- or at least a diamond like material.
These are large, arachnoid in shape and fast. They kill by piercing their pray with sharp, lance-like protrusions on the ends of their forelegs.

The problem is, thus far we've been thinking that the trick to killing these things has been 'shattering' what essentially equates to a CPU or central core suspended like the 'body' of your typical daddy long legs. Which makes sense if the material were diamond -- however we realised it might be more interesting if kinetic force like cannonballs and explosive shockwaves were less effective, and that the key was 'cutting them.'

Maybe we're bad writers, and we just can't drop the idea of our protagonist being forced to use a 'chain-blade' up close to combat one of these things, but visually we're very keen on just such a scenario. So if those among you more educated than us (i.e. all of you) could point us towards a material where 'cutting' or 'sawing' with what could be called 'super-chainsaws' is more feasible than explosives, large projectiles, or other methods of blunt force, it would be greatly appreciated.

We Do understand just such a material will probably not exist in crystalline form, thus losing the overall aesthetic we're aiming for -- but nonetheless we'd love to hear what you guys have to say in general regarding the question and context.

Thank you very much for your time.

The crystalline property that would allow to cut the "diamond creature" (or whatever those things are in your universe) you are looking at is cleavage (a writer might also use a term in a wordplay).
In crystallography, cleavage is the ability of crystals to crack or split in certain crystallographic directions.
This mechanical property of crystalline media is associated with their internal structure (the ratio of adhesion forces in the crystal lattice) and does not depend on the external shape of the crystals.
This property is also often utilized in the beginning of the process of fashioning the gemstones.
There is a fascinating demonstration with explanation: Crystals - Alan Holden 1958 (YouTube).
With diamonds you can do that too, just not always as easily as with mica: Diamond Cleavage (YouTube).
Diamond (as well as ﬂuorite, galena, topaz and some other minerals) possesses a so-called "perfect" cleavage, and the cleavage planes in diamond are parallel to the octahedral crystal faces [1, p. 30]:

Figure 4.1 Two examples of directional-dependant properties in diamond. The cleavage planes are parallel to the octahedral faces, which are also planes of near
maximum hardness. The easiest direction for sawing is at right-angles to the dodecahedral plane, which is also the plane of minimum hardness.

References

Read, P. G. Gemmology, 3rd ed.; Elsevier Butterworth Heinemann: Amsterdam; Boston, 2005. ISBN 978-0-7506-6449-3.


Work done by adiabatic expansion of gas The formula for the work done by adiabatic expansion of gas is
$$W = C_V\,\mathrm dT.$$
Why the volumetric heat capacity $C_V$ is constant and not the heat capacity at constant pressure $C_p?$ If the system is at constant volume, it is a isovolumetric process and hence no work can be done in that case.

Part 1. Let's consider first law of thermodynamics.

$\delta Q = dU + \delta W$ (classical sign convention)

or

$\delta Q = dU - \delta W$ (IUPAC sign convention)

In the case of adiabatic process, $\delta Q = 0$.

$0 = dU - \delta W$

$dU = \delta W$

$dU = C_VdT$ (by definition)

$C_VdT = \delta W$

After integration, assuming CV is constant.

$C_V\Delta T =  W$

Part 2. Why not Cp?

By definition, $\delta Q_p = dH = C_pdT$. Obviously, we don't have an isobaric process here.

P. S. Yes, those sign convention differences also make me feel bad.


What is responsible for the odor of gasoline? Gasoline as we know it is a blend of combustible hydrocarbons and has a characteristic odor. The natural gas we use to heat our homes and (for some of us) cook our food has its own odor, except...

Natural gas as piped to our homes consists of methane, which is odorless, plus an odorant so that we can be aware of its presence for safety reasons. So the odor we think of as "gas" isn't the fuel at all, but an odorant.

So this leads to my question: is the familiar odor we attribute to gasoline that of one or more of the components which are the fuel itself (heptane, octane, etc.) or an additive for the purpose of giving it an odor, again for safety reasons?

Many of the components of gasoline (petrol) have notable odours

A simple starting point for understanding the odour of gasoline is the typical composition of the product. That composition varies a lot and always has done as different refineries use different processing techniques and different crude oil sources have different components to start with. And regulators have changed the rules over time (for example to reduce the amount of benzene, a carcinogen). All that matters, in practice, is the octane rating which is what engines care about and this is achievable in a variety of ways by varying the composition and the additives that modify the combustion characteristics.

This (old) paper gives a crude overview of a typical composition.

  The typical composition of gasoline hydrocarbons (% volume) is as follows: 4-8% alkanes; 2-5% alkenes; 25-40% isoalkanes; 3-7% cycloalkanes; l-4% cycloalkenes; and 20-50% total aromatics (0.5-2.5% benzene).

Modern blends will now have far less benzene (but will still have a lot of other aromatics). Without getting too specific on the individual chemicals we can already tell some of the sources of the distinctive odour of the mixture. It is pretty meaningless to try to pin it to a single component.

Aromatics as a class are actually named because of their distinctive smell. even if we substitute other aromatics like toluene or xylene for the benzene, there will be a distinctive "aromatic" component in the product. But plenty of the other hydrocarbons also have notable odours. Cycloakenes (maybe 7% of the mixture) often have sharp smells (cycopentene is even said to have "a petrol-like odour"). Even linear and branched alkanes have some smell (though it is far less intense than the unsaturated components of gasoline).

In short, trying to pin down the typical smell of gasoline to a single compound is a little silly when there are often scores of components with well-known and strong odours and the specific composition varies a lot. aromatics are a big part with other unsaturated compounds providing some other odiferous components with a smaller but notable contribution even from unsaturated hydrocarbons.

Also, gasoline is not "odourised" as it doesn't need to be. But it sometimes has specific ingredients added to make tracing the source easier. But this is done with complex instrumentation not the human nose.


Forming hydrogen sulfide from electrolysis of sodium thiosulfate So I was doing electrolysis on two silver electrodes and sodium thiosulfate.
Some observations

The voltage drops extremely fast from 2 V to less than 1 V in 15 minutes, possibly because of the thiosulfate reduction.
Gases are liberated out of solution.
The solution turns black quickly.

Alert: I am using a fume hood.

Since I'm afraid of this spoiled eggs smell I just want to know if I am in any way creating hydrogen sulfide? Just want to be sure that the gas is tetrathionate and not hydrogen sulfide.

My other concern is with the voltage. How can I keep it from dropping so fast?

Yup, you're forming sulfide.

Render the thiosulfate ion as $(\ce {S-}-\ce {SO3-})$.  Then at the cathode:

$(\ce {S-}-\ce {SO3-})+2e^- \rightarrow \ce {S^{2-}}+\ce {SO3^{2-}}$

Both products are basic, with sulfide ion being strongly so,  and they will react further with protons; so the sulfide ion gives some hydrogen sulfide.  Sulfide ions can also combine with silver ions, from the anode, giving a black precipitate.  The other product, sulfite ion, can be reduced further, possibly forming still more sulfide ion/hydrogen sulfide.

So the reaction is not safe.  But kudos for using a fume hood.


Number of gauche interactions in an alkane The textbook states that (a) has 2 gauche interactions, (b) has 3 gauche interactions and (c) has 3 interactions.

But how do we determine this? I thought a gauche interaction would occur when two methyl groups are side by side, however I'm unsure where these answers are coming from.

When we sight down a carbon-carbon single bond and draw a Newman projection of a particular conformation, any two groups on the adjacent (vicinal) carbons that have a dihedral angle of 60° between them are said to have a gauche relationship.

In the following figure, the $\ce{X}$ and $\ce{Y}$ have a gauche relationship.

$\ce{X}$ and $\ce{Y}$ can be the same (they can both be methyl groups, other alkyl groups, other atoms such as halogen, etc.) or they can be different from each other. In most organic classes, the gauche interaction between methyl groups or alkyl groups is frequently discussed.

The answers you included (2,3,3) are correct.  Looking at the middle molecule (b) in your figure, there are 3 gauche interactions.  They are between

the front methyl and the back ethyl
the same front methyl and the back methyl
the other front methyl and the back methyl


Intramolecular Cannizzaro reaction product I came across the following question :

My reasoning was as follows :

It can't be option (c) because there is no chance of losing a carbon during Cannizzaro's reaction
It can't be option (b) because basically option (c) gives (b) finally
It can't be option (a) or option (d) because we can't have all carb acid groups or all alcohol groups. we need a combination of both.

So, are the options wrong?

Your reasoning is correct and the problem is with the publisher as they made  a mistake while compiling these previous year questions.
I found out the original 2003 question paper(pdf). The question you mentioned had the following options:

It is clear form the options that the answer is C(which is the product you expected) which can also be confirmed by the key attached at the end of the pdf.


What the difference in data if calibrate pH-meter with 2 and 3 solutions? pH-meter (12 bucks one) can be calibrated with 2 different solutions (let's say 4.0 and 6.86), but as an option 3 can be used (4.0, 6.86 and 9.18). If I use 2 instead of 3, what error I'm going to get.
For instance I need to determine if liquid higher or lower than 5.5.

You cannot "predict" any error by thinking that if we calibrate with two buffers the error will be x, and if we calibrate with three buffers the error will be y.
Ideally, in each case, the pH of the sample should be identical. Good quality pH meters tell you how close they are to the theoretical slope in terms of percentage. It should be close to 100%.
For example, at 25°C the theoretical slope is ~ -59 mV/pH. If your calibrate with two buffers and  is close to -58 mV/pH, the slope percentage is 98.3%. It is good to go. As electrode age, the percentages goes down.
As EdV said, three points might be better for the $12 pH meter.


Why do large cations stabilise large anions and small cations stabilise small anions? I see this trend from enthalpy change of decomposition data, though I struggle to find a decent explanation for why.

For more detailed info, search "cation-anion radius ratio" or "Pauling's rules". Below is my review of this idea.

This ratio is ${R_{C}/R_{A}}$ (C = cation, A=anion).

Small cation (I mean, too small) attracts anions in such a way that they come too close to each other and repulsive force comes into play. This happens when ratio is <0,155. So for larger anion, larger cation suits better. Ratio can be 0.155-1 for stable compounds, and this ratio also determines coordination number and type of void! (Check the picture below, recall that since <0.155 ratio is for unstable compounds, you have empty "example" line there).


How to calculate the dissociation constant of a weak acid from the titration with a strong base? Problem:
  A solution of an unknown weak acid of unknown concentration was titrated with a solution of a strong base of unknown concentration. During the titration, the pH after adding $\pu{2.00 mL}$ of the base was 6.912. An additional $\pu{14.00 mL}$ of the base was required to reach the equivalence point. Calculate the Ka of the weak acid.

I have decimated six sheets of paper trying different equations. Combining the two ICE tables you can make for each equation, I can still get no fewer than three unknowns. I'm sure I'm overlooking a super-simple solution but I cannot  think of it.

These images are what I have done so far.

This is a strong base weak acid titration. If the weak acid is of the form $\ce{HA}$, then at equivalence point, all the acid has been converted to its conjugate base.

Total volume of base required for equivalence point is $\mathrm{16~ml}$. Let the molarity of base used be $M$. Then, number of milli moles of base for equivalence point is $M\times 16$. This is also the number of milli moles of acid taken.

If only $\mathrm{2~ml}$ of base was used, there will be some acid and some salt of acid and base (A buffer, since the acid is weak and the base is strong).
$$pH=pk_a+\log\frac{M\times 2}{M\times 16-M\times 2}=pk_a+\log\frac{2}{14}$$
From this you will get $$6.912=pk_a-0.845$$
$$pk_a=7.757$$


Can anyone explain, why there are coefficients 2 in the equation for rate of change [A]? The reaction mechanism for the decomposition of $\ce{A2}$ is thought to be:
\begin{align}
\ce{A2 &<=>[$k_1$][$k_1'$] A + A} & &\text{ (fast)} \\[0.2cm]
\ce{A + B &<=>[$k_2$] P} & &\text{ (slow)}
\end{align}
The rate of formation of intermediate $\ce{A}$ is given by:
\begin{align}
\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = \color{red}{2} k_1[\ce{A2}] - \color{red}{2}k_1'[\ce{A}]^2 - k_2[\ce{A}][\ce{B}] \approx 0
\end{align}

In the equation above, why is the coefficient of $2$ present for both the forward and reverse reactions (highlighted in red)?

$$\ce{A_n ->[$k$] nA}$$
For the reaction given above, rate of formation of $\ce{A}$ is defined as:
$$\frac{1}{n}\frac{\mathrm d\ce{[A]}}{\mathrm dt} = k\ce{[A_n]}$$
In this case, we have three different reactions that involve A. They are:
\begin{align}
\ce{A_n &->[$k_1$] nA} \tag{1}\label{1}\\
\ce{nA &->[$k_1'$] A_n} \tag{2}\label{2}\\
\ce{A + B &->[$k_2$] P} \tag{3}\label{3}\\
\end{align}
Writing the rate equations for $\eqref{1}$, $\eqref{2}$, $\eqref{3}$ and summing them up to get total rate, we get:
\begin{align}
\frac{1}{n}\frac{\mathrm d\ce{[A]1}}{\mathrm dt} &= k_1\ce{[A_n]} \tag{4}\label{4}\\
-\frac{1}{n}\frac{\mathrm d\ce{[A]2}}{\mathrm dt} &= k'_1 [\ce{A}]^n\tag{5}\label{5}\\
-\frac{\mathrm d\ce{[A]3}}{\mathrm dt} &= k_2\ce{[A][B]} \tag{6}\label{6}\\
\frac{\mathrm d\ce{[A]_\mathrm{tot}}}{\mathrm dt} &= \frac{\mathrm d\ce{[A]1}}{\mathrm dt} + \frac{\mathrm d\ce{[A]2}}{\mathrm dt}+\frac{\mathrm d\ce{[A]3}}{\mathrm dt} \tag{7}\label{7} \\
\end{align}
Substituting values from $\eqref{4}$, $\eqref{5}$, $\eqref{6}$ into $\eqref{7}$, we get:
$$\frac{\mathrm d\ce{[A]_\mathrm{tot}}}{\mathrm dt} = \color{red}{n}k_1\ce{[A_n]} - \color{red}{n}k'_1\ce{[A]^n} - k_2\ce{[A][B]} $$
Here, $n = 2$, therefore
$$\frac{\mathrm d\ce{[A]_\mathrm{tot}}}{\mathrm dt} = \color{red}{2}k_1\ce{[A_n]} - \color{red}{2}k'_1\ce{[A]^2} - k_2\ce{[A][B]} $$


How to chose wavelength in spectrophotometry? How to use Beer-Lambert Law? How is the wavelength chosen to measure the concentration of an ion determined? and What does the wavelength correspond to in an atom/ion?

When light is shined on a solution what happens to the ions in solution? Include how and why light is absorbed or emitted from a substance.

Would a high or low concentration absorb more light?

How does the beer-lambert law relate to concentration and information from the spectrophotometer?

Working wavelength is chosen by analysing the spectrogram $A(\lambda)$. Such diagram is obtained by measuring absorbance $A$ in function of wavelength $\lambda$. Practically, you test a moderate concentration of your analyte and sweep the complete wavelength bandwidth of your analyser. Then you select wavelength where a maximum of absorbance occurs. If multiple maximum are present, you may prefer the highest one (best sensitivity) or a smaller one with less interferences (best selectivity). Both can happens simultaneously. The main criterion is:

$$ \underset{\lambda}{\max}\left\{A(\lambda)\right\} $$

For information: the wavelength is linked to photon energy $E$ by the following relation:

$$ E = h \nu = h \frac{c}{\lambda} $$

Where $\nu$ is the frequency of the EM wave, $c$ the speed of light and $h$ the Plank constant.

Absorbance is a property of matter, when an atom/ion/molecule absorbs a photon (visible or nearly visible IR/UV), it can promote electron to a highest energy level (UV, highest energy) or increase vibrational energy (IR, lowest energy) of the compound and the photon is destroyed.

A spectrophotometer assess the transmittance $T$ (ratio of transmitted $\phi$ and incident flux $\phi_0$ expressed as powers) of light beam through a sample cell.

$$ T = \frac{\phi}{\phi_0} $$

And transmittance (dimensionless) is linked to absorbance (dimensionless) in the following way (functional scaling to achieve linearisation between signal and physical quantity):

$$ A = -\log_{10}(T)$$

Beer-Lambert Law is a limit law conditioned by several hypothesis and it is only valid for low concentrations. When Beer-Lambert applies you can linearly link concentration $x$ with absorbance.

$$ A(x) = \epsilon(\lambda)lx $$

Where $\epsilon$ is a proportional coefficient (with suitable units) called extinction coefficient which describes how matter absorbs light at wavelength $\lambda$. The quantity $l$ stands for the optical path length through the sample cell.

Numerous substances validate Beer-Lambert Law but some does not. Therefore you must assess linearity before quantifying unknown concentrations. This is usually done when performing calibration curve. Then you know your dynamic range and sensitivity of your analytical method and you can regress concentration from absorbance.

I suggest you to read more about those concepts in a reference book such as Skoog & West, Fundamentals of Analytical Chemistry or Vogels, Textbook of Quantitative Chemical Analysis. There is also a free electronic book available on internet called Analytical Chemistry 2.0.


What are the products of a reaction between copper sulfate and sodium bicarbonate? I have been trying for a while to determine what products are produced when copper sulfate reacts with sodium bicarbonate at room temperature and standard pressure. However, while I was looking, I found conflicting information concerning what products are formed. Some sources claim that the products are copper hydroxide and sodium sulfate, while other sources state that the products are copper carbonate and sodium sulfate.

I believe the reaction is:

$$\ce{CuSO4 + 2NaHCO3 -> CuCO3 + CO2 + H2O + Na2SO4}$$

Is this correct?

Copper sulfate reacts with sodium bicarbonate reacts to form basic copper carbonate due to  the great affinity of the $\ce{Cu^2+}$ ion for the hydroxide anion $\ce{OH−}$. From its Wikipedia article:

  Basic copper carbonate can also be prepared by reacting aqueous
  solutions of copper(II) sulfate and sodium bicarbonate at ambient
  conditions. Basic copper carbonate precipitates from the solution,
  again with release of carbon dioxide:

  $$\ce{2 CuSO4 + 4 NaHCO3 → Cu2(OH)2CO3 + 2 Na2SO4 + 3 CO2 + H2O}$$

Here, the "ambient condition" is very important because if we alter the temperature and/or pressure , the reaction may tend to form copper(II) carbonate or any non-stoichiometric copper compound.

Heating the basic carbonate in an atmosphere of carbon dioxide (produced by the decomposition of silver oxalate at 500 °C) produces copper(II) carbonate but it is itself quite unstable and react with moisture to convert back to basic carbonate.

For advanced reading, you can refer to reference 2 below in the reference which describes Copper/bicarbonate equilibria in solutions of bicarbonate ion in details (which clearly shows formation of basic ion $\ce{CuCO3(OH)^2-}$ in the solution).

Notes and References

http://www.sciencemadness.org/talk/viewthread.php?tid=24056
https://www.sciencedirect.com/science/article/abs/pii/0043135471900492
http://science.wonderhowto.com/how-to/synthesize-copper-ii-carbonate-sodium-bicarbonate-302466/ (Although this site claims to make copper(II) carbonate, it is actually making the basic carbonate. You can clearly see its bluish green color of basic copper carbonate).


Calculating Compressibility factor from the Van der Waals' Gas equation So this problem has been bugging me for a long time. According to Wikipedia the compressibility factor $Z$ is defined as the ratio of the volume occupied by a real gas divided by the volume occupied by an ideal gas at the same temperature and pressure.

$$Z = \frac{V_\text{real}}{V_\text{ideal}}$$

Now the ideal gas equation is

$$P_\text{ideal} V_\text{ideal} = nRT$$

Now from here we can substitute the value of $V_\text{ideal}$ into the $Z$ expression to get $$Z = \frac{P V_\pu{m}}{RT}$$

Where $V_\pu{m}$  is the molar volume of the gas and $P$ is the value of $P_\pu{ideal}$ that was in the ideal gas equation.

OK so far so good.

Deriving the compressibility factor value from the van der Waals' gas equation:

We know for real gases the pressure exerted by them is always less than the pressure exerted by an ideal gas so we correct the pressure by the term
$an^2/V^2$

$$P_{\text{ideal}}= P_{\text{real}} + \frac{an^2}{V^2}$$

And the volume occupied by the gas molecules can be written as
$$ V_{\text{ideal}} = V_{\text{real}} - nb$$

On substituting the above values of $P_\text{ideal}$ and $V_\text{ideal}$ in the ideal gas equation, we get our van der Waals' Gas equation

$$\left(P+\frac{an^2}{V^2}\right)(V-nb) = nRT$$

Please note that in the above equation $P$ is $P_\text{real}$ and not $P_\text{ideal}$. Now I'm going a step further to calculate the value of $Z$ from this equation by shifting the terms and I got this $$Z = \frac{P V_\pu{m}}{RT} = \frac{V_\pu{m}}{V_\pu{m} - b} - \frac{a}{RTV_\pu{m}}$$

Now here's the problem that I'm facing: In the above expression the value of $P$ is not the $P_\text{ideal}$ but it is the value of $P_\pu{real}$ that we had in the van der Waals' gas equation. Doesn't that contradict the whole definition of Compressibility factor that we had learned in the first place? How this expression for $Z$ can be correct?
Thanks for reading.

$$Z = \frac{V^{\mathrm{r}}_{\mathrm{m}}}{V^{\circ}_{\mathrm{m}}}\tag{1}\label{1}$$

$V^{\mathrm{r}}_{\mathrm{m}} = \text{Volume of 1 mol real gas.}$

$V^{\circ}_{\mathrm{m}} = \text{Volume of 1 mol perfect gas.}$

The van der Waals equation is $$ \left( P^{\text{r}} + \frac{an^2}{(V^{\text{r}})^2}\right) (V^{\text{r}} - nb ) = nRT \tag2\label{2}$$

Now, consider you have a container containing a $1~\mathrm{mol}$ real gas. You know its pressure $P^{\text{r}}$, volume $V^{\text{r}}$ and temperature $T$ and you wish to find the compressibility factor of this gas.

So, for calculating $Z$, we know the real volume of the gas, i.e $V^{\mathrm{r}}$ and now we need to calculate  $V^{\circ}$ which is the volume it should have occupied if it behaved like a perfect gas, i.e. it obeyed the perfect gas law.

Thus, $$V^{\circ} = \frac{RT}{P^{\text{r}}}$$

Therefore, substituting this value of $V^{\circ}$ in $\eqref{1}$ $$Z = \frac{V^{\mathrm{r}} P^{\text{r}}}{RT}\tag3\label{3}$$

From $\eqref{2}$, by rearranging the terms, we get $$P^{\text{r}} = \frac{RT}{(V^{\text{r}} - b)} - \frac{a}{(V^{\text{r}})^2} \tag4\label{4}$$

Substituting this value of $P^{\text{r}}$ in $\eqref{3}$, we get $$Z = \frac{V^{\text{r}}}{(V^{\text{r}} - b)} - \frac{a}{V^{\text{r}}(RT)} \tag5\label{5}$$


Reaction between silver nitrate and aluminum chloride I googled a bit about what $\ce{AgNO3 + AlCl3}$ will yield and found out the following:

$\ce{3AgNO3 + AlCl3 -> Al(NO3)3 + 3AgCl}$

Why is that? I know that $\ce{Ag}$ is higher up in the reactivity series than $\ce{Al}$ but that does not make sense to me in this problem. Is it because $\ce{Cl}$ is just one when $\ce{NO3}$ is multiple?

Le Châtelier's principle only states that a system previously at equilibrium will want to stay at equilibrium - that is, if we perturb it, it will try to go back to equilibrium. This system is initially not at equilibrium, and therefore we don't need his principle.

I think the reaction proceeds because the formation of AgCl(s) is enthalpically favored. Molecular cohesion (enthalpy) is greater in AgCl than in AgNO3, thus promoting its formation. How do I know that? Well, it seems AgCl is less ionic than AgNO3 since it is insoluble in water.

This reaction is a type of metathesis reaction. It can be called double displacement reaction or salt metathesis reaction, depending on who you're talking with.

EDIT

The other answers make an error, I think. Both suggest that the solubility product of AgCl is of importance in determining why the reaction takes place. My opinion is that the argument, as presented, is circular, in the following manner:

The solubility product of AgCl in water is very low, therefore we observe formation of a precipitate.

vs

AgCl is insoluble in water, therefore its solubility product is low.

It's easy to see that the fact AgCl is insoluble in water implies that it has a low Ksp in water. If someone can predict and quantify the solubility product of AgCl in water from first principles, then that's another story, but the explanations offered only rely on observation (experiment) to quantify the Ksp.

My argument goes down the ladder. If you can answer why AgCl is insoluble in water, then you can definitely predict that Ag+ and Cl- ions from different sources (i.e. in different solutions) put into contact will spontaneously form a precipitate.

Now you maybe thinking this argument is also circular, but it's not.

We have access to experiment, which clearly informs us that AgCl is insoluble in water. We can confidently assume (but we might be wrong!) that the AgCl bond is less ionic than the AgNO3 bond, since it practically does not dissociate/has a low Ksp in water. A bond with a greater covalent character is more stable, since the electrons are shared and not transfered.

Perhaps the crystal structure of AgCl makes it especially stable to solubilization from H2O. I just think it's obvious that the enthalpy of formation of AgCl is favorable, since it has to counter an unfavorable entropy loss in the formation of an ordered crystal lattice.

As you can see, I cannot provide you with a thorough answer, it would be extremely time consuming (for me) and I'd have to conduct many many calculations. I really hope someone can clarify the why. Maybe you can do your own research from the answers provided here and come back with a more "elaborate" answer.


Separate less volatile component from distilling a mixture Is it a strict rule that when we boil a mixture of two liquids, with one being less volatile than the other, the more volatile vapour has to form the greater portion of the total vapours?

If it is so, how are we able to obtain a pure liquid of less volatility in the distillate and an azeotropic mixture by distilling a certain composition of a non-ideal mixture? Shouldn't we be getting ALWAYS the more volatile component in the higher plates and thus in the distillate?

And if it is not so, please enlighten me with a valid explanation. Thank you.

The more volatile compound forms greater vapour portion, compared to its portion in the distilled liquid.

It forms e.g. 15% molar fraction of vapour, if there is 10% molar fraction in liquid.

In case of deviation from the Raoult law, when there is interaction between compounds, an azeotrope with minimal  ( ethanol-water ) or maximal ( nitric acid water ) boiling point may be formed. This azeotrope acts like if it were the 3rd compound.

Within rectification columns , for each distillation stage upwards, the composition is slightly enriched with more volatile fraction.

And vice versa,  for each distillation stage downwards, the composition is slightly enriched with less volatile fraction.

If we are interested just in more volatile fractions, the column bottom is attached to the vapour source input, what is typical for labs.

If we are interested in all fractions, what is typical in oil industry, the input is injected somewhere in the middle of column, where the equilibrium composition is the most similar to the input.

The more or less volatile fraction may be an azeotrop. Components in ratio forming the azeotropic ratio cannot be separated by destination, unless other components are added.

The typical example is the azeotrope ethanol water, denying to get more than about 95% ethanol by distillation, as this azeotrope has lower boiling point than pure ethanol.

Pure ethanol can be gained by adding benzene,  forming the ternary azeotrope  benzene - ethanol - water, with even lower boiling point.


Differentiation between zinc, aluminium, and magnesium ions in solution If I have three aqueous ionic solutions in which I know that the cation is $\ce{Al^3+}$, $\ce{Mg^2+}$, or $\ce{Zn^2+}$, how do I find out which is which?

I was thinking to add $\ce{OH-}$ in the form of $\ce{NaOH}$, or something else to produce a precipitate, and testing these precipitates in other liquids, but I'm not sure what other liquids to try.

Apart from the methods, Ringo already described, you can do a few other tests.

Aluminium

This is loosely translated from the German chemgapedia.de. Look at the pretty pictures.

Probably the easiest test you can do is reacting it with Morin in ethanoic acidic medium. It forms a yellow-green chelate complex, which has strong fluorescence under UV light. The colour and UV activity should be persistent when adding glacial acidic acid (water free), but vanish when adding concentrated hydrochloric acid.

Unfortunately sodium is also fluorescent, so keep your equipment clean and if you use a base, use potassium hydroxide instead. Also perform blind tests with all the chemicals involved, to see if they are clean enough.

Other elements that can show fluorescence are $\ce{In^3+, Ga^3+, Sc^3+, Zr^4+, Be^2+}$, usually very pH dependent.

From my own experience in first year of university, this is a very efficient test. Another test, that can be performed from the salt, is reacting it to Cobalt blue. If you have a solution, you have to precipitate aluminium first: $$\ce{Al^3+ + 3 {}^{-}OH ->C[NH3/NH4Cl] Al(OH)3 v}$$
Put the precipitate on a magnesia groove and add some cobalt nitrate. Glowing the mixture will result in the characteristic cobalt blue:
\begin{align}
\ce{2 Al(OH)3 &->[\Delta] Al2O3 + 3 H2O ^}\\
\ce{Co(NO3)2 &->[\Delta] CoO + 2NO2 ^ + 1/2 O2 ^}\\
\ce{CoO + Al2O3 &->[\Delta] \color{\navy}{CoAl2O4}}\\
\end{align}

This is an almost fool proof method for aluminium and is not influenced by other elements. And it looks quite nice.

Zinc

Zinc is probably one of the only elements that form a white precipitate with sulfide ions.
$$\ce{Zn^2+ + H2S -> ZnS v + 2 H+}$$
This precipitate can be used to form Cobalt green. The procedure is the same as for the above Cobalt blue.
$$\ce{ZnS + Co(NO3)2 + O2 ->[\Delta] \color{\green}{ZnCo2O4} + 4 NO2 ^ + SO2 ^ }$$

The only interference with this test is, that using too much cobalt nitrate will produce the black cobalt oxide, which can make it hard to see the green colour.
$$\ce{3Co(NO3)2 ->[\Delta] \mathbf{Co3O4} + 6 NO2 ^ + O2 ^}$$

Again you can have a look at the pretty pictures at chemgapedia.de.

Magnesium

A common problem with magnesium is, that all tests only work well and specific if you have a pure solution. Most heavier elements react more readily with the chemicals, than magnesium itself does. Another problem is, that salts from magnesium have no flame colouration at all.

I am again loosely translating from chemgapedia.de.*

You can precipitate magnesium with 8-hydroxyquinoline:

It forms a yellow-green complex. This is only a specific test, if you know you have no heavier elements in your test tube. For example, aluminium forms the complex Tris(8-hydroxyquinolinato)aluminium: $$\ce{Al^3+ + 3 C9H6NOH -> Al(C9H6NO)3 + 3 H+}$$

Another possibility is precipitation as magnesium ammonium phosphate (struvite). And yes, you guessed it, this happens with all other alkali and earth alkali elements (except lithium). But if you are certain you only have got magnesium, you can grow nice crystals. The pH for this reaction should be buffered at around 8-9 with ammonia/ ammonium chloride.
$$\ce{Mg^2+ + NH4+ + {}^{-}OH + HPO4^2- ->C[H2O][NH3/NH4Cl] Mg(NH4)PO4⋅6H2O v}$$

There are two more tests, that I would consider best suitable, because of the pretty colour.

The reaction of magnesium hydroxide with titan yellow in presence of sodium hydroxide will produce a red precipitate. It is important to do a control/ blind experiment, as older solutions of titan yellow will form a precipitate with sodium hydroxide alone.

The reaction of magnesium hydroxide with azo violet (Magneson) or Magneson II (Magnezon II) in presence of sodium hydroxide will produce a blue precipitate. Again test with sodium hydroxide to observe the violet solution. This reaction also occurs with $\ce{Ca^2+, Al^3+}$.

Testing for magnesium is as far as I remember is some kind of a lottery when in combination with other elements (qualitative analysis). Good luck.

* Please note that the first reaction equation has depicted hydroxyquinoline with two circles for the aromatic system, which is wrong. In benzene one circle stands for 6 $\pi$ electrons. So that would lead to hydroxyquinoline having 12 $\pi$ electrons, two too many.


What is this blue crystal? I was sorting out some old boxes in one of the rooms in my house, when I found a small container with about 10 grams of this blue crystalline powder. I assume that it contains some form of copper due to the colour, but the label's all ripped and I can't see the name of it.

So what compound is it?

It appears to glow in the image, but this is just the reflection from the camera flash and it doesn't look like this in real life.

As orthocresol also mentioned, it's probably copper sulfate. As you said, the blue color essentially gives it away.

(picture taken from Science Madness Wiki)


What gas is released during dissolving NaOH What gas is released during dissolving solid $\ce{NaOH}$ in a high purity water? Water more than 10 M$\Omega$/cm and standard quality $\ce{NaOH}$.

I don't think any gas is released.  You are probably experiencing dust or mist that contains NaOH itself, or tiny amounts of gas phase NaOH.

There is related discussion at a soapmaking website

http://www.tikvah.com/medical/lye.html

See also "Pneumothorax following inhalation of caustic soda fumes" Archives of Emergency Medicine, 1988, 5, 45-47


Why are there 14 electrons in the third shell of iron atom? The third shell must contain at least 18 electrons. But in iron there are 14 electrons in third and 2 electrons in 4th. Why are there 14 electrons in the third shell of iron atom?

Your doubt is - Total number of electrons in 3rd shell  (i.e. electrons from 3s + electrons from 3p + electrons from 3d) is not yet 18 but still how we filled up the 4th shell?

You have a general misconception than atomic orbitals always follow the 2,8,18... electron filling rule. According to this 2,8,18.. electron filling rule, 2 electrons should get filled up in 1st shell, then 8 electrons in second shell, then 18 electrons in 3rd shell........

Remember this rule mentioned above is completely elementary! Just remember 2,8,18... electron filling rule is actually a waste. Simply I would recommend forget about that 2,8,18... rule. It is very elementary rule... simply forget it. In 2,8,18... electron filling rule, "lower energy level orbitals gets filled up first"- this statement is not considered in that elementary 2,8,18..rule. But this statement which that elementary rule doesn't consider is very important statement to be followed while filling up of electrons. So remember 2,8,18... electron filling rule is an elementary rule for beginners who  just start to read about atoms and orbitals! Forget about that rule.

Its not necessary that 3rd shell has to have 18 electrons for 4th shell filling up to start.
The electron filling depends upon the energy levels of the orbitals. Lower energy orbitals get filled up first.

Iron's configuration is: 1s2 2s2 2p6 3s2 3p6 4s2 3d6. You can note from the configuration, 4s gets filled up before 3d . The reason behind this is lower energy orbitals are filled up first. Even though 4s belongs to 4th shell, still its energy is less than 3d which can be reasoned out by (n+l) rule. 4s energy is less than 3d, hence the two electrons which you expected to be going to 3d goes to 4s before 3d filling starts. Hence you end up having 14 electrons in 3rd shell and 2 electron in 4s. Remember 4s got filled up before 3d and the reason is 4s energy is less than that of 3d.


Why is the boiling point of homologous primary > secondary > tertiary alcohols whilst their solubility in water is the reverse? From my understanding, for homologous alcohols, primary alcohols have the highest boiling points, because the OH group is the most accessible for other molecules to hydrogen bond to, therefore stronger intermolecular forces result in higher mp/bp.

One the other hand tertiary alcohols have the highest solubility in water because the packed structure of branches means the non-polar part of the molecule is condensed and therefore water near any part of the molecule can receive these hydrogen bonds. Doesn't this mean tertiary alcohols have the most accessible hydrogen bonds and therefore should have the highest melting point too?

Alternatively does this have to do with water being small and alcohols being larger therefore 'access' requirements to hydrogen bonding are different as larger alcohols need to be even closer than water to make a difference therefore the packing is less relevant as even being slightly away from the molecule means failure to access hydrogen bonds but water will be ok with being slightly away.

The key concept here is that for solubility, you need to compare alcohols with the same number of carbon atoms. For example, the solubility order of isomers of butanol is as follows:
$$\ce{CH3CH2CH2CH2OH < (CH3)2CHCH2OH < (CH3)3COH}$$

This difference in solubility follows the order 3°>2°>1° and can be explained by the branching concept that you have used in your question. However, If you compare alcohols with different number of carbon atoms, say butanol and methanol, then the alcohol with a smaller hydrocarbon part will be more soluble in water.

The boiling point of alcohols increases with the increase in the number of carbon atoms, as stronger van der waals forces start acting in between the molecules. For example, Phenol boils at around 180°C whereas methanol boils at around 60°C. For alcohols with the same number of carbon atoms, a more exposed hydroxyl group will result in stronger intermolecular H-bonding.

TL;DR: your comparision can only be applied for alcohols with the same number of carbon atoms


What is the range of photon energies can be effectively used to split a diatomic molecule like hydrogen? I saw photo-ignition of hydrogen-chlorine mixture using UV light on YouTube.

Chlorine molecule bond dissociation energy is $\pu{2.51 eV}$ (bluish colour).
Bromine molecule bond dissociation energy is $\pu{2 eV}$ (orange colour).

Can I use any photon which has energies greater than $\pu{2 eV}$ to split bromine molecule,  like green or blue or UV light? Or should I use only photons which has energies around $\pu{2 eV}$?

My concern is about absorption spectrum of gases. And this question is not just for bromine gas, but for any diatomic molecule like hydrogen or Chlorine.

If the photon's $E=\hbar\omega=h\nu=\frac{hc}{\lambda} $ energy is greater than the energy of the bond which you want to break up it will break it up. It's always best to use higher energy photons but not too high because it may break up bonds which were unwanted to break up in your reaction . So the best for you is to use blue or higher energy UV light for your experiment.You need to deal with the rate of the reformation of bromine molecules, there will be an equlibrium between the production and decomposition of bromine molecules depending on the intensity, not the energy of the light. If the photons has higher energies than a given bond they will always break it up, this is why gamma radiation is so dangerous. The absorbtion spectrum doesn't have a lot connections to this phenomena. You can use the Maxwell-Boltzmann distribution
$$f_{v}(v_{i})=\sqrt{\frac{m}{2\pi k_{B}T}}exp\left[-\frac{mv_{i}^{2}}{2k_{B}T}\right]$$
to determine the Doppler-effect's impact on the absorption spectrum depending on the temperature and to construct an absorbtion function explicitly. The doppler effect will affect the percieved frequency and energy of the light by gas particles so if you want to be as accurate as possible, you need to calculate it.The Maxwell-Boltzmann distribution is only suitable in low temperatures (below several thousand Kelvins).


Is the Fischer Esterification mechanism very simplified? Example mechanism:

Is that very simplified, or is that pretty much what is very likely happening at the molecular/atomic level when you do the reaction in real life?  (Of course besides the free $\ce{H+}$, that is obviously hydronium, protonated alcohols, or the acid itself.)

That is the wrong question to ask. Mechanisms are putative pathways. Accepted mechanisms are mechanisms that we have not found evidence to refute. There is no way to determine whether a mechanism is the true mechanism.


pH change with temperature I understand that the pH of solutions decreases as the temperature increases but is there a reason as to why some solution's pH decreases more than others even if the temperature change is the same?

The pH of a solution is dependent on acid/base equilibria of the substances in the solution. Each of these equilibria will be affected differently by temperature changes. If reactions are endothermic, the equilibrium will shift towards the products upon increasing temperature, otherwise towards the reactants. How much the equilibrium shifts depends on how exo/endothermic the reaction is.

One of the equilibria is the autodissociation of water, which is endothermic. Increasing the temperature will increase product (H+ and OH-) concentration in pure water. Per definition, the pH will decrease as the H+ concentration (or activity) increases.

More complex solutions will either show an increase or a decrease in pH when temperature is raised. For example, some buffers are notorious for changing pH a lot while others are pretty stable, see this table.


Dissociation of ligand If a ligand K[Ag(CN)2] is dissociated.. Which one is correct dissociation equation?

$$\ce{K[Ag(CN)2] -> K+ + Ag+ + 2CN^-}$$

or

$$\ce{K[Ag(CN)2] -> K+ + [Ag(CN)2]^-}$$

Three different types of dissociation can occur here, the two mentioned by the OP:

$$\ce{K[Ag(CN)2] <=> K+ + Ag+ +2CN-}$$
$$\ce{K[Ag(CN)2] <=> K+ + Ag(CN)2^-}$$

And one where only one cyanide ligand dissociates from the silver:

$$\ce{K[Ag(CN)2] <=> K+ + AgCN + CN-}$$

However, as Klaus points out in his comment dissociation $\ne$ ionization. Ionization is a chemical reaction that generates ions where there were none before. Consider the ionization reaction of ammonia $\ce{NH3}$ in water:

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

I'm not sure if this type of reaction is possible, although the cyanide anion is the conjugate base of a weak acid $(\ce{HCN}, \ \text{pK}_a =9.21)$. Thus, any liberated cyanide ion could react with water (assuming this is an aqueous solution!), though I would hardly call this an ionization (as there is no net increase in ions).

$$\ce{CN- + H2O <=> HCN + OH-}$$


why is the pressure exerted by ideal gas on the walls of the container more than that of real gas I have learnt that ideal gases have no intermolecular forces. So even at high pressure the volume gets completely compressed.
But in case of real gases,  we have both attractive and repulsive forces. At far distance,  attractive forces dominate.
But at high pressure , when the particles are closer to each other,  the repulsive forces dominate over the attractive forces.
So my question is,  if the repulsive forces dominate over the attractive forces at close range,  then real gas should exert more pressure on the walls of the container.  With this one more doubt arises,  how do gases liquify if repulsive forces are dominant?

For neutral gases, the intermolecular forces that we observe tend to be attractive. Significant repulsive forces don't really show up until your particles have charge.

The other difference between ideal gases and real gases are that real gases take up space, but the volume deviation from the ideal gas law is usually swamped out by the attractive intermolecular forces. This causes a real gas to exert less pressure than an ideal gas, and also is the reason matter undergoes phase changes.


Can we equate nucleophilicity and Lewis basicity? Please note that it is specifically Lewis basicity, not any other definition of basicity.

Consider the interesting trend in nucleophilicity and basicity of the fluoride and iodide anions. The reason claimed for the weak nucleophilicity of the fluoride ion in polar protic solvents is due to the hydration effects which it experiences as it interacts with solvent molecules. This hydration effect effectively prevents it from reacting with electrophiles. In a sense, the hydration effect severely decreases the availability of the electron pair for donation.

Now, although it is claimed to be a weak nucleophile in aqueous solution, the fluoride ion still behaves as a strong Lewis base. This is logical because the interacting species are different in the two different situations. In the case of nucleophilicty, we are considering its nucleophilicity of the species towards another electrophile (e.g. a haloalkane) placed in solution.

In the case of basicity, we are considering it, in terms of its hydrolysis reactions with the water molecules in the hydration shell, producing the hydroxide ions. Thus, the solution becomes an alkaline one and we say that it is basic.

Now, if we use the same perspective to view nucleophilicity, we can thus say that the water molecule is the electrophile and that the fluoride ion is indeed a strong nucleophile because it has managed to attack the electron-deficient hydrogen atoms on the water molecule.

Consider this: What if I remove all the hydroxide ions produce from the hydrolysis of water from an aqueous solution of fluoride ions and add in an acidic species? Would we still observe a reaction between the fluoride ions and the acid species?

My point is that when we consider Lewis basicity and nucleophilicity of reacting species we are not making a fair comparison and thus, we say that fluoride ions are strong bases but weak nucleophiles in polar protic solvents. This may be true based on the Arrhenius definition of acids and bases. But if we were to look purely at Lewis basicity, it would be certainly be inaccurate.

From the perspective of interactions with water molecules, the fluoride ions are definitely good Lewis bases and also good nucleophiles. From the perspective of interactions with other reacting electrophiles placed into solution, the fluoride ions would be both weak Lewis bases and weak nucleophiles, neglecting the hydroxide ions (because even though the acidic species may react with the hydroxide ions, it is not reacting with the fluoride ions, thus we would see those reactions of the acids with the hydroxide ions as a side reaction).

Therefore, is it appropriate to equate these two terms because they both refer to the electron-donating ability of a chemical species?

If I have overlooked something (i.e. some other differences between the two terms), please inform me via the comments.

I appreciate your perseverance to think so deeply about this!
But...Yes, you have overlooked the main difference between Basicity & Nucleophilicity ,i.e., Basicity (i.e. Lewis basicity) is a Thermodynamic term whereas Nucleophilicity is a Kinetic term. But here you have compared them under the belt of only thermodynamics.
Basicity is measured on the basis of the stability of the species, whereas Nucleophilicity is mainly aimed at the 'donating ability' of the species.
e.g. In polar protic solvents, the order of nucleophilicity in case of halide ions is:

  $\ce{F-}$ < $\ce{Cl-}$ < $\ce{Br-}$ < $\ce{I-}$

Which contradicts with the basicity order.

Whereas in polar aprotic solvents, the order is:

  $\ce{F-}$ > $\ce{Cl-}$ > $\ce{Br-}$ > $\ce{I-}$

Which follows the basicity order.

I hope this might help you facing your confusion a li'l more logically!


Heitler-London hydrogen model As I am learning about the valence bond theory, the model of hydrogen that Heitler and London created, seems to make sense. But, the theoretical and the experimental results are not the same.

Experimental results:

For the molecule to form, the distance between atoms should be $r=74.1~\mathrm{pm}$

and the energy on the other hand $E= -4.474~\mathrm{eV}$.

The theoretical results are this way:

$r= -\frac{1}{4}~\mathrm{pm}$ and $E= 89.9~\mathrm{eV}$

My question is, is it known why these deviations occurred?

Or is this considered as a correct calculation with approximation?

I think you have some typo or flaw in your question.  "E = 89.9 pm" doesn't make sense.  You are saying energy equals a length. The edited version "E = 89.9 eV" has energy units, but the theory never predicted such as large value.  And "r = -1/4 pm" doesn't make sense either.  Distance can't be negative. Also, even it were positive 1/4 pm, the theory never predicted such a small magnitude value.

The initial calculations of Heitler and London were:

r = 80pm
E = -2.4 eV

A major source of error was inability to compute integrals.  Additionally, the calculations neglect ionic contributions.

See Computational Imperatives in Quantum Chemistry for a historical overview.


In the nitration, why fluorobenzene has less products in the ortho position than Clorobenzene? Is it because, due to the eletronegativity of Fluorine, his mesomeric effect will be less pronounced, which means the electronic density will be lower, than for the clorobenzes, in the ortho positions?

Halogens are deactivating groups even though they show $+M$ mesomeric effect. The inductive effect is stronger than mesomeric effect. As the halogens show $-I$ inductive effect, they are deactivating groups.

When it comes to selectivity, the mesomeric effect plays a role. As the halogens show $+M$ mesomeric effect, the resonance structure has higher electron density at ortho and para positions. Therefore, halogens are ortho-para directing.

When you take inductive effect into account for its directing properties, the para position is more favorable for electrophilic reagents because it has a higher electron density compared to the ortho position. The strength of the inductive effect decreases with distance. As the ortho position is closer to the halogen, the inductive is stronger at this position. Therefore, the major product for electrophilic substitution reactions for haloarenes is the para-substituted product.

Fluorine is more electronegative than chlorine and shows a higher tendency to withdraw electrons towards itself than chlorine. Therefore, fluorine shows a stronger inductive effect than chlorine. As fluorine reduces the electron density at the ortho position more than chlorine, you get a lesser amount of ortho substituted product with fluorine.


A few questions about the conditions of the Diels Alder Reactions I'm attempting to extract the diels alder reaction into a generic reaction and am trying to gain a better understanding of it.

Can a conjugated diene react with another conjugated diene via the diels alder reaction? And is the reaction limited to only dienes or can it be expanded to polyenes?

Does anyone happen to have a link as to the details on hetero diels alder reactions and the structures that can undergo such reactions and limitations on chemical species that can react via this reaction.

Yes, conjugated dienes can react with other conjugated dienes in Diels-Alder reactions. A well known example of this is the dimerization of cyclopentadiene. All the standard caveats, restrictions, and other considerations still apply, namely:

The diene must not be rotationally locked in such a way that it cannot adopt an s-cis/synperiplanar conformation. As a logical corollary, cyclic dienes that are already locked into the correct s-cis conformation often have lower activation barriers.
The kinetics of the reaction are accelerated by electron-withdrawing substituents on the dienophile and electron-donating substituents on the diene.
The stereochemistry of the major product is highly substrate-dependent. A preference for the endo isomer exists in cases where the dienophile has unsaturated substituents. In those cases, the transition state is stabilized by an overlap of the $\pi$-electron system between said unsaturated substituents and the $\pi$ orbitals of the newly forming double-bond at the back of the diene. (Just as a side note, Carey & Sundberg refer to this as the "Alder rule" in Advanced Organic Chemistry, but I've never seen that term used anywhere else.)

In principle, there's no reason why two conjugated $\pi$-bonds of a polyene cannot react similarly, provided they satisfy the orbital symmetry and conformational requirements and the reaction kinetics are favorable. Even aromatic molecules can undergo Diels-Alder reactions under the right set of conditions (see, for example, the Diels-Alder reaction between benzene and furan via a benzyne intermediate). There are also numerous instances of well-documented Diels-Alder reactions involving heteroatomic dienophiles, alkynes as dienophiles, polymerization reactions, etc.

If you can get your hands on the aforementioned Carey & Sundberg book, they discuss the Diels-Alder reaction at length (including relatively rigorous explanations of the stereospecifity and reaction kinetics via molecular orbital theory). If you're looking for a comprehensive reference, I doubt there's anything superior to the venerable March's Advanced Organic Chemistry. There's a chapter on Diels-Alder reactions (along with virtually every other organic chemistry reaction known to science), as well as a massive list of accumulated primary literature references if further detail is required.


Why isn't Cahn Ingold Prelog principle used to name the isomers of complex compound having square planar geometry & different ligands attached to it? For a complex compound with coordination number 4, say having square planar geometry, and say all the ligands attached to the central metal atom are different, then we know that we will get 3 geometrical isomers.

Now my doubt is like in organic chemistry if we had 4 different groups around a carbon atom, we still named those isomers as Z and E isomers (by Cahn Ingold Prelog principle) . Similarly why can't we name the isomers as Z and E for the case of complex compound with square planar arrangement having four different ligands attached to it?

That is why can't we name those 3 geometrical isomers as Z and E isomers of the complex compound by Cahn Ingold Prelog principle?

Part I

The CIP rules in fact are used to determine the priority. However, a different notation system has been developed for inorganic nomenclature of coordination compounds.

For the square planar geometry, SP-4-x descriptor (an instance of ‘polyhedral symbol’) is used, where x is a CIP-based priority number of the ligand in trans (opposite) position relative to the ligand with the highest priority number (1).

In $\ce{[Mabcd]}$ complexes (ligand priority a > b > c > d, priority numbers 1, 2, 3, 4, respectively), for the 3 possible isomers, the descriptors are (forgive me the use of left/right superscript/subscript typography to depict square planar geometry, for the sake of simplicity):

$\ce{^\style{font-weight:bold}{a}_cM^b_\style{font-weight:bold}{d}}$    SP-4-4
$\ce{^\style{font-weight:bold}{a}_dM^b_\style{font-weight:bold}{c}}$    SP-4-3
$\ce{^\style{font-weight:bold}{a}_dM^c_\style{font-weight:bold}{b}}$    SP-4-2

For $\ce{[Ma2b2]}$ (priority numbers 1, 1, 2, 2) and $\ce{[Ma2bc]}$ (priority numbers 1, 1, 2, 3) complexes, where only 2 isomers are possible, the cis/trans notation can be used as well. The E/Z notation from organic chemistry is not used at all for coordination compounds central atom configuration.

Examples:

$\ce{^{\style{color:grey}{(1)}{ }I}_{\style{color:grey}{(3)}{ }Cl}Pt^{Br{ }\style{color:grey}{(2)}}_{F{ }\style{color:grey}{(4)}}}$
(SP-4-4)-bromidochloridofluoridoiodidoplatinate(2−)
(hypothetical)
$\ce{^{\style{color:grey}{(1)}{ }Cl}_{\style{color:grey}{(2)}{ }NH_3}Pt^{NH_3{ }\style{color:grey}{(2)}}_{Cl{ }\style{color:grey}{(1)}}}$
(SP-4-1)-diamminedichloroplatinum(II)
trans-diamminedichloroplatinum(II)
$\ce{^{\style{color:grey}{(1)}{ }Cl}_{\style{color:grey}{(1)}{ }Cl}Pt^{NH_3{ }\style{color:grey}{(2)}}_{NH_3{ }\style{color:grey}{(2)}}}$
(SP-4-2)-diamminedichloroplatinum(II)
cis-diamminedichloroplatinum(II)
cisplatin
$\ce{^{\style{color:grey}{(1)}{ }Cl}_{\style{color:grey}{(2)}{ }Ph_{3}P}Ir^{PPh_{3}{ }\style{color:grey}{(2)}}_{CO{ }\style{color:grey}{(3)}}}$
(SP-4-3)-carbonylchloridobis(triphenylphosphane)iridium(I)
trans-carbonylchloridobis(triphenylphosphane)iridium(I)
Vaska's complex (a different IUPAC name descriptor claimed there is disputed)

(References: see Loong's comment below the question.)

Part II

Now, your question is self-contradictory. You would like to use E/Z (two) descriptors for three different isomers.

Maybe you wanted to ask something like
Why are there 3 geometrical isomers of square planar [Mabcd] complexes, but only 2 isomers of tetrasubstituted ethene C2abcd?
(Note the latter statement is not true)

It's because for $\style{font-weight:bold}{\ce{[Mabcd]}}$, the four different ligands are at the vertices (corners) of a square. Which can be abstracted as a ‘closed loop’, i.e. a cycle graph. The combinatorics can be abstracted to free circular permutations, $P'_{n}={1 \over 2} (n-1)!$,
for $n=4$: $P'_{4}=3$.

For $\style{font-weight:bold}{\ce{C2abcd}}$, the situation is somewhat different. This is a combination of positional isomers $\ce{xyC=Czw}$, where the set $\{x,y\}$ is $\{\text{a}, \text{combination}_2(\{\text{b},\text{c},\text{d}\})\}$, and $\{z,w\}$ is a complement. There are (concidentally?) 3 of them (${n \choose k} = {{n!} \over {k! (n-k)!}}, {3 \choose 2} = 3$):

$\ce{abC=Ccd}$
$\ce{acC=Cbd}$
$\ce{adC=Cbc}$

(Here the different sortings of substituents at each carbon does not matter.)

However, the isomerism one is talking about when using the E/Z descriptors applies to geometrical isomers of a single positional isomer (take e.g. $\ce{abC=Ccd}$). Now, the sorting of substituents at each carbon does matter. What makes the isomers different is matching vs non-matching sort order at each carbon:

$\ce{abC=Ccd}$    ≡    $\ce{baC=Cdc}$    (ascending-ascending, descending-descending – matching)
$\ce{abC=Cdc}$    ≡    $\ce{baC=Ccd}$    (ascending-descending, descending-ascending – non‑matching)

which are 2 possibilities, and correspond to E and Z configurations, respectively (provided that the lexicographical order corresponds with the CIP priority order).

So, the total number of C2abcd isomers is 6 (3 positional isomers, each has 2 geometrical isomers). One think of the alkene as an object similar to the square planar metal complex, where the C=C can be arranged in the center in two orientations, horizontal and vertical, hence 2× as many possibilities than ${\ce{[Mabcd]}}$.

(I hope exactly zero mathematicians and chemists were harmed during the reading.)


Can solid acids and bases have pH values? If not, how are they classified as acids or bases? This answer states

  $\mathrm{pH}$ is the aqueous concentration of $\ce{H3O+}$ or $\ce{H+}$ ions in soution. I would not say that ice lacks $\ce{H3O+}$ and $\ce{OH-}$ ions as ice's structure would allow for such, however, since the ions are not in aqueous solution, the material cannot rightfully have a "$\mathrm{pH}$" as we know it.

How does that generalize to solid acids and solid bases? Can solid acids and bases have pH values? If they don't have a pH, how would the solid form of a new compound then be classified as an acid or a base?

I think we should not mix the concept of pH, which is purely an analytical measurement, with the concept of acid and bases. It is a common misconception that pH does not exist in organic solvents.  As you already know, there are several views of an acid or bases. The current IUPAC version is "A molecular entity or chemical species capable of donating a hydron (proton) (see Brønsted acid) or capable of forming a covalent bond with an electron pair (see Lewis acid)." This definition does not require the presence of water or any pH value.

If we take pure dry HCl gas and ammonia gas and mix them. The reaction is a classic acid base reaction. Solid ammonium chloride is formed instantly. One may then ask a question of what is the pH of a gas? However gas phase acid base chemistry does exist and it can be studied by mass spectrometry but pH does not need to be invoked.

Now instead of asking the pH of solid acids/bases, a more realistic question is what is the surface pH of a given solid? I was interested in this type of problems sometime ago but didn't find much literature. Assume an ion-exchanger which consists of $\ce{SO3-H+}$ groups on a solid polymeric matrix, basically sulfonate styrene divinylbenzene. Assume that it is in equilibrium with 1 mM HCl solution. The solution pH is 3, but what is the surface pH? It is certainly way below zero, because sulfonic acid is a very strong acid and the surface concentration of sulfonic acid groups is pretty high yet the groups are not mobile in the solution!

Similarly, solid acid catalysts exist. Also see how solid acids are titrated [1].

References

Chai, S.-H.; Wang, H.-P.; Liang, Y.; Xu, B.-Q. Sustainable Production of Acrolein: Investigation of Solid Acid–Base Catalysts for Gas-Phase Dehydration of Glycerol. Green Chem. 2007, 9 (10), 1130–1136. https://doi.org/10.1039/B702200J.


Disproportionation of silver(I) in aqueous solution Use the following equations to predict whether or not $\ce{Ag+}$ ions will disproportionate in solution:
$$
\begin{align}
\ce{Ag+(aq) + e- &-> Ag} &\qquad E^\circ = \pu{+0.80 V}\\
\ce{Ag^2+(aq) + e- &-> Ag+} &\qquad E^\circ = \pu{+2.00 V}
\end{align}
$$

I used the method $E_\mathrm{cell} = E_\mathrm{red} - E_\mathrm{ox}$ and thought that $\pu{0.80 V}$ would be more negative. Therefore, it is oxidised and the reaction would be thermodynamically feasible.
However, my textbook and another webpage has a different answer. I am a bit stuck on how to get the right answer. Please someone explain me.

It is convenient to solve problems like that with a Latimer diagram, which is a great tool for predicting conditions for the reactions of disproportionation and synproportionation.
A generic Latimer diagram
$$\ce{A ->[$E_1$] B ->[$E_2$] C}$$
posesses the following properties:

If $E_2 > E_1,$ then $\ce{B}$ is thermodynamically unstable and disproportionates to $\ce{A}$ and $\ce{C}.$
If $E_2 < E_1,$ then the mixture of $\ce{A}$ and $\ce{C}$ is thermodynamically unstable and synproportionates to $\ce{B}.$

Now we can visualize your problem using a Latimer diagram
$$\ce{Ag^2+ ->[\pu{+2.00 V}] Ag+ ->[\pu{+0.80 V}] Ag}$$
and the condition of disproportionation resulting from the application of Nernst equation:

if the potential to the right of the species is higher than the potential on the left, it will disproportionate.

Since $E^\circ(\ce{Ag+(aq)/Ag}) < E^\circ(\ce{Ag^2+(aq)/Ag+(aq)}),$ disproportionation is thermodynamically unfavorable, and silver(I) can be considered stable in solution.
You can get the same result from a linear combination of both equations written for the disproportionation:
$$
\begin{align}
\ce{Ag+(aq) + e- &-> Ag} &\quad E^\circ_1 = \pu{+0.80 V} & \tag{1}\\
\ce{Ag^2+(aq) + e- &-> Ag+(aq)} &\quad E^\circ_2 = \pu{+2.00 V} &\quad|\cdot (-1)\tag{2}\\
\hline
\ce{2 Ag+(aq) &-> Ag + Ag^2+(aq)} &\quad E^\circ = \pu{-1.20 V}
\end{align}
$$
Since resulting $E^\circ = \pu{-1.20 V} < 0,$ free Gibbs energy $Δ_\mathrm{r}G^\circ = -nFE^\circ > 0,$ and the disproportionation of silver(I) in solution can be considered a thermodynamically unfavorable process.


What makes Citrate or Citric Acid an acid? As far as I know, an acid is something that gives off a proton or hydrogen ion $H^{+}$.

But when I look at Citric Acid,

There are three $COO^{-}$, which I think is a carboxyl group, that lacks $H^{+}$.

They seem to rather demand a hydrogen ion from its environment than donate one, which make it basic and not acidic. But the name is still citrate or citric acid. Can someone explain?

It depends on the pH of the solution.  At higher pHs the carboxyl groups will be dissociated (as drawn).  At lower pHs the equilibrium will shift and most of the COO- will become protonated.

In general, carboxylic acids are relatively weak compared to stronger acids that dissociate almost completely when added to an aqueous solution.  For example, acetic acid, or vinegar, will not burn a hole in cotton clothing, whereas sulfuric acid (the kind in an automobile battery) will easily burn a hole in cotton cloth.

Also, don’t confuse carboxylic acids with their carboxylate salts.  Sodium citrate, where the three protons are replaced with Na+, will act as a base when added to water, and raise the pH.


How effective is a level A hazmat suit for immersion in deadly substances? I know that level A type hazmat suits can protect against minor splashes, but could somebody in one survive without harm after complete immersion for a few minutes in deadly chemicals such as sodium cyanide, high concentration (100%) hydrogen peroxide, and 80% hydrochloric/nitric acid?

Note that I'm used to standard lab safety measures including gas masks and masks mith independent air supply but I never used a Hazmat level A suit in research. You might want to take the following with a grain of salt!

    Sodium cyanide

($\ce{NaCN}$) is a solid with a melting point around 560 °C. While it is a solid, jumping into a pool of it shouldn't do any harm. Level A suits are like "your personal bubble" with independant air support. They are however not that heat resistant: Do not jump into a pool of molten sodium cyanide!

    High concentration (100%) Hydrogen Peroxide

3% solutions of hydrogen peroxide ($\ce{H2O2}$) in water are used for wound disinfection, solutions up to 18% are used to bleach hair. (In German, the term wasserstoffblond was used for bleach blonde).

Much higher concentrations or waterfree (100%) ($\ce{H2O2}$) is another league.

Hydrogen peroxide can decompose into water and oxygen:

$$\ce{2 H2O2 ->[\textrm{cat}] H2O + O2}$$

The reaction releases some heat, may happen spontaneously and is catalyzed by some metals. Add some burnable material that can be vapourized and you have a rocket propellant. Think Messserschmitt Me 163.

If the hazmat suit is tight, and the material withstands "bleaching" by the hydrogen peroxide there's still the question whether hooks, valves, visor frames or any other metal parts, that might be part of the whole gear will catalyze the decomposition.

    80% hydrochloric acid

That does not exist in water. Water won't take up that much hydrogen chloride gas under normal conditions. Concentrated hydrochlorid acid (~ 38%) should be fine.

    80% nitric acid

That's close to fuming nitric acid and again a pretty strong oxidant. I'd say that you're safe, but don't take my word for it ;)


Why does an acid exist when dissolved in water? Acid is a substance that releases $\ce{H+}$ ions when mixed with water But if it releases $\ce{H+}$ ions the  acid will ultimately not exist. For e.g we take the acid $\ce{H2SO4}$. If we add water then it becomes $$\ce{H2SO4 + H2O -> H+ + SO4^{-2} + H2O}$$ So the bond between hydrogen and sulphate will be broken therefore the acid will not exist, but we see that diluted acids are possible. Why?

Your doubts are legitimates, but there are some aspects that you should take in account:
1-Not all acid are fully dissociated
Not all acids fully dissociate, so if you take acetic acid you have in fact an equilibrium with some acetic acid molecules not dissociated. So your question holds only for strong acid like sulfuric acid you mentioned.
2-You are using pseudo-Arrhenius definition, why don't you use Lewis acid definition?
If you think the acid as a 'substance that that releases $H^{+}$' this is still an acid even if it dissociate, and even here you should notice that acidity is linked with $H^{+}$. However if you want to dig deeper there are many definitions of acid if you use Lewis acid theory and so you think the acid as a specie that accepts electrons you can easily see that in fact that the acidity is linked mainly with $H^{+}$!


Energies of Virtual Molecular Orbitals (LUMO) I just had a question about molecular orbitals. I was just wondering if the energies of the LUMO (virtual orbitals) derived from CCSD(T)/cc-pTVZ calculations could be trusted. I remember reading some time ago that these orbital energies could not be trusted since they contained no electrons. If they cannot be trusted, could you suggest any calculations I could perform to obtain the LUMO molecular orbital energies (perhaps TD-DFT or some other method)?

When thinking about orbital energy levels, it is not a good idea to think of them as fixed energy 'levels' that can be filled with electrons. This kind of picture is often found in introductory textbooks, but it is wrong. For example, it is known that even for the Hartree-Fock theory, the energy of a molecule is not a sum of the orbital energies. In this sense, you can never really consider orbital energies as the energies of the given electrons - which is really a meaningless quantity, if you think about it. Basically the only interesting quantity that you can get from an orbital energy is the first ionization energy based on Koopmans' theorem, which also is just approximately true.

So can you trust orbital energies? It really depends what you want to calculate. One thing that people often want is to calculate the UV-VIS spectrum of a molecule, which is, of course, the energy required to move an electron form an occupied orbital to a virtual orbital. But this energy should not be calculated as the difference of the energies of the two orbitals. The reason for this is not that virtual orbital energies can not be trusted, but the fact that when you move an electron, you now have a different electron configuration, that is, a different wavefunction. And with a different wavefunction, of course, you would get different energy levels. You should not use the energy levels of the original wavefunction to describe the energy levels of an excited state.

So in order to calculate UV-VIS spectra, you should rather use a multi-reference method with a large enough active space that includes both the ground and the excited states and also some 'buffering states' to expand the difference in the electron correlation. If you can afford a CCSD(T) calculation, then you might also be able to afford a CAS-SCF calculation with a formidable active space, maybe even add a bit of NEVPT2 to it. I also have some good experience with the SORCI method of Orca, if you want to check that out. And if you can not afford to do such large calculations, you might also fall back to TD-DFT - some good double hybrid methods have been made available this year.


Identifying unknown solution with indicators. why is one pKa value ignored and how to treat negatives values? Three test tubes contain an identical solution of unknown pH.

  The first one is tested with thymolphthalein and turns colourless
  The second is tested with a-naphtholphthalein and turns blue

  What colour would thmyol blue appear in the third tube?

  Note: If the ratio of one form to the other form (e.g. acid to base form or vice versa) is greater than $10:1$, the colour observed will be that of the predominant form. However, if between $10$ and $0.1$, the colour will be a mixture.

  \begin{array}{lccc}
\text{Indicator} &
\mathrm{p}K_\mathrm{a} &
\text{Colour of acid form} &
\text{Colour of base form} \\
\hline
\text{$\alpha$-naphtholphthalein} & 8   & \text{yellow}     & \text{blue}   \\
\text{thymol blue}                & 2   & \text{red}        & \text{yellow} \\
                                  & 8.8 & \text{yellow}     & \text{blue}   \\
\text{thymolphthalein}            & 10  & \text{colourless} & \text{blue}   \\
\hline
\end{array}

Steps to solve this question:

Find pH of solution via Henderson-Hasselbalch equation:
\begin{align}
  \mathrm{p}K_\mathrm{a}
  &= \ce{pH} - \log_{10} \frac{[\ce{In-}]}{[\ce{HIn}]}
\end{align}
For α-naphtholphthalein, as it appears blue, its basic form dominates and so
\begin{align}
\frac{[\ce{In-}]}{[\ce{HIn}]} &> 10:1 \\
  \mathrm{p}K_\mathrm{a}
  &= 8 =  \ce{pH} - \log_{10} (10)
   =  \ce{pH} - 1\\
\ce{pH} &= 9
\end{align}
For thymol blue
\begin{align}
  \mathrm{p}K_\mathrm{a}
  &= \ce{pH} - \log_{10} \frac{[\ce{In-}]}{[\ce{HIn}]}\\
  8.8
  &= 9 - \log_{10} \frac{[\ce{In-}]}{[\ce{HIn}]}\\
  \log_{10} \frac{[\ce{In-}]}{[\ce{HIn}]}
  &= -0.2
\end{align}

Where I get stuck:

How do I handle the negative sign in negative 0.2?
If it was positive 0.2, this would tell me that the ratio of $\log_{10} ([\ce{In-}]/[\ce{HIn}])$ is between $10$ and $0.1$, hence a mix between the acid colour and basic colour would result, giving a green solution. But I don't know how to treat the negative.
In my solution manual they ignore the $\mathrm{p}K_\mathrm{a}=2.0$ value of the thymol blue and only use the $\mathrm{p}K_\mathrm{a}=8.8$ value. Why is this the case?

As you noted, the $\mathrm{pH}$ is fairly close to 9.

It's probably quite different from the $\mathrm{p}K_{\mathrm{a}}$ value of $\alpha$-naphtholphthanlein because if you have any of the yellow acidic form around, the solution will take on a greenish tinge. One order of magnitude seems sufficient so, $\mathrm{pH} \geq 9$

The sample with thymolphtalein follows the same logic in that since we don't see any blue, there must be little of the base form, so the $\mathrm{pH} \leq 9$.

9 is very close to 8.8, the $\mathrm{pH}$ where we would expect equal forms of thymol blue. Indeed, $10^{0.2}\approx 1.05$, so both forms are present in solution in roughly equal amounts.

Further assuming that intensity of the two colors is similar, you would have a mix of yellow and blue, also known as green.


Oxidation of acetic acid in wine? When wine is exposed to oxygen, acetic acid forms because of an oxidation reaction.

The wine then loses acetic acid. Why? Is it because it is oxidised itself to form carbon dioxide?

The oxidation/reduction reactions of

ethanol <-> ethanal <-> acetic acid (ethanoic acid)

are at equilibrium. Oxidation (presence of $\ce{O2}$ or other oxidizing agent) and reduction (addition of $\ce{H}$ to the $\ce{C-O}$ carbon) are in constant flux. The system can be pushed to the right if wine is exposed to air for long periods of time, as you found, but at a certain point it will reach equilibrium, in which case the excess of acetic acid will reduce (marginally) back into ethanal and/or ethanol.

A better explanation may be that since the B.P. of ethanol (78 dec. C) is significantly lower than acetic acid (118 deg. C), the vapor pressure of ethanol is greater and it has a higher affinity to convert to a gaseous state. This also would shift equilibrium to the left to produce a lower post-equilibrium concentration of acetic acid.

Acetic acid does have the possibility of decarboxylating, but this is usually in excess heat. In this case, a $\ce{CO2}$ molecule breaks off of the acetic acid leaving $\ce{CH4}$, which would also be generated as a gas.

Finally, it is possible that you're mistaking "loss of acetic acid" for an actual increase in pH, irrelevant to the presence of whole acetic acid molecules. It could be that other materials in the wine "soak" protons to increase pH and leave ethanoate counter ions ($\ce{CH3COO-}$). You should clarify how you're actually quantifying the concentration.

I wouldn't expect the "decrease" of acetic acid to be significant, compared to the amount that was produced in the first place. I would also expect your system to re-equilibrate after a while longer. It would be interesting to see what happens next.


Na+ or Ca2+ . Which ion has the larger radius? Which ion has the larger radius? Some sites say $r(\ce{Na+}) > r(\ce{Ca^2+})$. Any explanation would be appreciated.

The thing is that you need to know the coordination environment in the first place as ionic radii are C.N.-dependent. From Shannon's canonical paper "Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides"[1]:

\begin{array}{cc}
\hline
\text{Ion} & \text{C.N.} & \text{C.R., Å} & \text{I.R., Å} \\
\hline
\ce{Na+} & 4 & 1.13 & 0.99 \\
& 5 & 1.14 & 1.00 \\
& 6 & 1.16 & 1.02 \\
& 7 & 1.26 & 1.12 \\
& 8 & 1.32 & 1.18 \\
& 9 & 1.38 & 1.24 \\
& 12 & 1.53 & 1.39 \\
\hline
\ce{Ca^2+} & 6 & 1.14 & 1.00 \\
& 7 & 1.20 & 1.06 \\
& 8 & 1.26 & 1.12 \\
& 9 & 1.32 & 1.18 \\
& 10 & 1.37 & 1.23 \\
& 12 & 1.48 & 1.34 \\
\hline
\end{array}

where C.N. - coordination number; C.R. - crystal radius; I.R. - ionic radius. The data is also available online for free. Comparing similar coordination environments, it's easier to notice that overall $\ce{Na+}$ is indeed larger than $\ce{Ca^2+}$:

Reference

Shannon, R. D. Acta Cryst A, 1976, 32 (5), 751–767 DOI: 10.1107/S0567739476001551.


Atomic number ranges in "integral units"? My textbook says the following:

  For an electrically neutral or complete atom, the atomic number also equals the number of electrons. This atomic number ranges in integral units from 1 for hydrogen to 92 for uranium, the highest of the naturally occurring elements.

I'm wondering what is meant by "integral units" in this context?

I would greatly appreciate it if people could please clarify this.

This is only a way of saying that the atomic number will never be a decimal number like 1.5 or 16.8

One consequence is that if you carry out an experiment which gives you a decimal number as an atomic number, then you must have a mixture or something is wrong.


Rate law of the iodine clock reaction I am dealing with the iodine clock reaction, which has the half-reactions:
$$\begin{alignat}{2}\ce{H2O2 (aq) + 2I- (aq) &-> I2 (s) + 2 OH- (aq)} &&\quad (\text{slow}) \\ \ce{I2  (s) + 2S2O3^2- (aq) &-> 2I- (aq) + S4O6^2- (aq)} &&\quad  (\text{fast}) \end{alignat}$$

Now, I have to find the rate law. What I did was 4 solutions – one with standard amounts, another with double the $\ce{I-}$, one with double the $\ce{H2O2}$, and one with double the $\ce{S2O3^2-}$. I recorded the amount of time for the ppt ($\ce{I2}$) to appear. Now I have to calculate the rate law.

The thing is, I am not quite sure how. Usually you take the "slow" step and set up an equation based on that, like $\text{rate}=\ce{[H2O2]$^x$ [I- ]$^y$}$. However, the reaction is not just dependent on those chemicals, it's also dependent on the $\ce{S2O3^2-}$. When it was doubled, the reaction was significantly slower (for obvious reasons).

How do I set up the rate law for this reaction? Or, if I can't, what else can I do with my data?

IMO, the only reaction that you are studying is the first reaction:  the conversion of $I^-$ to $I_2$ by hydrogen peroxide.  The thiosulphate reaction is simply a dramatic way to indicate when a particular amount of iodine has been produced.

The reaction was not slower when you doubled the amount of thiosulphate.  Twice as much iodine needed to be produced (and reconverted to iodide ion) before the thiosulphate was consumed and the first permanent iodine changed the starch indicator blue.


Is there a general / simple formula for Slater's rules? I just learned how Slater's rules work on Wikipedia. These rules really are very simple. But the presentation of the rules seemed not very efficient. I would think there would be someway to set up and equation of some sort such that all you have to do is plug in the relevant quantum numbers and the element number and use some sort of formula to just compute Slater's rules more easily.

Is there such an equation / formula? If so, I'd like to see it. It would be more convenient than memorizing the rules in verbal form.

Sadly, no. It wouldn't be too hard to program a "get Slater shielding" script. But it wouldn't be terribly useful.

First off, let's back up briefly. Slater's rules allow you to estimate the effective nuclear charge $Z_{eff}$ from the real number of protons in the nucleus and the effective shielding of electrons in each orbital "shell" (e.g., to compare the effective nuclear charge and shielding for say $3d$ vs. $4s$ in transition metals).

Slater wanted to use these for quantum calculations, such that the radial wavefunction would be something like:

$$\psi_{ns}(r) = r^{n-1}e^{-\frac{(Z-\sigma)r}{n}}$$

So the rules he devised were fairly simple and produced fairly accurate predictions of things like the electron configurations of transition metals and ionization potentials.

Later, others performed better optimizations of $\sigma$ and $Z_{eff}$ using variational Hartree-Fock methods. For example, Clementi and Raimondi published "Atomic Screening Constants from SCF Functions." J Chem Phys (1963) 38, 2686–2689. (That article covers elements up to Kr.)

It's pretty obvious that no simple formula will cover these curves.

These days, basis set optimization is done by computer, and I doubt many people even glance at the exponents.

In my mind, Slater's rules serve two main purposes at this point:

They're relatively simple to teach and learn and allow quick qualitative predictions, particularly for electron configurations in the transition metals.
They emphasize that exponents for multi-electron atoms must be empirically derived.

I still teach them, particularly, for the first reason. To a lesser degree, they also reinforce the idea of effective nuclear charge and provide some semi-quantitive basis for that concept.


In which cases does the 3d subshell take precedence over the 4s subshell? I've learnt in Chemistry that for example in Chromium, the 4s subshell only has one electron in favor of making the 3d subshell half-full. Also, for Copper the same happens.

What other circumstances does this happen under? Do ions obey the same pattern, e.g. Mn+?

If we take the definition of "3d subshell takes precedence over the 4s subshell" as being that the configuration has zero 4s electrons, yet at least one 3d electron, then all such cases are ions.

+1

V: 3d^4
Cr: 3d^5
Co: 3d^8
Ni: 3d^9
Cu: 3d^10

2+

All elements from Sc to Zn

3+

All elements from Ti to Ga

Etc.


Show that there can be 18 electrons in a "g" subshell I'm learning about quantum numbers right now, and I can't figure out how to calculate the number of electrons in each shell without memorizing it. Is there a formula to find the possible numbers of electrons in the s, p, d, f, and other (theoretical) subshells? If it helps, this problem is put under the "Quantum Numbers, Exclusion Principle" topic.

Simple mnemonic: each orbital, s, p, d and f, has double the consecutive odd numbers of electrons, i.e. 1, 3, 5 & 7 --> 2, 6, 10 & 14.

BTW, from spectroscopy, s, p, d & f refer to splitting of lines, the sharp, principal, diffuse and fine; see Atomic orbital. As Tom Lehrer stated, "These are the only ones of which the news has come to Hahvahd, and there may be many others, but they haven't been discovahd." Although element 121, unbiunium, begins the theoretical g-block...


How does adding HCl convert AgCl into a sol? According to the text I follow, in peptization, the ions of the added electrolyte (peptizing agent) get adsorbed by the particles of the precipitate and turn the precipitate into a sol. An example of this is, silver chloride can be converted into a sol by adding hydrochloric acid (chloride being the common ion)

What is the mechanism behind this?
And why does having a common ion matter?

What is the mechanism behind this?

During peptization, the precipitate adsorbs one of the ions of the electrolyte on its surface. This causes the development of positive or negative charge on precipitates, which ultimately break up into smaller particles of the size of a colloid. Note that the presence of equal and similar charges on colloidal particles is largely responsible in providing stability to the colloidal solution, because the repulsive forces between similarly charged particles prevent them from coalescing or aggregating when they come closer to one another. Thus, they remain in dispersion medium as stable colloidal sols.

  And why does having a common ion matter?

The sol particles acquire positive or negative charge by preferential adsorption of positive or negative ions. When two or more ions are present in the dispersion medium, preferential adsorption of the ion common to the colloidal particle usually takes place.


The effect of reactant concentration on when equilibrium is established Take the equation
$$\ce{A(g) <=> B(g)}$$
where the reaction is first order with respect to A. The rate law is thus
$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k[\ce{A}]$$
Say you run the reaction, starting with $x$ moles of A. Equilibrium is established at time $y$. If you run the reaction again, this time with $2x$ moles of A, won’t equilibrium be stablished at time $y$ as well? You double the rate of reaction, but you have twice as many molecules to react, so wouldn’t time remain the same? I’ve been told that the equilibrium effect dominates over the kinetic effect; that is, it will take a time that is greater than $y$ for the reaction to reach equilibrium. I cannot fathom the mathematical basis of such a conclusion. Can someone elucidate?

I you have an equilibrium with significant amounts of $\ce{A}$ and $\ce{B}$ you need to consider the backwards reaction. If it is first order in $\ce{B}$, the rate law is

$$\frac{\mathrm d[\mathrm{A}]}{\mathrm dt}= -k_1 \cdot [\mathrm{A}] + k_2 \cdot [\mathrm{B}] \tag{1}$$

If the initial concentration of $\ce{A}$ is $\mathrm{[A]}_0$ and the initial concentration of $\ce{B}$ is $0$ the solution of the differential equation is

$$\frac{\mathrm{[A]}}{\mathrm{[A]}_0} = \frac{1}{k_1 + k_2} \cdot \left(k_2 + k_1 \cdot \mathrm e^{-\left(k_1 + k_2\right) t}\right) \tag{2}$$

The equilibrium concentration of $\ce{A}$ is obtained as the limit for $t \to \infty$:

$$\frac{\mathrm{[A]_{eq}}}{\mathrm{[A]}_0} = \frac{k_2}{k_1 + k_2} \tag{3}$$

From $(2)$ and $(3)$ it follows that

$$\frac{\mathrm{[A]} - \mathrm{[A]_{eq}}}{\mathrm{[A]}_0} = k_1 \cdot \mathrm e^{-(k_1 + k_2) t} \tag{4}$$

or, logarithmically,

$$t = -\frac{1}{k_1 + k_2} \left(\ln\left(\mathrm{[A]} - \mathrm{[A]_{eq}}\right) - \ln\left(\mathrm{[A]}_0\right) - \ln\left(k_1\right)\right) \tag{6}$$

The criterion for equilibrium $\left|\mathrm{[A]} - \mathrm{[A]_{eq}}\right| \ll 1$ lets us neglect the term $\ln\left(\mathrm{[A]}_0\right)$ in equation $(6)$.

This shows that the time it takes to reach the equilibrium is independent of the initial concentration of $\ce{A}$.


Arranging different oxygen-containing organic compounds by decreasing acidity I am facing difficulty in the following problem.

Arrange the following compound in decreasing order of acidity.

It is clear that conjugate base of (Q) is more stable.hence it is most acidic.in R) there is much steric hinderence so my view is that it should be less acidic.however I am not sure about it.

And this is the type of question for which Professor Mayr made us chemistry students rote memorise typical $\mathrm{p}K_\mathrm{a}$ values. Here are some.

\begin{array}{lc}\hline
\text{compound type} & \text{typical }\mathrm{p}K_\mathrm{a}\\ \hline
\text{alcohol} & \approx 15 \\
\text{ketone} & \approx 20\\
\text{carboxylic acid} & \approx 5\phantom{0}\\
\text{secondary }\ce{C-H} & \approx 45\\ \hline\end{array}

Therefore, the order is $\mathrm{Q > P > R > S}$


Why does one bicyclic diastereomer react faster than the other one? I have been asked to explain the following observation:

I have a number of issues with this:

1) The fast reaction must be a syn-elimination. But how can syn E2 elimination occur? If an E2 elimination is stereospecific because it must go through an anti-periplanar transition state surely syn-peri-planar elimination cannot happen.

2) How does the slow reaction even happen? The H and the Cl are never going to be in the same plane (syn or anti).

As a side question: does the boat conformation (which the bicyclic molecule must adopt) have equatorial and axial sites like the chair conformation does?

Yes, the fast reaction is a syn-periplanar E2 reaction. If we are ranking the rates of E2 based on the orientation of the C-H and C-LG bonds, anti-periplanar would be the fastest, syn-periplanar would be slower, and perpendicular would be the slowest. The anti-periplanar arrangement is the fastest because it provides the best overlap of the (filled) bonding C-H orbital with the (empty) anti-bonding C-LG orbital. The syn-periplanar arrangement is slower because there is much less productive overlap between these orbitals. However, the small lobe of the bonding C-H molecular orbital does provide some overlap with the anti-bonding C-LG orbital. A perpendicular arrangement results in no productive overlap.

The slow reaction must occur by some other mechanism, such as an E1 or E2-conjugate base mechanism.


What is a "period" of the periodic table? I'm familiar with the periodic table / periodic system, but I wonder why it's called "periodic" since there seems not much periodic about (there seems to be little or no predictability of which elements are stable) and from physics I know Z-value of elements and that is not periodic either. So what is a "period" in this context?

I agree that the "periods" in the periodic table are not mathematically regular.  The simplest definition is that a period begins when a new s-subshell starts to fill.  Recall that in the s, p, d, f subshells there are 2, 8, 10, and 14 electrons, respectively, so the periods have to get bigger over time and thus cannot be regular.  As to the lack of predictability of stability, that may be true, but the periodic table is pretty good at predicting the chemistry of elements by collecting together similar elements into groups (the columns).  This similarity is mostly expressed by the s and p subshells, which is where a lot of the reactivity occurs.


Which etching process is better, photochemical or electrochemical? My boss and I are experimenting with several different ways to micro machine .01" and .005" stainless steel. When I look up how it's done professionally, I see an overwhelming majority of companies use photochemical etching (acid etching). I have also seen electrochemical etching as an option, but not as a "precision" method. So far, both methods have achieved sub-millimeter precision and electroetching currently takes a slight lead over photochem, but neither have achieved the incredible accuracy that I've seen online that appears to have resolution finer than the human eye can distinguish.

Electrochem etching has also proven to be much simpler and safer since we don't have to be involved with potentially dangerous chemicals. My suspicion was that acid etching would be more precise since it is so widely used, but our results are not supporting that hypothesis. I am wondering what advantages does acid etching have over electroetching that makes it so much more popular? I'm curious to know more so we can further refine our etching process.

While I have no expertise in this subject I did a web search for "advantages acid etching has over electroetching" which led to the following which I copy verbatim, specifically the part that seems relevant to your comments:

  Electro-etching offers a great alternative to etching with acid, but
  as with everything, it also has some disadvantages:

  • A feature of electro-etching is that it will bite isolated lines deeper than those that are close together, or cross hatched lines.
  Also it tends to bite harder in the external areas than at the center
  of plates (edge-effect).
  In addition, electrolytic etching bites harder on the edges of wide
  lines and edges of resists of large areas of open bite. This is more
  noticeable in prolonged bites. However, all these drawbacks can be
  compensated for by using a cathode grid in the electro-etching unit
  instead of a cathode plate.


What's the voltage of a battery? The definition of voltage I'm familiar with is the amount of electric potential energy per unit charge at a location. My prof defined the voltage of a battery to be the amount of potential energy difference an unit charge would have at the positive terminal vs. the negative terminal.

But from what I remember from high school, the "terminals" aren't points, their rods of metal:

Since the voltage is defined at a point, not cylinders or rods what does the voltage of a battery mean? Also, how does the definition connect to the "pushing" power of a battery?

I took a look at Electrode Potential but it was a bit over my head.

I'm looking for an intuitive explanation, not one using rigorous chemistry concepts. Thanks!

Voltage is indeed something that is measured at a particular point, but that doesn't mean its a property of only one point and in fact every point will have some voltage. In particular you can have a series of points that map out an "equipotential" where the potential is equal all along these points. A terminal of a battery is an equipotential so no matter where an electron "is" it will have the same potential energy U = e V.

Intuitively this is just like potential energy due to gravity. The battery is like a mountain with a flat top and level plane at the base. Everywhere along the top the altitude is equal and a ball will have the same potential energy. Similarly everywhere along the level base of the mountain the ball will have the same potential energy.