What is the effect of adding salt to ice? What will happen to the melting point temperature of ice if some common salt is added to it? How to justfy the answer?

From http://antoine.frostburg.edu/chem/senese/101/solutions/faq/why-salt-melts-ice.shtml:

Consider that ice, a crystalline solid, forms via weak partial-polar forces between water molecules, known as hydrogen bonding. As liquid water cools, it loses the energy that keeps these forces from attracting and capturing neighboring water molecules, and so these bonds form in a crystalline structure. Exactly what that structure is depends on the temperature and pressure at which the ice forms; your everyday freezer ice as well as most of the stuff outside in January is Ice Ih, which is a very spacious hexagonal arrangement that makes ice less dense than liquid water.

Also consider that even within what looks like solid ice, individual molecules "melt" and "freeze" all the time as they transiently gain enough energy to break free of the lattice, and then lose it and are recaptured. When the rate at which molecules melt and freeze are equal, the ice is stable. This happens at 0.01*C; below this temperature, more water will freeze, while above this temperature, more water will melt. Right at 0.01*C, water remains water and ice remains ice (the other half of this coexistence, besides temperature, is that water like most substances requires more energy gain or loss to change its phase, known as the Latent Heat of Vaporization, and for water that's 334J/mL, 80 times what's needed to cool liquid water from 2*C to 1*C).

Introduce salt, and these dissolved ions interfere with this hydrogen bonding within the ice, preventing it from happening at the normal freezing point of water. Now the melting molecules don't refreeze, because their partial polar charges are more strongly attracted to the full charges of the sodium and chloride ions than to other water molecules. That upsets the balance; more water is melting than freezing at the normal freezing temperature, and so the ice melts. It takes more energy loss within this system before the water is too unenergetic to stay liquid and settles into crystal, releasing the salt, which then finds other liquid water at the boundary layer between liquid and ice.

This creates a vicious cycle; as the salt concentrates within the remaining liquid water, its freezing point is lowered even further as the sodium and chlorine, which won't resolidify as salt crystals until there's just too little water to remain dissolved, will become even more attracted to the remaining water, even further lowering the remaining liquid's freezing point as the water must free itself from even more ions before it can be captured by the ice crystal.

So, an aqueous salt solution, in nearly any concentration, will lower the boiling point of the solution below 0*C, but the minimum freezing point is at -21.1*C, which is the freezing point of a fully-saturated salt solution. Below this temperature, salt that loses its water to freezing can't stay dissolved, and precipitates out as a separate crystal solid suspended within the ice.


Differences between miscible and soluble What are the differences between soluble and miscible? So far I found this (page 3 http://faculty.sites.uci.edu/chem1l/files/2011/03/D01MANmiscibility.pdf):

soluble=solid dissolves into a liquid

miscible=two liquid give rise to a homogeneous solution

Are there other differences?

Thank you.

Your definition is basically correct, except using two liquids might provide for a clearer example than a solid and a liquid.  For example, diethyl ether is soluble in water to about 7% at $\mathrm{20^oC}$. This does not mean that diethyl ether is partially miscible in water, but that it it partially soluble. This Wikipedia article provides a good distinction for miscibility:

  Miscibility is the property of substances to mix in all
  proportions (that is, to fully dissolve in each other at any
  concentration), forming a homogeneous solution.


Choosing a vacuum pump for fractional distillation I have a fractional distillation setup. I would like to purify a small  mixture ($\pu{\approx 100-150 mL}$) of high boiling point compounds (between $200$ and $\pu{300^\circ C}$). The one that interests me has a b.p. of $80\rm~^\circ C$ at $\pu{13 mmHg}$. I don't know what this mixture is composed of.

Now the tricky part and the one that give me the most headaches is the vacuum pump. I need one that has enough vacuum power to be able to distill at $\pu{13 mmHg}$ and one that is not expensive too.

I also need a system to be able to monitor and control the vacuum pressure too, so the vacuum pressure remains constant.

I figured that a diaphragm pump is not suitable since the ones that have that kind of vacuum power are way more expensive than rotatory vane pumps. Can anyone help?

If you are keeping your hands on you can't beat a water aspirator for cost, simplicity, and chemical compatibility.

Here is an aspirator assembly with a gauge and valves I put together for about $50.  Connecting it to a standard hose bib it can pull up to 27mmHg.  The water valve controls the vacuum rate, and if you have a closed system once you reach the desired vacuum you can shut one of the other valves to hold it.

Note: This was inspired by Lysander's answer to my similar question last year, which has additional useful details.


What are the differences and similarities between an aliphatic and an alicyclic compound? I read that alicyclic compounds are compounds that are both aliphatic and cyclic and i know that aliphatic compounds are acyclic while alicyclic compounds are cyclic but are there any other differences and similarities between them and what is exactly an aliphatic compound?

Aromatic compounds (wiki) (about.com) (study.com) (chemguide) (khan)

Aromatic compounds are compounds that contain one or more aromatic rings.

Each aromatic ring, in its unsubstituted form, has the chemical formula $\ce{C_nH_n}$ (e.g. benzene), or with one or more of the $\ce{CH}$ groups replaced with an atom from Group V or Group VI, usually $\ce{N}$ (e.g. pyridine) or $\ce{S}$ (thiophene) or $\ce{O}$ (furan). They are usually stabilized by resonance.

Aliphatic compounds (wiki) (about.com) (study.com)

Aliphatic compounds are hydrocarbons (wiki, compounds containing only $\ce{C}$ and $\ce{H}$) which are not aromatic.

An example would be hexane (wiki).

Alicyclic compounds (wiki)

Alicyclic compounds are aliphatic compounds that contain one or more rings.

An example would be cyclohexene (wiki).

Gallery

Benzene, aromatic (courtesy Introduction to Organic Chemistry):

Furan, aromatic (courtesy Wikipedia):

Cyclohexane, alicyclic (courtesy Jarad Chemicals):

Cyclohexene, alicyclic (courtesy Exporters India):


Is the activity coefficient always a constant? The activity of a species $i$ is defined as $a_i = e^\frac{\mu_i-\mu_i^\ominus}{RT}$, with $\mu_i$ the chemical potential, and $\mu_i^\ominus$ the chemical potential under standard conditions.

Very often, for a species in a non-ideal solution, the activity is written $\gamma_i x_i$, where $x_i$ is the mole fraction and $\gamma$ is a constant known as the activity coefficient for that species. This implies that $\mu_i = \mu_i^\ominus + RT\log \gamma_i x_i$.

What I don't have a good idea of is whether the $\gamma_i x_i$ formula is exact (under some assumptions) or an approximation. Are there situations in which the activity depends on the mole fraction in some non-linear way, or upon the concentrations of other species in addition to species $i$? Is the formula $a_i = \gamma_i x_i$ an exact one under some circumstances, or are there always (in principle) nonlinear correction terms that would make it more accurate?

In my copy of Atkins Physical Chemistry, 6th edition (it's old!) there is a chart that shows a calculation of the activity and the activity coefficient for an example system (chloroform and acetone). It shows the results using a calculation based on Henry's law (approximating ideal-dilution) and again with one based on Raoult's law (approximating a pure substance).

In both cases, the activity coefficient is not constant over the range of concentrations.

For some evidence from a less-reputable but more easily accessible source, check out the wikipedia article on activity coefficients:

  $\gamma_B$ is the activity coefficient, which may itself depend on $x_B$

($x_B$ is the mole fraction of the solute).

Now this means that the activity coefficient is not constant for a given substance in a given mixture at all concentrations, but that isn't the same as saying that the definition of activities is not exact - it is exact, because it is defined that way.

In other words,

$$
\mu_i = \mu_i^\ominus + RT\log \gamma_i x_i
$$

is an exact relationship, even though $\gamma_i$ is not constant.


What is the density of sodium trimethylsilylpropanesulfonate (DSS)? I tried to look for the density of sodium trimethylsilylpropanesulfonate (DSS) online but could not find anything. Perhaps someone here measured it/knows where it may be found?

The reported crystallographic density for sodium 3-(trimethylsilyl)-1-propane-sulfonic acid monohydrate is $\pu{1.32 g cm^-3}$ [1].

The title compound which is used as reference for some NMR spectra in $\ce{D2O},$ was obtained casualy during an attempt to crystallize a cephalosporin of a serie we are interesting for […].
The X-ray analysis was undertaken because the expected cephalosporin and the two independent molecules of the title compound which were in fact present in the unit cell have nearly identical molecular weights.
The compound crystallizes from water in the triclinic space group $P\bar{1}$ with unit cell dimensions $a = 5.914(3),$ $b = 10.816(7),$ $c = \pu{18.869(9) Å},$ $α = 97.96(5),$ $β = 95.46(4),$ $γ = 89.76(5)^\circ,$ $V = \pu{1190(1) Å^3}.$
The calculated density is $\pu{1.32 Mg m^-3}$ for $Z = 4.$

Figure 1. Unit cell of sodium trimethylsilylpropanesulfonate monohydrate (CSD: BEHNIL).
Reference

Tinant, B.; Coene, B.; Declercq, J. P.; Germain, G.; Meerssche, M. van. X-Ray Crystal Structure of 3-(Trimethylsilyl)-1-Propane Sulfonic Acid Sodium Salt Hydrate $\ce{C6H15O3SSi^-Na^+ · H2O}.$ Bulletin des Sociétés Chimiques Belges 1982, 91 (1), 95–96. DOI: 10.1002/bscb.19820910116.


Which element should be placed in axial position bulky one or the smaller one? In one question for finding $\ce{P(CH3)2H3}$ structure? Which one should be placed in axial position? I think $\ce{-CH3}$ should be axial position to minimise repulsion.

No, it's the other way around. Axial positions are less spacious, so you'd want to place your smaller substituents there and the bulky ones to the equatorial.


Predicting the temperature of an object using the heat of its surroundings after a certain amount of time? how would you predict what the temperature of something would be if it spends a certain amount of time exposed to heat from its surroundings? is this possible?

ex. what temperature would a hot bottle of water at initial temp 48 C be after being put in a 4 C refrigerator for 2 hours?

You're basically looking at radiation, conduction, and convention as methods to dissipate heat.

Heat loss by radiation can be computed from Wein's law or something similar.

Conduction requires knowing the thermal properties and dimensions of the materials in contact.

Convection is the trickiest one since it requires some model for how the air or other fluid around the object is moving and coupling this to conduction.

All three are subject to significant errors in computation if you're not careful though if you're trying to estimate rough order of magnitude, you might not care.

All in all, this is not an easy problem to compute from first principles.

Note also that your example is not nearly well-defined enough to solve this problem. What is the surface area of the bottle and the size of the refrigerator. How well does the refrigerator maintain its internal temperature? Is there a fan? Is the water inside a thermos bottle?


Does saltwater damage electronics? newbie here.

I'm asking a question that is related to hygrometer calibration.

I read in sensitive digital hygrometers, it is stated that saltwater test should not be used to calibrate as it will damage the electronics (resistor?) that is used for measurement.

But how does this ring true? Saltwater does not evaporate, or do they?

Salt water is an excellent electrolyte, and will accelerate galvanic corrosion reaction between two different metals especially in oxygen rich environments (including dissolved oxygen in water). This may create metal oxide deposits on circuits. The longer the exposure time the more corrosion can occur.

If your device is on, that is, there are circuits with electricity flowing, clearly adding a salt water could also short circuit your device as well.

As to whether any of the above results in irreversible damage is hard to say with certainty as some devices have been dried out and appear to function at least following exposure to common tap water.


Finding the mM concentration of capric acid (decanoic acid) in coconut oil Im trying to figure out the mM concentration of capric acid in coconut oil. I know that There is 8% capric acid in coconut oil, but I would like to know how much capric acid is in coconut oil in a mM ammount, not a percent. I know the molecular weight of capric acid (decanoic acid) is 172.26

The reason being: I read a study that capric acid solutions are effective against microbes and fungus at a mM concentration of 20 mM. Im wondering if coconut oil's 8% of capric acid converts to roughly around a 20mM concentration of capric acid within the oil.

Thanks very much for your answers!

Thank you very much for a well-specified question.  If 8% of coconut oil were free capric acid, then as noted in the comments the units of "%", i.e. w/w, w/v, or v/v, would affect the calculation somewhat, but since capric acid likely has a density that is very close to that of bulk coconut oil, the difference will be slight.  8% w/w, and assuming a density of 0.8 g/mL for coconut oil would yield $0.08~{\frac{\rm g~capric~acid}{\rm g~oil}}\times \frac{\rm mol~capric~acid}{172~\rm g~capric~acid}\times0.8\frac{\rm g~oil}{\rm mL}\times\frac{1000~\rm mL}{\rm L}=0.372\frac{\rm mol~capric~acid}{\rm L}=372~\rm{mM}$.

However, I suspect that the capric acid in coconut oil is not free acid.  It is part of triglycerides.  Instead of containing capric acid, the oil contains an ester of capric acid with glycerol.  Triglycerides are molecules where all of the available 3 hydroxyl groups of glycerol are esterified with a fatty acid.  Because triglycerides are large complex molecules, they were historically difficult for chemists to analyze directly.  The method of choice for analysis of plant oils was transesterification or ester hydrolysis, converting the triester triglyceride molecules into three smaller monoester molecules that were easier to detect by analysis.  The 8% figure you are using is probably the result of such an analysis.

The bottom line is that coconut oil that is "8% capric acid" actually contains very little capric acid, but contains a large amount of glycerol-esterified capric acid.  But only free capric acid is likely to be antimicrobial.  So the capric acid content of coconut oil is unlikely to make it antimicrobial.

The presence of free fatty acids in plant oils is usually a sign of degradation.  In olive oil, an acidity above 0.5% (i.e. of all the fatty acid chains in the olive oil, 1 in 200 are in the free form and 199 in 200 are in the esterified form) is a sign of lower quality.  I would imagine the for coconut oil the situation is similar, although the specific numbers used as quality thresholds may be different.


Does resonance destabilize a molecule? According to this table, more heat is released in breaking a single double bond in 1,3-butadiene than in trans-2-butene. Does this mean that the resonance of 1,3 butadiene leads to a destabilization of the compound?

None of the comparisons you might make suggests any sort of deatabilization by conjugation.

Let's look at the four-carbon compounds.  Depending on which source you believe for 1,3-butadiene, you could release up to 27.8 kcal/mol by hydrogenating one double bond (the figures given for this compound refer to hydrogenating both double bonds).  But then you release 30.3 kcal/mol hydrogenating the remaining double bond in 1-butene.  The combination of two double bonds in 1,3-butadiene has less energy to be released by hydrogenation than two independent bonds like the bond in 1-butene has.

Similarly for the five-carbon compounds.  When you hydrogenated both double bonds in the pentadiene isomers, a pair of conjugated double bonds in 1,3-cyclopentadiene has less energy to release than the non-conjugated double bonds in 1,4-cyclopentadiene.

Finally, something a little more subtle is going on between 1-butene and 2-butene.  There can no conjugation like that in 1,3-butadiene when only one double bond is present, yet 2-butene has clearly less energy to release by hydrogenation than 1-butene.  Thinking of carbocation nomenclature, the former is a like a "secondary carbication by having an alkyl group on only one end of the double bonds, while the latter is "tertiary" with an alkyl group on both ends.  Hyperconjugation, like that in carbocations, is at work here.


Is there a phase diagram available for hydrogen fluoride? Has a phase diagram for hydrogen fluoride been published in an article on the internet or in a book ?

Or is there a table for HF for different pressures up to 100 atm. and temperatures up to 500$⁰$ C ?

Or, even better, is there a calculator for HF, like this one for steam ?

Linde Gas contains a vapor pressure curve.  Note that this does not extend to Venusian surface temperatures which are above the critical point.

Hydrogen fluoride may not lift as well as you think.  Its gas phase may associate through hydrogen bonding so molecules could be heavier than just HF.  See the note near the end of the safety information here and note that the association could persist to higher temperature if you apply higher HF pressure than one atmosphere.


pH of solution after adding SiCl4 Why is it that adding $\ce{SiCl4}$ will cause the pH of a solution to become acidic? I thought that only salts (ionic bonds break in water) can change the pH once dissolved in water. $\ce{SiCl4}$ has covalent bonds, so why does the pH change?

Silicon is very oxophilic and $\ce{SiCl4}$ irreversibly hydrolyzes in water as follows:
$$\ce{SiCl4_{(l)} + 4 H2O_{(l)} -> 4 HCl_{(aq)} + Si(OH)4_{(s)}} $$

The release of $\ce{HCl}$ creates hydrochloric acid and lowers the pH.


How to determine if a complex has a crystal field stabilization energy of zero? I have encountered with the following question:

  The complex having zero crystal field stabilisation energy is:

  (A) $\ce{[Mn(H_2O)6]^3+}$ (B) $\ce{[Fe(H_2O)6]^3+}$ (C) $\ce{[Co(H_2O)6]^2+}$ (D) $\ce{[Co(H_2O)6]^3+}$

Zero crystal field stabilisation energy will be the case when the metal has either $d^{10}$ configuration or $d^5$ configuration with every orbital having one electron. For Iron($\ce{Fe^3+}$) the d orbital is half filled. Now the problem is, water is in the middle of the spectrochemical series, i.e. it is moderately strong ligand. So, how to decide whether the d orbital will have the 'every orbital half filled' condition or not ?

In the given example it is pretty simple to chose B as the correct answer, as it is the only option that could possibly create a spin forbidden transition $d^5$ or a $d^{10}$ configuration. For most situatons I have encountered it is sufficient to assume high-spin filling, until you encounter very stable configurations: completely full or half full orbitals (like $d^6$)


Why isn't 'chemically-strengthened glass' made with potassium carbonate to begin with? Instead of making potassium-strengthened glass by creating ordinary soda-lime glass first, then replacing the sodium atoms/ions with potassium by putting the glass in a bath/solution of potassium nitrate, why not replace the 'soda' (sodium carbonate) in the initial process(es) with potassium carbonate?

The potassium is not added first because the potassium does not intrinsically make stronger glass, it is the substitution of a larger ion for a smaller one that does.

To understand why ion exchange strengthens glass, you have to understand why the ion exchange makes the glass harder. The process of ion exchange hardening is done at a temperature that allows ion diffusion, but disallows a reconfiguration of the glass structure or relaxation of bonds (i.e. below $T_{g}$, the glass transition temperature). When sodium atoms are replaced by potassium in glass, the potassium ions occupy a site that is sized for sodium; this mismatch creates the compression that in turn gives the glass its scratch resistance. Thus the only way to chemically harden the glass is to substitute larger ions into the glass.

If you make a glass with potassium, you have potassium ions occupying potassium sites, thus no strengthening occurs.


How to separate a 5 atom thick layer of Cu from Au I've become very curious about (electrochemical) atomic layer deposition (ALD) via surface limited redox replacement.

Since the technique allows for deposition of metals in atom thick layers it's possible to create films of nearly exact atomic thickness. However, I haven't found any reference that describes how one can remove the film from the (usually Au) substrate that it's deposited on.

Hypothetically speaking, if one used ALD to generate a film of $\ce{Cu}$ that is 5 atoms thick (arbitrary length and width), how would one go about removing the film in-tact? Is it even possible?

In most cases I've seen, ALD is performed on the intended substrate - that is, there's no need to separate it. In other cases, you use a sacrificial substrate that you can remove easily.

If you wanted to separate the film from the substrate, there are various techniques, but nothing comes to mind for separating Au from Cu. I find it incredibly unlikely in the specific case of a 5-atom thick layer of Cu, which would likely not survive typical techniques (vs. bending, mushing, forming a larger packing, etc.)

Some typical techniques for separating two films:

Adhesive tape (as mentioned in a comment) - works for many films, e.g., graphene, other layered materials, where you physically pull apart layers
Swelling (e.g., separating Au from mica) - including various intercalation methods used in graphenes and graphene oxides and other layered materials, so that the layers separate more easily
Dissolving, etching, or eroding the substrate (e.g., back-etching a Si wafer to expose an $\ce{SiO2}$ or thin metal film deposited on the wafer)

In my group, for example, we make "template-stripped gold" where Au is deposited on cleaved mica substrates. Then the thin gold layer is glued to glass substrates and the mica is separated from the Au by swelling / degredation, leaving a very smooth Au surface.

This technique can work with fairly thin layers of gold (~10-30 nm) but there's a large difference between Au and mica, making separation possible. Not true with Au/Cu.


Why does temperature affect the dissolved oxygen concentration? I know that an increase in temperature decreases the solubility of oxygen in water, but I don't know why. Could anybody explain?

The solubility of a many gases in a variety of liquids is observed to decrease with increase in temperature, for example oxygen, nitrogen and methane in water.  This follows the qualitative predictions of Le Chatelier’s principle, which in thermodynamics is supplanted by the Clapeyron equation. As one of the two phases involved is a gas the Clausius–Clapeyron equation can be used.

This assumes that the molar volume of any gas/vapour is far greater than that of the solution. Assuming that all values correspond to one mole the Clausius–Clapeyron equation is

$$\frac{\mathrm d\ln(p)}{\mathrm dT}=\frac{\Delta H}{RT^2}$$
which in its integrated form between pressure and temperature $p_1, T_1$ to $p_2, T_2$, and by assuming that $\Delta H$ is independent of temperature gives

$$\ln\left( \frac{p_2}{p_1}\right)=- \frac{\Delta H}{R}\left( \frac{1}{T_2} - \frac{1}{T_1}   \right)$$

Despite the approximations involved in its derivation this equation does describe well the measurements of the vapour pressure over many solids and liquids, for example depression of freezing point, elevation of boiling point and solubility of solids and gases in liquids, although the latter is not as frequently described as the other two.

In the elevation of the boiling point the solution at a temperature $T_1$ has a lower vapour pressure $p_1$ than the pure solvent. The solution only boils when the temperature is increased to $T_2$  where its vapour pressure has increased to  $p_2$ (normally 1 atm ), which is the pressure at which the pure solvent boils.  By Raoult’s law  $p_1=p_2x_{\mathrm s}$, where $x_{\mathrm s}$ is the mole fraction of solvent
$\ln(p_2/p_1)=\ln(1/x_s)$
As $x_\mathrm s=1-x_\mathrm t$ where $x_\mathrm t$
is the mole fraction of solute,
$$\ln\left( \frac{1}{1-x_t}\right)= -\frac{\Delta H_\text{vap}}{R}\left( \frac{1}{T_2}  - \frac{1}{T_1}   \right)$$
and as $x_\mathrm t$ is far smaller than unity
$\ln(1-x_t) \approx -x_t -x_t^2/2 \ldots$
from which the second term can be ignored as it is insignificant compared to the first then,
$$x_\mathrm t= \frac{\Delta H_\text{vap}}{R}\left( \frac{1}{T_2}  - \frac{1}{T_1}   \right)$$

Thermodynamically the elevation of boiling point and solubility are very similar but we must swap solute and solvent over so what we formerly called solvent has now to be solute and vice versa.

Considering now solubility, the variation of the mole fraction dissolved $x_\mathrm t$ can be written as
$$x_\mathrm t=\frac{\Delta H_\text{vap}}{R}\left( \frac{1}{T_2}  - \frac{1}{T_\mathrm B}   \right)$$

where what was $T_2$ is now the normal boiling temperature of the liquefied gas. Also $\Delta H_\text{vap}$ (a positive quantity) is now the (molar) heat of vaporisation of the liquefied gas and we approximate this to be the heat of vaporisation from solution.

This last equation indicates that the solubility decreases with increase in temperature and should be the same in all solvents in which it forms ideal solutions.

In the solution a gas molecule has an attractive interaction energy between itself and the solvent. (Experiment shows that $\Delta H_\text{vap}$ is positive thus the heat to dissolve is negative). The solvent molecules also have an attractive interaction between themselves otherwise they would not form a liquid.  Thus for a gas-solvent interaction a solvent-solvent one has to be destroyed. Overall the difference in energy is probably small.  However, there is an entropic effect to consider. Clearly this will have increased when a gas dissolves as now there is another species in the solvent and thus the number of ways the molecules can be arranged has increased and this will have increased the entropy. Together these effect make the free energy negative as expected for a spontaneous process.

As the temperature is increased the molecules gain in average energy and so are more able to overcome the attractive potential that keeps them in solution and thus the vapour pressure of the solvent increases and the solubility of the gas decreases. It is also worth noting that the average kinetic energy $(3/2)kT$ is the same in the solution and in the gas phase, in solution it is merely restricted to a smaller range on motion. Thus the notion that it is the kinetic energy that is acquired that causes a molecule to enter the gas phase in incorrect.


Synthesis of phenethyl alcohol from benzene How can I make phenethyl alcohol $(\ce{C6H5CH2CH2OH})$ from benzene?

I have got information that phenethyl alcohol is made from benzene by Friedel-Crafts reaction between benzene and ethylene oxide in the presence of aluminium trichloride as follows:

$$\ce{C6H6 + CH2CH2O + AlCl3 → C6H5CH2CH2OAlCl2 + HCl}$$

But I have no idea how it reacts and proceeds to work.

Can anyone help me understanding the mechanism of synthesis and reaction of making phenethyl alcohol from benzene?

$\ce{AlCl3}$ is a Lewis acid and will bind to the oxygen of the ethylene oxide giving a species that functions as $\ce{[R2O-AlCl2]+}.$

This cationic species will be attacked at carbon by the electron-rich aromatic ring in much the same way as a standard Friedel-Crafts reaction intermediate generated from an alkyl halide to give $\ce{[Ph-CH2CH2-OAlCl2]+}.$ This then loses $\ce{HCl}$ to give $\ce{PhCH2CHOAlCl2}$ which is stable until aqueous workup which gives phenethyl alcohol.


Can you differentiate between Kc and Kp in this problem? I know how to do this problem. We must find the value of K to compare to the value of Q, which can easily be calculated.

I know that the equation we use to get a K out of delta G is as follows:

However, does this equation give us a Kp or a Kc value? What would it give me in this problem if I plugged in the provided delta G standard value?

I calculated the value of this K (whatever it is) to be 1.00, and I also calculated the Qc and Qp for this problem to be 0.00005 and 0.36, respectively. So clearly, Qc and Qp are both less than K. But how can we determine of Qc is really < Kc in this problem?

The equation yields Kp.

$$\Delta G^{o}=-RTln(K_p)$$

There is, however, a relationship betweek Kp and Kc.

$$K_p = K_c(RT)^{\Delta n} $$

From the balanced equation there are 3 moles of gas on the right (2+1) and 1 mole of gas on the left. In chem, delta means final minus initial and we arrive at $$\Delta n = 3-1=2$$

and you now can find Kc. Note that you should use 0.08206 L•atm/mol•K for R.

Sources:

Zumdhal example 13.5
https://www.quora.com/In-a-gas-equilibrium-which-K-constant-is-used-in-the-equation-%CE%94G-RTlnK-Kp-or-Kc


How to find mole fraction of solvent and solute in a vapor pressure lowering problem? At $25$ degrees celsius, the vapor pressure of pure benzene ($\ce{ C6H6}$) is $93.9$ torr. When a non-volatile solute is dissolved in benzene, the vapor pressure of benzene is lowered to $91.5$ torr. What is the concentration of the solute and the solvent, expressed in mole fraction?

Through Raoult's formula, I found that the mole fraction of the solvent, benzene, is:

$$P_{solution}=x_{\ce{C6H6}}\cdot P^{o}_{\ce{C6H6}}$$

$$91.5=x_{\ce{C6H6}}\cdot 93.9$$

$$0.974 = x_{\ce{C6H6}}$$

Now the only thing I am having trouble finding is the mole fraction of the solute:

$$x_{\ce{C6H6}}=\frac{n_{\ce{C6H6}}}{n_{\ce{C6H6}}+n_{\ce{solute}}}$$

I was wondering, is it even possible to calculate the mole fraction with just the given info? I think that it might not be possible, since I need the grams of solute dissolved or grams of solvent to do this.

$\displaystyle \ce{x_{solute}} = \ce{\frac{n_{solute}}{n_{solute} + n_{solvent}}}$

Thus, we have that $\displaystyle\ce{x_{solute}} + \ce{x_{solvent}} = \ce{\frac{n_{solute}}{n_{solute} + n_{solvent}}} + \ce{\frac{n_{solvent}}{n_{solute} + n_{solvent}}} = \ce{\frac{n_{solute} + n_{solvent}}{n_{solute} + n_{solvent}}} = 1,$ so $\ce{x_{solute} + 0.974 = 1\implies x_{solute} = 0.026}$


Are the units of mole of oxygen molecules the same with the units of mole of nitrogen molecules? Suppose we have in a box $3 \,{\rm mol}$ $\ce{O}$ and $3 \,{\rm mol}$ $\ce{N}$. Is the addition of these two quantities meaningfull? We do it all the time when we want to find the molar fraction. I can't understand if it dimensionally correct to add these two things. If a mole is like a dozen then a mole of eggs and a mole buildings can't add up.
I think a crucial step that is omitted when we want to find the molar fraction is that we should convert these moles of arbitrary entities (eggs, buildings, molecules) to same entities. I can't find anything else to justify such a process (addition of different kind of moles). Would be the following way a proper way to find the mole fraction?
$$n_{_{T}}= \frac{\,{\rm mol} \, \text{of molecules}}{\,{\rm mol} \, \text{of O molecules}}\cdot n_{_{O}} + \frac{\,{\rm mol} \, \text{of molecules}}{\,{\rm mol} \, \text{of N molecules}}\cdot n_{_{N}} $$
So $n_{_{T}}= 6\,{\rm mol}$. What about the molar fraction of oxygen? $$\chi_{_{O}}=\frac{n_{_{O}}}{n_{_T}}= \frac{\text{3 mol of O molecules}}{\text{6 mol of  molecules}}=0.5 \text{ of what?}$$
We know the ratio must be dimensionless but why? We deviding moles of $\ce{O}$ molecules by moles of molecules. Are these units different or not?

It is a good question which has already been somewhat addressed in the 1880s. This "field" was called quantity calculus. Calculus here is not the integration / differentiation, but rather the Latin calculus implying a method of calculation.
There is a very nice article "Quantity Calculus: Unambiguous Designation of Units in Graphs and Tables" by Mary Anne White in the Journal of Chemical Education. Please read this if you are seriously interested. Search on Google Scholar and it is free to download from there.
In quantity calculus Each physical quantity as the product of a numerical value and a unit:
physical quantity = numerical value × unit
This approach was introduced by British scientists and many leading physicists used it. Now there is there is nothing less or nothing more. Therefore your ambiguity arises from introducing another factor such as "oxygen" or "nitrogen". The unit mol does not know whether it belongs to oxygen or nitrogen.
As explained in the comments, suppose we write L symbolizing the height of a tree, then I can only write, L = 10 m. For mathematical purposes, I will not introduce "tree" anywhere in this equation. The tree is already incorporated in L (in your mind) but not in the mathematical equation. One can also write L/m =10. Now you have a pure number on both sides.
For fun: Think about taking the derivatives or log of quantities with units. What is the unit then?


Why is phosphorous acid more acidic than phosphoric acid? Phosphoric acid has basicity of 3 i.e it can loose 3 $\ce{H+}$ while phosphorous acid has basicity of 2. Why is phosphorous acid more acidic than phosphoric acid? Acidity refers to the ability to liberate protons. Phosphoric acid liberates more protons than phosphorous acid (as the basicity of phosphoric acid is 3 and that of phosphorous acid is 2). Why then is phosphorous acid more acidic than phosphoric acid?

I'm pretty sure it is because phosphorus acid is more polar than phosphoric acid. This is because in the phosphoric acid there are more OHs and an O so it kind of balances out more towards the center. In phosphorus acid, there is a H on the main carbon. Hydrogen doesn't pull as much on electrons as Oxygens do and hence that part of the molecule becomes more positive causing a stronger dipole moment than that in phosphoric acid.


Are all PP(Polypropylene), i.e. resin code 5 plastics, safe for microwave and steaming? If a plastic is marked PP as in Polypropylene (i.e. resin code 5), is it automatically safe for steaming and microwaving, such that it wouldn't leach chemicals into the food(besides not melting)?

No.

Just because the polymer is of one or the other kind does not let you say what industrial process actually synthesized it. Subsequently it is impossible to know what trace chemicals are left in the polymer packaging.

If if isn't labelled as microwave-safe, it isn't!

They test the package for it's microwave compatibility, not one type of polymer each! It seems that resin code 5 polymers are often used for microwave-proof packaging, but again: You don't know how it was made, you have no idea what's in there, don't microwave it (together with food or something else that you then ingest).


How does the production of pure, crystalline silicon via the Czochralsky or Siemens process? I have a few questions regarding the production of pure, crystalline silicon:

As far as I know Silicon usually appears in nature in the form of sand or $\ce{SiO2}$ and then is split apart and brought into the form $ \ce{Si + 2CO} $ ;
I am imagining now a hot glowing molten mass.
However then I don't understand the following methods to separate the Si from the elements you don't want to have, like iron, carbon etc.
How do you remove the impurities, why do those methods work?
For example you use a few other chemical reaction to bring Si into several other forms like $ \ce{HSiCl_3} $ and split them apart later on.
I don't understand at the moment, how this helps to separate those materials as in: There is still this big mass with lots of different chemical connection you don't want to have.

I also lack an actual understanding of the Czochralski process to further purify your mass of silicon. I guess want has to accept that silicon likes to join pure crystalline silicon and adapt to its structure. However I don't see, why one is not using this method beforehand for example, if one already has raw silicon you get from splitting up $ \ce{SiO_2} $ if it works so greatly. And if it not does work so greatly, why don't you have other impurities ?

How does trichlorosilane $\ce{HSiCl3}$ help in the purification of silicon?

Trichlorosilane is a highly volatile compound with a boiling point of 32 °C. This means that it can be destilled off the raw material and separated from impurities.

The purified trichlorosilane can either be decomposed thermally

$$\ce{4 HSiCl3 ->[\Delta] Si + 3 SiCl4 + 2 H2}$$ or reduced with hydrogen

$$\ce{HSiCl3 + H2-> Si + 3 HCl}$$

This does give you rather pure silicon, but it is, at best, polycrystalline.

What's the Czochralski process good for?

Note that this in't just a melting/cooling cycle!

The crucial point in the method is the use of a rotating, monocrystalline seeding crystal, dipped into (previously purified) molten silicon. The goal is to obtain huge, monocrystalline silicon rods of defined diameter, that can be used in the production of wafers.


Computational Chemistry: calculation of sterical effects Is there a way to calculate sterical forces between atoms/groups of atoms (preferably with free and open source tools)? I have heard that this is possible within the framework of NBO. I am using ORCA, nwchem and Gaussian09 mostly. Sterics are obviously one of the most important aspects of organic chemistry but still I could find close to no resources on the computational chemistry aspects of it. It would be nice to have a way of quantifying sterics and also visualizing it somehow.

Does anyone have some experience on that?

I think that you're searching for non-covalent interactions; if so, here's the usual Wikipedia link, https://en.wikipedia.org/wiki/Non-covalent_interactions

But more importantly here's the paper describing the software, http://pubs.acs.org/doi/abs/10.1021/ja100936w and here's Prof Rzepa's blog with 'non-covalent interaction' searched, http://www.ch.imperial.ac.uk/rzepa/blog/?s=non-covalent+interaction

That's the start, post again when you want to know how to implement it.


How does the presence of chlorine atoms in PVC make it flame resistant? In a paper that I'm reading, it states that:

  The presence of chlorine atoms in polyvinyl chloride make it flame resistant

and I'm just wondering why the presence of chlorine atoms will make it flame resistant. To the best of my knowledge, burning something is just providing a material with enough activation energy such that it can react with oxygen to produce products such as carbon dioxide and water. If so why does the chlorine prevent this process in any way?

PVC is not truly flame resistant, it does decompose on relatively gentle heating (around 150 °C) However, the process produces significant volumes of hydrogen chloride, that is not flammable and isolates the plastic from the flame. This makes makes PVC self-extinguishing or fire-retardant.


Usage of platinum in flame test Why should we use platinum wire or loop in flame test, can we use any other material ? Why can't we use glass it something?

Platinum is chemically inert ("noble") meaning it resists forming oxides or other compounds. This stability is desirable when doing a flame test because it ensures that we only see the spectrum of what we are burning. (Think how much noise we would have in our spectrum if we used a Magnesium wire to hold our sample)
Of the chemically inert metals (e.g. Au, Ag, Pd, Pt), Platinum has the highest melting point (2041º C), so it's the most useful for holding in a flame.
Glass, on the other hand, melts at a temperatures ~500º C lower than Platinum.


What is the meaning of cmmol in chemistry? I came across a unit called $\text{cmmol dm}^{-3}$ in buffer solutions. What does this unit mean?

Upon researching this, I found that this a measurement of charge:

$$\mathrm{cmol=10~\mathrm{meq}}$$

and

$$1~\mathrm{eq}= \mathrm{F}$$

where $\mathrm{F}$ is Faraday's constant and is:

$$\mathrm{F}=96 485.3365~\mathrm{C}$$

So:

$$\mathrm{cmmol}=\frac{\mathrm{cmol}}{1000}=\frac{\mathrm{eq}}{100000}=.96 4853365~\mathrm{C}$$

Sources:

http://en.wikipedia.org/wiki/Equivalent_%28chemistry%29

https://www.agronomy.org/files/publications/jnrlse/pdfs/jnr014/014-02-0084.pdf


Synthesis of Aspirin, made unknown fluid When I tried synthesising aspirin, after quenching and cooling it I noticed a translucent, milky, white fluid floating in the solution.

The method used was:

Salicylic acid (2 ml) was added to acetic anhydride (2 ml) in a flask with 8 drops of concentrated sulfuric acid.
Warmed the solution in a water bath until all solid was dissolved.
Added 5 ml of cold water to the solution to quench.
Placed solution in water bath to dissolve the solids again. (I noticed that not all the solid dissolved)
Added 10 ml of water to solution and placed in an ice bath (This was when I noticed the fluid forming)

When the solution was filtered only that fluid remained as the residue. Another person also had the same problem but said they got rid of the fluid by scratching the side of the flask with a stirring rod after taking it out of the ice bath. Could someone please explain what it is?

Often, acetylizations are performed with acetic anhydride being deployed both as reagent, as well as solvent.  Consequently, acetic anhydride is used in excess when referring to the number of $\ce{-OH}$ to react.

The addition of water you describe in step 3) intends to hydrolyse this excess of acetic anhydride to form acetic acid.  Apart from acetic acid equally representing a good solvent for many organic products, it takes time to complete this hydrolysis of acetic anhydride. You may accelerate the hydrolysis of the acetic anhydride if the beaker is immersed into an ultrasound bath, or just continue stirring the mixture at room temperature as usual.

The second point is, albeit its high melting point (136 $\mathrm{^\circ{}C}$), the crystallization of aspirine equally takes some time, too. Crystallization of organic materials from water may demand some more attempts prior to success than from organic solvents, too.  Cooling the mixture often will lower the solubility of your product in the solvent, however, if cooled too quickly to too low temperatures you may obtain either just a glassy gel / syrup instead of a solid easy to filter off, or a solid that has to be recrystallized to purify it further.  Scratching the inner walls of the beaker with a glass rod intends to create fresh sites where crystallization may start.

For peracetylating glucose with acetic acid, for example, it is not too strange to stirr the quenched reaction mixture with an excess of water over night at room temperature to convert the initially "honey-like" product into a good solid.


Why are complexes of type MA3B3 not optical active? Complexes of type $\ce{MA3B3}$ have two geometrical isomers, namely:

fac-isomer
mer-isomer

If we look closely at the mer-isomer, it has a plane of symmetry, so it is optically inactive.
But the fac-isomer has no plane of symmetry. Still, it is optically inactive, why?
Being a 12th grader, I wonder where am I wrong in my understanding of optical isomerism in coordination complexes.

Orient the fac- isomer so one A atom is facing towards you and a space between two B atoms is below it.  Then the left half of this view is a reflection of the right half so you are actually looking down a mirror plane.
Because you could have chosen any of three A atoms in your view there are three mirror planes, which pass through a common rotation axis.  The axis passes through the face that has all A atoms at its corners and the opposite face which has all B atoms.  This symmetry is called $C_{\mathrm{3v}}$ in the language of point groups.


How to calculate what temperature change in entropy for the reaction equal to the change in entropy for the surroundings given Enthalpy and entropy A reaction has $\Delta H_\textrm{rxn} = \pu{-107 kJ}$ and $\Delta S_\textrm{rxn} = \pu{285 J/K}$. At what temperature is the change in entropy for the reaction equal to the change in entropy for the surroundings?

My thoughts:
I wasn't really sure if there was an equation that related the entropy and the enthalpy and would like to know if there is such an equation.

You're looking for the equation of Gibbs free energy:
$$\Delta G^\circ =\Delta H^\circ - T\Delta S^\circ$$
Per Wikipedia's Gibbs free energy page:

The equation can be also seen from the perspective of the system taken
together with its surroundings (the rest of the universe). First
assume that the given reaction at constant temperature and pressure is
the only one that is occurring. Then the entropy released or absorbed
by the system equals the entropy that the environment must absorb or
release, respectively. The reaction will only be allowed if the total
entropy change of the universe is zero or positive. This is reflected
in a negative ΔG, and the reaction is called exergonic.

So effectively you're trying to deduce the $T$ that produces $\Delta G = 0$ which is the "tipping point" (then the reaction will occur for $\Delta G < 0$)
So you solve for $T$ in:
$0 = \Delta H - T \Delta S$
Thereafter as $T \rightarrow \infty$ you see that $\Delta G$ keeps decreasing, making the reaction more favorable.
Thus $S_\textrm{system} = S_\textrm{surroundings}$ is a fancy prerequisite for reaction occurrence, which we commonly associate with $\Delta G \le 0$.


Bicycle tube inflation with 16g CO2 cartridge How much pressure will a 16 gram cartridge of $\ce{CO2}$ (carbon dioxide) provide in a bicycle tire? Remembering my high school chemistry (forty years ago) and googling, I came up with this answer:

Figure the volume of the tube (a torus): area of a cross section times the circumference, for a $30\:\mathrm{mm}$ tube on a $70\:\mathrm{cm}$ wheel, I got $1,554\:\mathrm{cm^3}$, or $1.554\:\mathrm{L}$. One mole is $24.47\:\mathrm{L}$. $1.554 / 24.47 = .06353$. One mole of $\ce{CO2}$ is $44\:\mathrm{g}$ (using atomic weights of carbon, 12, and oxygen, 16). $.06353 \cdot 44 = 2.795\:\mathrm{g}$. So the volume of the tube at one atmosphere is $2.795\:\mathrm{g}$ of $\ce{CO2}$. $16 / 2.795 = 5.7\:\mathrm{atm}$. One atmosphere is $14.7\:\mathrm{psi}$. $5.7 \cdot 14.7\:\mathrm{psi} = 84\:\mathrm{psi}$.

Is this correct?

Assumptions. The $\ce{CO2}$ will cool down a bit as it expands out of the cartridge (due to the Joule-Thomson effect). But let's assume that you're interested in the pressure in the tire when it warms back up to room temperature, $25~^\circ\mathrm{C}$, so there is essentially a zero temperature change before and after filling. Let's also assume that the full $16~\mathrm{g}$ of $\ce{CO2}$ is contained by the cartridge and tire when they are connected; none of it is vented during filling.

Estimating the volume of the tire and cartridge. A $30 \times 700$ bicycle tire has a $30~\mathrm{mm}$ outside tube diameter and a $70~\mathrm{cm}$ wheel diameter. Let's say the rubber in the tire is $3~\mathrm{mm}$ thick. You then have an inside diameter of about $24~\mathrm{mm}$. If we assume the tube has a circular cross-section, the volume inside the tire when fully inflated is.

$$V = \pi \left(\frac{2.4~\mathrm{cm}}{2}\right)^2\cdot (\pi \cdot 70~\mathrm{cm}) \cdot \frac{1~\mathrm{L}}{1000~\mathrm{cm^3}} = 0.99_5~\mathrm{L}$$

Let's assume the $\ce{CO2}$ cartridge has a volume of $21~\mathrm{cm^3}$. The total volume of the $\ce{CO2}$ in the attached cartridge and tire will then be about $1.0_2~\mathrm{L}$.

If you have $16~\mathrm{g}$ $\ce{CO2}$, and the tire is fully deflated to begin with, the moles of gas is.

$$n = 16~\mathrm{g} \cdot \frac{1~\mathrm{mol}}{44.01~\mathrm{g}} = 0.36_4~\mathrm{mol}$$

Estimating the pressure. Since we're dealing with a gas with significant intermolecular interactions, which is also at pressures above $1~\mathrm{atm}$, let's compare an ideal gas calculation of the pressure
with a van der Waals calculation to see if nonideality makes any difference.

$$\begin{align}
P_{ideal} &= \frac{n R T}{V}\\
          &= \frac{(0.364~\mathrm{mol})(0.082059~\mathrm{L~atm~mol^{-1}~K^{-1}})
              (298~\mathrm{K})}{1.02~\mathrm{L}}\cdot
              \frac{14.696~\mathrm{psi}}{1~\mathrm{atm}}\\
          &\approx 129~\mathrm{psi}\\
P_{vdw}   &= \frac{n R T}{V - n b} - \frac{n^2 a}{V^2}\\
          &= \frac{(0.364~\mathrm{mol})(0.082059~\mathrm{L~atm~mol^{-1}~K^{-1}})
             (298~\mathrm{K})}{1.02~\mathrm{L} - (0.364~\mathrm{mol})
             (0.04267~\mathrm{L~mol^{-1}})}\\
          &~~~~~~ - \frac{(0.364~\mathrm{mol})^2(3.592~\mathrm{L^2~atm~mol^{-2}})}
                    {(1.02~\mathrm{L})^2}\\
          &= 8.43~\mathrm{atm} \cdot \frac{14.696~\mathrm{psi}}{1~\mathrm{atm}}\\
          &\approx 124~\mathrm{psi}\\
\end{align}$$

A tire gauge reads (gas pressure - atmospheric pressure), so you'd read about about $109~\mathrm{psi}$ if you started with a fully deflated tire. If you started with a tire filled with air at $1~\mathrm{atm}$, though, you'd get about $124~\mathrm{psi}$.


Enthalpy (H) formula: what is the pressure volume product $$H=U+pV$$

Above is the formula for enthalpy or the sum of the internal energy and pressure volume product.

$-p\Delta V$ is work done by or on the system. But $pV$, yet very similar to work, I think, calculates the whole “cylinder”. Instead of just calculating the height difference of the cylinder before and after the air is condensed, it calculate the height of whole cylinder.

But anyway this is just my guess, what is it really?

Enthalpy $H$ is defined so that $\Delta H = q_P$, the heat at constant pressure. This is convenient for constant-pressure calorimetry---it makes it possible to track energy changes without considering volume changes in the system.

$$\begin{align*}\Delta U &= q + w & \text{(first law)}\\
\Delta U_P &= q_P + w_P & \text{(constant pressure)}\\
\Delta U_P &= \Delta H + w_P & \text{(definition of enthalpy)}\\
\end{align*}$$

...so the difference between $\Delta H$ and $\Delta U$ can be interpreted as minus the work for a constant pressure process.

But this doesn't answer your question for a general process. If the process is not constant pressure, we can still apply the definition of enthalpy, but the difference isn't minus the expansion work anymore. But what is it, then? How can we use something we've defined for a constant pressure process for all processes in general?

We'd like a new thermodynamic function $H$ that includes all of the information in $U$. We have $$dU = TdS - PdV$$ so the natural variables of $U$ are $S$ and $V$. We want this new function $H$ to have $P$ as a natural variable in place of $V$, so it will be convenient for constant-pressure calorimetry. Consider the following plot of $U$ vs. $V$:

The equation for the tangent line at some value of $V$ is
$$ U = \left(\frac{\partial U}{\partial V}\right)_S V + intercept$$

If we define the new function as the intercept of the tangent line shown in the graph, it will have the properties we need:

$$H \equiv U - \left(\frac{\partial U}{\partial V}\right)_S V$$

You can see that the derivative $\left(\dfrac{\partial U}{\partial V}\right)_S = -P$ from the equation for $dU$ above. Then

$$H \equiv U - (-P) V = U + PV$$

so

$$dH = dU + d(PV) = TdS - PdV + PdV + VdP = TdS + VdP$$

By adding $PV$ to $U$, we've transformed the information in $U$ (with natural variables $S$ and $V$) to a new function $H$ (with natural variables $S$ and $P$).

This trick is called a Legendre transformation, and it can be used to understand the definitions of other thermodynamic functions like $G$ and $A$ as well.


Gases produced by bacterial decomposition of human urine I know that when urine (more specifically, the urea within) is metabolised by bacteria, ammonia is released. However, because of the high solubility of ammonia, I assume it simply dissolves into the water.

A container of urine will develop gases inside. What are they?

Carbon dioxide and ammonia when treated with bacteria along with water vapour


How must elemental symbols be ordered in the formula for a compound? An online problem asked me to find the molecular formula of a certain substance given its mass composition. I found that it was $\ce{N_5P_5Cl_{10}}$. It was incorrect and told me the ordering of my symbols was wrong. I switched it to $\ce{N_5Cl_{10}P_5}$ because that's the order that they appear on the periodic table. It marked me wrong again and told me that the answer was $\ce{P_5N_5Cl_{10}}$, but didn't tell me why it must be ordered this way. Is there a convention that must be followed here?

Yes there is.  The most general rule is that elements are listed in increasing order of electronegativity, and in your case the Pauling ENs are 2.19 (P), 3.04 (N) and 3.16 (Cl), so that's why the application wanted it that way.

Most of the time we approximate that rule by saying elements are listed in increasing order of column of the Periodic Table, so e.g. OF2, CO2, and BF3, and when elements appear in the same column, in decreasing order of period, so e.g. SiC, BrCl and SO3.  The column rule takes precedence over the row rule, because EN varies more strongly left-to-right than up-and-down, generally.

The place where this approximation fails most obviously is with hydrides, since H has an EN close to C but appears at the far left of the Table.  Normally we skip the approximation and mostly use the underlying rule for hydrides, so e.g. BeH2 and GeH4 even though H is above and to the left of Be and Ge  (Pauling EN is 1.57 for Be, 2.01 for Ge, 2.2 for H), but H2O and HF. However CH4 (Pauling En 2.55 for C) doesn't quite follow that rule and NH3 (Pauling EN 3.04 for N) is wildly contradictory.  These may just be historical accidents.


What is the standard heat of formation of the nitrate ion, based on the standard enthalpy of formation of silver nitrate? Given the standard enthalpy of formation of $\ce{AgNO3(aq)} = \pu{-99 kJ/mol}$ and the standard enthalpy of formation of $\ce{Ag+(aq)} = \pu{+105 kJ/mol}$, what's the standard heat of formation of $\ce{NO3-(aq)}$?

How can I know this if the standard enthalpy of reaction is not known? This is self test 10.1 in Atkins, PC.

Silver nitrate, $\ce{AgNO3}$ is a strong electrolyte. Based on the information given, I expect you are meant to assume that it dissociates completely when dissolved in water:

$$\ce{AgNO3(aq) -> Ag+(aq) + NO3- (aq)}$$

Thus, we can make the assumption that $$\ce{AgNO3(aq)} \equiv \ce{Ag+(aq) + NO3- (aq)}$$

Then, we can state:

$$\Delta_\mathrm{f} H (\ce{AgNO3(aq)}) = \Delta_\mathrm{f} H (\ce{Ag+(aq)})+\Delta_\mathrm{f} H (\ce{NO3- (aq)})$$

$$\implies \pu{-99kJ/mol} = \pu{+105kJ/mol} +\Delta_\mathrm{f} H (\ce{NO3- (aq)})$$

Thus $$\Delta_\mathrm{f} H (\ce{NO3- (aq)})=\pu{-204kJ/mol}$$


Particle in a box with different barriers between two molecules I'm trying to understand different ways to solve a PIB model.

I'm given a molecule with a tetrahedral geometry and there are two related ground states - electron pair pointing up or down. I'm being asked to draw the n=1 and n=2 states in three separate cases of a finite potential barrier in the middle of the box -> VBarrier = 0, VBarrier = V0, and VBarrier = ∞. So that left of the barrier are the wavefunctions of the up orientation and right is the down orientation.

I know that for the case of VBarrier = ∞, it'll just be the same sin(n pi x/L) wavefunctions on both sides of the barrier. I don't understand how they would change for V=0 and V=V0, and why these barriers would have different heights? I'd be grateful for any help.

for the infinite barrier I'd put an electron on just one side of the box with length L/2, meaning that electron is trapped on one side/face of the ammonia.

For the no barrier part treat the electron as a box with length L meaning it can easily go from one side to the other without barrier (for example maybe the ammonia is at the planar transition state).

For the finite barrier, I'd show tunneling from one side to the second, however, normally for PIB the tunneling region has finite length.

Assuming you have a tetrahedral geometry, according to the Born-Oppenheimer separation of electron and nuclear motion the finite barrier would extend from L/2 to L (or equivalently from 0 to L/2). Therefore the tunneling would be from the middle to the edge.

Image source


In ionic solids such as NaCl, do the cations fill interstital holes in the cation's cubic close pack structure? Are binary ionic solids like $\ce{NaCl}$ comprised of a cubic close packed structure of the anions with the cations filling interstital holes or do the cations and the anions contribute to the cubic close packed structure equally? I am really struggling with solid state chemistry - any help would be gratefully received. Also any suggestions for good learning resources on this topic would be great. Thanks.

The relative radii of the ions, and the stoichiometric ratio of the ions needs to be considered.

For NaCl, the radius of Na+ is 0.695 that of Cl-.

Na+ ions occupy octahedral holes in the cubic close packed lattice of Cl-.

If the radii are more equal (ratio greater than 0.732) then each type of ion will form a simple cubic lattice, with each ion of one lattice filing a cubic hole of the other lattice. For example, CsCl.

If the radii are very different (ratio less than 0.414) the smaller ions will fill tetrahedral holes in the cubic close packed lattice of the larger ions. For example ZnS.

See http://depts.washington.edu/chemcrs/bulkdisk/chem162A_sum06/handout_Lecture_11.pdf for more information.


Explaining a biochemical reaction (protein labelling, e.g. with fluorescent probes) I am an immunology master student without prior knowledge of biochemistry.
Could someone explain to me the following 6 reactions?
What I want to understand is why exactly the reactions happen at this site ( in terms of donor/acceptor ).  What is the donor, what is the acceptor?

All reactions you gave are typically used in protein labelling, e.g. with fluorescent probes.

In order to allow further measurements on the labeled protein, the following conditions have to be met:

The bonds formed have to be stable like the peptide bonds in your protein. You don't want the probe flying around somewhere.

Proteins tend to be a bit sensitive to degradation; boiling in acetic acid is obviously not an option to run the labeling. As a consequence, the functional group to be attached needs to be activated, e.g. with good a leaving group, to allow the labelling of the protein under mild conditions. The succimidyl ester in the first reaction is a good example.


Find the maximum temperature attained by the gas during the change The main question is:

  2 moles of an ideal gas is changed from its initial state $(16\ \mathrm{atm},\ 6\ \mathrm L)$ to final state $(4\ \mathrm{atm},\ 15\ \mathrm L)$ in such a way that this change can be represented by a straight line in $P{-}V$ curve. Find the maximum temperature atained by the gas during the above change. Take $R=\frac{1}{12}\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}$.

My approach :
I plotted the initial and final conditions on a $P{-}V$ curve using a Cartesian coordinate system. Joining them gives me the temperature transition.

I also made two isotherms (hyperbolic curves) on the same graph, passing through the points a previously plotted. The values of initial and final temperatures I calculated are

$$\begin{align}
T_\mathrm f&=360\ \mathrm K\\
T_\mathrm i&=576\ \mathrm K
\end{align}$$
I tried finding a function in $T$ which I could maximise, but in vain.

The ideal gas law $pV = nRT$ allows us to relate one of $p$, $V$, and $T$ to the two other variables, and is the main equation to work with here.

Rephrasing the original question to make it more suggestive, we are interested in maximizing $T(p,V)$ subject to the constraint $p(V)$, a linear function in $V$ readily found from the conditions specified in the problem. Substituting $p(V)$ for $p$ to get $T = T(p(V),V) = T(V)$, we have reduced the problem to one amenable to simple calculus techniques, and the answer follows from taking a derivative.

Remark. In general, extremization problems are not so easy, and we will need to use the more powerful tool of Lagrange multipliers.


How can we define a non-spontaneous reaction? Consider a reaction:$$\ce{$aA + bB$ <=> $cC + dD$}$$

The value of reaction quotient at a certain time $t$, $${Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}}$$
where the concentrations $[A], [B], [C]$ and $[D]$ are at time $t$.

Let the reaction start initially at $t = 0$, with only reactants $i.e.$ $[A]$ and $[B]$ equal to say $1$ mol and $[C]$ and $[D]$ equal to $0$ mol. Hence, $$Q_c = 0$$

We know that the value of equilibrium constant $K_c$ must be such that, $$K_c > 0$$
Thus, $$Q_c < K_c$$
Which is the only condition for the advancement of reaction in forward direction. This condition does not consider the value of change in Gibbs energy $\Delta G$.

Now, considering the relation $$\Delta G = \Delta G^o + RT~\mathrm{ln}~Q_c$$
When $Q_c = 0+$ then $\mathrm{ln}~Q_c \to -\infty$, which means $\Delta G << 0$ and reaction is spontaneous in forward direction.

Hence, can it be concluded that every reaction is spontaneous in forward direction if it starts with only reactants?

If so, how can we define a non-spontaneous reaction?

It sounds like your confusion arises from not making a distinction between $\Delta G$ and $\Delta G^\circ$ when describing a reaction as spontaneous or not.

The $\Delta G^\circ$ is the free energy change for the reaction at the defined "standard" conditions of 1 M solute concentrations and/or 1 bar gas partial pressures of both the reactants and products. When we make the general statement that a reaction is spontaneous or not, we are usually referring to whether this $\Delta G^\circ$ is greater or less than 0.

If $\Delta G^\circ < 0$, then the reaction is described as spontaneous, and it will proceed in the forward direction when starting from a mix of 1 M reactants and 1 M products until equilibrium is reached.

In contrast, $\Delta G$ is the free energy change for the reaction in whatever state is of interest. In your case, you have chosen a state with no product present at all. When we say that a reaction is spontaneous based on $\Delta G$, we mean whether it will go in the forward or reverse direction from this specific state of interest, not the standard state.

Obviously, if there is no product present, the reaction cannot go in the reverse direction (forming reactant from product), so all reactions will go in the forward direction if started in a state of $0$ product. That is what you have shown quantitatively using $Q=0$ so that $\Delta G <<0$.

So you are absolutely correct that every reaction in a state in which no product is present is spontaneous. That does not mean, however, that the reaction will necessarily happen anytime soon. Very often, the activation energy barrier is high enough that the reaction rate is extremely slow. Or it is possible that the equilibrium constant may favor reactants so much that even one molecule of product is sufficient to reach equilibrium and the reaction stops very quickly.


Nernst equation for a fuel cell when gases are stored separately I have run some tests with a PEM fuel cell and the voltage we have found was $1.44\ \mathrm{V}$. Given that the standard potential for a fuel cell is $1.229\ \mathrm{V}$ what is causing the difference in our measurement?

Initially we provide power to the fuel cell so it can generate oxygen and hydrogen into separate containers. These containers are submerged in water so there is water pressure keeping them contained.

All measurements were done in standard conditions, except that the pressure of each container is slightly different (hydrogen displaced twice the amount of water since there are twice the moles of atoms present).

I am not sure how to calculate this using the Nernst equation since the conditions of both the oxygen and hydrogen are slightly different (each gas is of a different concentration in its respective container since there is different water pressure and water vapour pressure present).

My equation is:

$$E_\text{cell} = 1.229\ \mathrm{V} - \frac{RT}{2F}\cdot\ln\left(\frac{[\ce{H2O}]}{[\ce{H2}][\ce{O2}]^{0.5}}\right)$$

For $[\ce{H2}]$ I am using the molar concentration of the gas in its containing cylinder (I believe this accounts for water vapour).

I also do the same thing for $[\ce{O2}]$

I'm not sure what to use for $[\ce{H2O}]$

The answer I am getting is $1.24\ \mathrm{V}$. What am I missing?

Given that the standard potential for a fuel cell is 1.229 V what is causing the difference in our measurement?

Multiple sources (e.g. here, here and there) estimate the voltage to be much less than the standard potential, especially when there is a load.

Individual fuel cells produce relatively small electrical potentials, about 0.7 volts, so cells are "stacked", or placed in series, to create sufficient voltage to meet an application's requirements.

So it is surprising that the voltage is higher than the standard potential. Perhaps two cells were stacked to provide about 1.4 V, similar to the voltage of a standard alkaline battery, and about twice what you would expect from a single cell under load.

For [H2] I am using the molar concentration of the gas in its containing cylinder (I believe this accounts for water vapour).

For the standard hydrogen electrode, the standard state is defined as a partial pressure of 1 atm. This is much lower than a concentration of 1 M. In any case, the Nernst equation does not predict a voltage that is significantly higher than the standard potential.


"Tie-breaker" rules for IUPAC nomenclature of organic compounds I have three questions:

To find the parent chain of a saturated hydrocarbon, we have to find the longest chain and then if we have a choice, we choose the one with the more substituents. But what if we still had a choice? What would we do next?
Similarly, for numbering the chain, the last step is to number the chain such that the substituent which comes alphabetically first gets the lower number. But what if we again had a choice? I'm not sure here, but I think we can't possibly have another choice. Can anyone confirm this?
If an alkyl substituent is attached to the parent chain by a double or triple bond, how would we name it?

Examples: Here are some examples which demonstrate the above 'tie-breakers'.

For question 1:

The (green + blue) and (green + magenta) chains are the longest and both have two substituents.

The (green + blue) and (green + yellow) chains contain the maximum double bonds and are also the longest. They both have one substituent.

For question 2: As I said, I'm not sure if there can possibly be a 'tie-breaker' here.

For question 3:

The longest chain is the green one. I'm not sure how to name the substituent attached by the double bond.

For Question 1

Example 1

You should select the (green + blue) chain because, no matter which among the blue and yellow you pick, the numbering would not start from the green chain rather the yellow or blue.
Now if you pick the blue chain the substituent ($\ce{methyl}$) would get number 4, whereas if you picked the green chain the substituent($\ce{isopropyl\ or\ methylethyl}$) would get the number 5, so in accordance to the First Point of Difference Rule you should select the green + blue chain.
Also Worth noticing is the fact that no matter which chain you pick, the other chain which becomes the substituent would always get the number 7

So locants (4,7) are better than (5,7).
$$\begin{matrix}
\text{Better name:}& \text{7-(2-isopropylhexyl)-4-methylheptadecane}&\\
\text{Rather than:}& \text{5-isopropyl-7-(3-methylhexyl)heptadecane}&
\end{matrix}$$

Example 2

In this case according to the above mentioned rule, you should check the position of the double bond, now since there is a double bond already in the green chain at the first position it would get number 1 and the numbering would start from there.
If you choose the green + blue chain the other double bond would get number 5 and for the green + yellow chain the other double bond would get number 6. So you should go for the green + blue chain in this case.

So locants (1,5) are better than (1,6) with reference to double bonds.
$$\begin{matrix}
\text{Better name:}& \text{(5}\textit{Z}\text{)-5-ethenylhepta-1,5-diene}\\
\text{Rather than:}& \text{(3}\textit{Z}\text{)-3-ethylidenehepta-1,6-diene}
\end{matrix}$$

For Question 2

If the chain has to be numbered and two substituents have the same locant the n you see the alphabetical order of the names of the groups/substituents and if they are also the same it only means that they are the same substituent and hence are written together with there locants separated by a (comma),.

For example in you would name the below compound as
$\text{2,4-dichloropentane}$

For Question 3

Example 1

The substituent in this example would be named $\ce{ethylidene}$.
$$
\begin{matrix}
\text{Final name:}& \text{5-ethylidenenona-1,8-diene}
\end{matrix}
$$
Similar substituents can be named as

$$
\begin{array}{|c|c|c|c|}
\hline
\ce{Substituent& Saturated\ form& Name\ of& Name\ of \\
&& saturated\ form& substituent \\
\hline
R=CH2& R-CH3& Methyl& Methylidene\\
\hline
R=CH-CH3& R-CH2-CH3& Ethyl& Ethylidene\\
\hline
R-CH=CH2& R-CH2-CH3& Ethyl& Ethenyl\\
\hline
R=CH-CH2-CH3& R-CH2-CH2-CH3& Propyl& Propylidene\\}
\hline
\begin{align}
\ce{R=C&-CH3\\
|& \\
C&H3}
\end{align}&
\begin{array}{rl}
\ce{R-C&-CH3\\
|\ &\\
C&H3}
\end{array}&
\ce{Isopropyl& Isopropylidene}\\
\hline
\end{array}$$


Why is CH3CH2NH3Cl acidic? Why is $\ce{CH3CH2NH3Cl}$ acidic?

I know that it has something to do with how it is formed from a weak alkaline, but I cannot identify which acid and base it is formed from.

In short

$\ce{CH3CH2NH3Cl}$ is acidic because it is formed from a strong acid ($\ce{HCl}$) and a weak base ($\ce{CH3CH2NH2}$).

Elaborated

Notice that this compound is ionic, comprising $\ce{CH3CH2NH3+}$ cation and $\ce{Cl-}$ anion.

I mentioned that $\ce{CH3CH2NH2}$ is weak, meaning that this reaction is reversible:

$$\ce{CH3CH2NH2 + H2O <=> CH3CH2NH3+ + OH-}\tag1$$

In an acidic environment, we can rewrite it as:

$$\ce{CH3CH2NH2 + H3O+ <=> CH3CH2NH3+ + H2O}\tag2$$

Therefore, when $\ce{CH3CH2NH3Cl}$ dissolves in water, there would be initially an equal amount of $\ce{CH3CH2NH3+}$ and $\ce{Cl-}$ in terms of number of moles.

Using equation $(2)$, we know that $\ce{CH3CH2NH3+}$ will react with water reaching an acidic equilibrium.

Note that there is no $\ce{OH-}$ to start with (very little in reality), so equation $(1)$ is not applicable.

Therefore, the following species would be present:

$\ce{CH3CH2NH3+}$
$\ce{CH3CH2NH2}$
$\ce{H2O}$
$\ce{H3O+}$
$\ce{Cl-}$

Now it is apparent that $\ce{H3O+}$ makes it acidic.


How does a catalyst affect the rate equation? If I determine the order of the reactants in a reaction without using a catalyst, and then use a catalyst, will the order of the reactants then be different?

Yes. It is possible for a catalyst to change the order of a reaction since it shows different path to the reaction i.e. reaction occurs through different mechanism. In the new path, rate determining step (slow step) may involve different number of reactant molecules.

For example, in case of reaction catalyzed on metal surfaces, most of the times, the rate does not depend on the concentration of the reactants at higher concentration since now it is surface area that governs the rate.


How is a reduction in the number of molecules associated with a decrease in Entropy? How is a reduction in the number of molecules associated with a decrease in Entropy? For example, stoichiometric equations like $\ce{H2 + O -> H2O}$ are enthalpy driven if I am correct. But it also implies that entropy is decreasing because the system is becoming "more orderly." Is this intuition correct?

Interpreting entropy as "disorder", which comes from thermodynamics applications, is only one way to interpret entropy. Entropy can also be described according to statistical mechanics as the statistical distribution of possible degenerate microstates:

$$S=k_B \ln\Omega$$

where $k_B$ is Boltzman's constant and $\Omega$ is the number of microstates. Thus it is a measure of uncertainty. The more equivalent/degenerate microstates there are, the less likely we know exactly which one of them the system is in, and thus, the higher the entropy. This probabilistic interpretation provides a more intuitive and satisfactory understanding of the relationship between number of molecules and entropy of the system.

  A decrease in the number of molecules leads to a decrease in the total number of degenerate microstates, which leads to a decrease in entropy. There are more degenerate ways to distribute N objects in M spaces than there are degenerate ways to distribute (N-1) objects in M spaces.

Let's look at a very simple example, in which we have three molecules $\ce{2H2 + O2}$ in a container with room for four molecules. How many ways can we fill four spaces with these three molecules? There are four spaces in which we could put the $\ce{O2}$ molecule:

$$\begin{pmatrix}
\ce{O2} & ... \\
... & ... \\
\end{pmatrix}
\begin{pmatrix}
... & \ce{O2} \\
... & ... \\
\end{pmatrix}\begin{pmatrix}
... & ... \\
\ce{O2} & ... \\
\end{pmatrix}
\begin{pmatrix}
... & ... \\
... & \ce{O2} \\
\end{pmatrix}
$$

How many ways can we put two $\ce{H2}$ molecules in one of these arrangements? In order to answer this, we need to temporarily distinguish one $\ce{H2}$ from the other $\ce{H2^{\circ}}$ (or to distinguish the one we place first from the one we place second) in order to generate all of the degenerate states.

$$\begin{pmatrix}
\ce{O2} & \ce{H2} \\
\ce{H2^{\circ}} & ... \\
\end{pmatrix}\begin{pmatrix}
\ce{O2} & \ce{H2} \\
... & \ce{H2^{\circ}} \\
\end{pmatrix}\begin{pmatrix}
\ce{O2} & \ce{H2^{\circ}} \\
\ce{H2} & ... \\
\end{pmatrix}\begin{pmatrix}
\ce{O2} & ... \\
\ce{H2} & \ce{H2^{\circ}} \\
\end{pmatrix}\begin{pmatrix}
\ce{O2} & \ce{H2^{\circ}} \\
... & \ce{H2} \\
\end{pmatrix}\begin{pmatrix}
\ce{O2} & ... \\
\ce{H2^{\circ}} & \ce{H2} \\
\end{pmatrix}$$

There are six arrangements of two $\ce{H2}$ molecules for each location of $\ce{O2}$ for a total of $\Omega=24$ arrangements. The entropy $S=k_B\ln(24)=4.39\times10^{-23}) \frac{\text{J}}{\text{K}}$. Note that this happened to be $\Omega=4\times3\times2=12$, or 4 ways of placing the first molecule, 3 ways of placing the second, and two ways of placing the third. We will use this shortcut next time we do this.

Now, let us run the reaction, converting hydrogen and oxygen to water.

$$\ce{2H2 + O2 -> 2H2O}$$

And let us repeat the exercise. How many ways can we fit two $\ce{H2O}$ molecules in four spots? There are 4 ways to place the first one, and 3 ways to place the second for each of those 4 placements of the first. Thus, $\Omega=4\times3=12$ and the entropy is $S=k_B\ln(12)=3.43\times10^{-23}) \frac{\text{J}}{\text{K}}$.

  Note that there is a subtle flaw in my explanation. The molecules themselves have an intrinsic entropy hidden in their structures as well. According to this site the standard molar entropies of $\ce{O2},\ \ce{H2},$ and $\ce{H2O}$ are (in$\frac{\text{J}}{\text{K}\cdot\text{mol}}$) 205.1, 130.7, and 188.8, respectively. Thus, we have an entropy change of reaction:

$$\Delta_rS^\circ=\sum S^\circ_{products} - \sum S^\circ_{reactants}$$ $$ \Delta_rS^\circ =\left(2\times188.8 - (205.1+2\times130.7)\right)\frac{\text{J}}{\text{K}\cdot\text{mol}}=-40\frac{\text{J}}{\text{K}\cdot\text{mol}}$$

By comparing standard entropies, we also get a decrease in entropy. Standard entropies are easier to work with because we are frequently working on the molar scale. If we changed our simplified example to be moles of $\ce{H2}, \ \ce{O2},$ and $\ce{H2O}$, in 4 moles of spaces we would have the original number of microstates be:

$$\Omega=(4N_A)(4N_A-1)(4N_A-2)=64N_A^3-48N_A^2+8N_A$$

When you plug in $N_A=6.022\times10^{23}$, all but the most staunch calculators and computer programs are going to give up. Wolfram Alpha can do it because it does so symbolically, and we get

$$\Omega=1.398\times10^{73}$$

$$S=8.314\frac{\text{J}}{\text{K}\cdot\text{mol}}\ln(1.398\times10^{73})=1.400\times10^3\frac{\text{J}}{\text{K}\cdot\text{mol}}$$

(At the molar scale we replace $K_B$ with $R$, the ideal as constant.)

Another apparent weakness of the statistical approach is that it assumes no phase change. As F'x mentioned, there is an entropy change associated with a phase change. Roughly, the number of available spaces increases or decreases.


Does the term 'Cation' always refer to a positively charged particle? From what I was taught in middle school, cations are those ions that move towards the cathode, likewise anions are those ions which move towards the anode.

I didn't have issues with this back then, since all we studied were electrolytic cells. But now that we've crossed over to electrochemical cells, I'm having doubts.

Every piece of chemistry literature I've come across so far, always deals with positive ions as cations and negative ions as anions; even my school textbook does it! That makes sense with respect to electrolytic cells, since the cathode's negative and anode's positive therefore positive ions would move to the cathode (hence 'cations') and negative ions would move to the anode (hence 'anions').

But in an electrochemical cell, the cathode's positive and the anode's negative. So if ions are to be classed according to the electrodes they move over to, then positive ions would be anions and negative ions would be cations, which is exactly opposite to the first case (on the basis of electrolytic cells).

This is really confusing....

So should I refer to positive and negative ions as cations and anions respectively or as anions and cations respectively? Or are both acceptable, depending on the scenario?

Yes, cations always have a positive charge and anions always have a negative one.

The difficulty is that the term cathode and anode do not always correspond to the same pole. The cathode is that pole of an electrolytic/electrochemical cell where reduction takes place (cathodic reduction) while the anode is where oxidation takes place (anodic oxidation). Since it depends on the type of cell which will be oxidised and which will be reduced, the terms cathode and anode mean different things in different cells.


Volume change in positive deviation from Raoult's Law Let us take a mixture of ethanol and acetone as an example. When ethanol is mixed with acetone, the hydrogen bonding between ethanol molecules gets disturbed as actone molecules get in between the ethanol molecules. This results in $\Delta_\text{mix} H\gt0 $ as heat must be supplied to make the solution from its constituents (as attraction is less) but I am not sure about $\Delta_\text{mix} V$. This page says it is greater than zero because volume expands on mixing. But I think it should be less than zero because the observed vapour pressure is more than that predicted on the basis of theoretical calculations, so a greater amount of solution will be present in the form of vapour, hence volume of solution will be less (than the sum of the volumes of its constituents). Where am I wrong?

I'm a little late to the party here, but wanted to chime in.

The link you posted isn't working anymore, but you appear to be making too many assumptions about the system being measured.  Nothing about the measurement of the volume change implies there is necessarily an open space above the liquid for vapor to form.  Imagine an open beaker of ethanol with a small disk floating on top so that no molecules can escape upwards, but it can move up and down with the change in liquid level.  This would also keep the pressure of the liquid constant (although it doesn't really matter since the liquid is essentially incompressible) .

If you were able to inject a certain amount of acetone into this beaker without opening it to the atmosphere for very long, the resulting volume of liquid in the beaker would be greater than the sum of its parts. Hopefully it makes sense that if there is no space for liquid to form, the positive volume change of mixing can show itself.

As far as the idea that more vapor would form to decrease the volume of liquid, realize that evaporation is a time-dependent process, so likely the mixed liquid would reach an effective equilibrium state where it's volume change could be observed before the extra liquid has time to evaporate.  It is true that it would evaporate faster if the effective vapor pressure (technically "fugacity" or "chemical potential") were higher, as the driving force for mass transfer would be greater, but this is also very dependent on the way the system is specified.  Technically if you left the container open in a large room forever, it would all evaporate in either case.


Why are long oxygen chains unstable? H-O-H (water), H-O-O-H (hydrogen peroxide), H-O-O-O-H, H-O-O-O-O-H, etc.

This is analogous to hydrocarbon chains, methane, ethane, propane, butane, etc.

Note the Hydroxygen Chains satisfy 2 covalent bonds for every oxygen atom, just as every Hydrocarbon Chain satisfies 4 covalent bonds for every carbon atom. Yet the oxygen chains seem unstable. Why?

I can't imagine it has anything to do with shells deeper than p. And carbon and oxygen are very close together on the same row, so it would appear the valence shells are the same and thus governed by the same laws.

I'm also aware of those weird exceptions, Ozone and Red Oxygen. Maybe that has something to do with it, but I don't know how those bonds work either.

I also saw this related question. It just asks if they're possible. I didn't see any explanation about why they're unstable.

The top of the p block is unusual in the atoms being so small they can readily form strong multiple bonds, because (from a VB point of view) they can get close enough for good side-by-side p orbital overlap.  Hence they prefer to form multiple bonds if they can, and both N and O can do that to form diatoms.  So extended chains of either N or O atoms are generally less stable than a collection of N2 or O2 molecules.  Or if you have extra capping atoms around, like H, then a collection of N2/O2 plus some H2O/NH3, et cetera.

Carbon is the exception because it has four valence electrons, and a quadruple bond to form C2 is too high energy.  (From a VB point of view, there are no valence atomic orbitals left to overlap to form the 4th bond.)  The next smallest molecule it can form with itself with lots of double bonds is buckminsterfullerene (C60).  So without another element around, a long chain of O atoms can form a more stable collection of O2 molecules, while a long chain of C atoms can't form anything more stable at all.  (When other elements are around, all bets are off: long chains of carbon tipped with hydrogen are notoriously unstable to CO2 and H2O if there's plenty of oxygen around.)


Is enthalpy of a reaction meaningful at variable pressure? Why should the enthalpy of a reaction be defined at constant pressure. Why can't it be defined at variable pressure?

Why should the enthalpy of a reaction be defined at constant pressure? Why can't it be defined at variable pressure?

Not sure "should" is the right choice of words: It can be defined for a reaction either at constant volume (variable pressure) or constant pressure (variable volume). You might choose one measurement over the other, but you'll get two different quantities.

At constant volume, there's no work done by the system on the surroundings, so you're just getting the change in internal energy of the system (this is typical of a bomb calorimeter lab setup); in the other case, you get the internal energy change as well as the work, and that is how enthalpy is defined (mathematically). See this Wikipedia article for a little more detailed discussion.


Are the odours added to natural gas completely flammable? I live in Australia, where both electricity and natural gas are common forms of household energy. An odour is added to the natural gas as a safety precaution since it is otherwise almost undetectable. I haven't been able to find what gas/mixture of gases is used in Australia, with all the sites I've checked just saying "an additive" or "an eggy smell".

The Wikipedia article on natural gas says that both tert-Butylthiol and tetrahydrothiophene can be used as an additive, and the article on tetrahydrothiophene says:

  It is also used as an odorant for natural gas, usually in mixtures containing tert-butylthiol.

Whenever I've used natural gas (i.e. burning it for heating or cooking) I've noticed that there is no smell afterwards. Are all the additives completely flammable? Presumably, their combustion products are odourless, but I haven't been able to find any information on that either.

On looking it up on Google and a few more places, it turns out that natural gas is primarily methane, along with a few other hydrocarbons.
Well, I have found out through some searches online that in the case of propane based fuel, like LPG (Liquid Petroleum Gas) in India, ETHYL MERCAPTAN is a popular option. It's smell is described as a

  strongly disagreeable odor that humans can detect in minute
  concentrations

And that it's smell resembles that of

  leeks, onions, durian or cooked cabbage, but is quite distinct.

(source :Wikipedia)

And also, most important point of all, it is highly explosive.

This is what the New Jersey Department of health and safety has to say:

(Source : This document)

So yeah, You're right. Based on this example, I think it logically follows that these "additives" are flamable and result in products that are ODOURLESS.


Very slow chemical clock reaction I'd like to preface this with a declaration of my understanding that this is not necessarily possible.

I was wondering if anyone knew of a chemical clock reaction where the period before the first change (the induction period?) is very long (between 3 hours and a day) and the change is à la Landolt or Old Nassau reactions (i.e they cross a concentration threshold to create sudden color change).

Bonus Question:
Is there any remote possibility that a reaction like this can happen in pigments or dyes once they are absorbed by a cloth?

I assume you are aware of the common clock-reactions, such as the iodine clock, etc. By varying concentration of reagents and pH of the solution, you could run several repetitions to generate a trend-line, and from that extrapolate the estimated pH and reagent concentration to provide you with the time delay you are looking for. The iodine clock is a two-step reaction as outlined below:

In the first, slow reaction, iodine is produced:

$$\ce{H2O2 + 2I- + 2H+ -> I2 + 2H2O}$$

In the second, fast reaction, iodine is reconverted to 2 iodide ions by the thiosulfate:

$$\ce{2S2O3^2- + I2 -> S4O6^2- + 2I-}$$

By decreasing pH in reaction 1 and increasing the thiosulfate level in reaction 2, you should be able to lengthen the ‘clock time’. Let me know how things turn out!


Implicit dependence of reaction rate through reduction potential and activation energy On one hand, the larger the reduction potential of a chemical reaction and the smaller its activation energy, the larger the reaction reaction. on the other hand the rate of a reaction depends on thermodynamic conditions, such as temperature and pressure, and the concentration of the species involved.

Now given that both the reduction potential and the activation energy of the reaction depend on thermodynamic conditions, can we deduce that the whole thermodynamic dependency of the rate of the reaction lies inside the reduction potential and the activation energy?

For the sake of clarification, that I said the larger is the reduction potential and the smaller is the activation energy the greater will be the rate of the reaction is a conclusion based on the analogy of a redox reaction and an electric circuit, the reduction potential being in place of the voltage difference, the activation energy as the energy consumer being in place of the electric resistance and the electron transferred during the reaction being in place of the electric current. It is clear that the greater the electric current, the greater the rate of reaction should be.

The rate of a reaction depends on the thermodynamics of its transition state.
It would be better to look at it in terms of Gibbs activation energy. Therefore there two factors affecting the rate of a reaction:

the activation enthalpy $\Delta^‡ H $, that corresponds to the Arrhenius activation energy
the activation entropy $\Delta^‡ S $ that takes into account the probability of a transition state.

The whole dependency of the kinetic constant on these parameters is expressed by the Eyring and Wynne-Jones equation:
$$ k = \frac{k_\text{B}T}h \operatorname{e}^{\frac {\Delta^‡ S} R }\operatorname{e}^{-\frac{\Delta^‡ H} {RT} }$$

Where $T$ is the temperature, $k_\text{B}$ is the Boltzman constant and $h$ is the Planck constant.
Sometimes a reflection coefficient $\kappa$ is added to the equation to take into account some other dependencies. It is near to 1 for most of the simple reactions.
$$ k = \kappa \frac{k_\text{B}T}h \operatorname{e}^{\frac {\Delta^‡ S} R }\operatorname{e}^{-\frac{\Delta^‡ H} {RT} }$$


Can chloride attack the carbonyl bond in formaldehyde in the reaction with hydrochloric acid? I already know that a cyclic trimer and polymer of formaldehyde is formed at lower temperatures, but would chloromethanol and dichloromethane be possible at higher temperatures?

We can both immediately agree on the first step of the reaction between formaldehyde and hydrochloric acid which is the protonation of the carbonyl oxygen. The mechanism of formaldehyde oligo- or polymerisation then suggests the attack of the oxygen atom of another formaldehyde moiety to form a hemiacetal. This would be a nucleophilic addition onto a π system.

One possible course this species can now take is the back reaction to reform formaldehyde. Another one is a subsequent attack of the newly-formed hydroxy moiety of the hemiacetal function onto another carbonyl group to form a full acetal. The latter reaction creates a product which is inert enough to not undergo back reactions.

You are indeed correct that chloride, which is also a nucleophile, can equally attack the π $\ce{C=O}$ bond of formaldehyde. And this attack would indeed result in chloromethanol. However, this reaction is also reversible as is the attack of oxygen. Furthermore, this molecule has no real way of stabilisation: chloride will always be a good leaving group while a formal alcoholate is not. Thus, while the attack of chloride may occur it is not productive and it is rapidly reversed.

Therefore, we do not see any chlorinated product — which would decompose on isolation at the latest anyway.


What are the fundamental parameter by which lubricants are graded? I am into hobby machining using an old beaten CNC mill in my garage. Being novice my confidence is rather low and I have genuine fear of breaking something (even ancient CNC machines like mine cost a fortune), so when I got to the part of the users manual that listed recommended lubricants for the different parts of the machine I felt out of my league completely (especially since I just basically submerged the whole machine in a bath of general purpose lubricant spray).

So I would like to know, how are lubricants usually classified/graded?

Which classification/grade matters to which kind of use?

What are the risks of using the wrong lubricants?

Lubricants are usually graded for a host of different properties, but ultimately what you are looking for is it's thermal stability, and a viscosity index related to the temperature your mills parts run at. These properties are optimized for different parts of the machine because they run at different speeds and as a consequence temperatures. I could delve into more, but I would need more specifics about the machine itself. If you wish to pursue the problem further on your own I would suggest This website as a reference. Also take a look at this cautionary article regarding accurate measurements on viscosity index as it can be a bit tricky.

The consequence of using the wrong lubricants varies based on how wrong you are... But ultimately the lubricant will thin out to the point that is does not decrease friction and or cool the parts it is lubricating if it degrades at the temperature the part runs at.


How to explain molecular geometry without the help of VSEPR, valence bond, or hybridization theories? I was taught, at the high school level, how to rationalise molecular geometries with the help of VSEPR, valence bond, and hybridization theories.

However, I have recently also come to know that these theories are outdated and have many limitations. So, how can I rationalise these geometries without the help of these theories? Can I use molecular orbital theory for that, and if so, how?

Short Answer: No, you can't. Reason being that so far this is the only way devised, which works most of the time.

Long Answer:
The Molecular Orbital Theory is, in practice, a complement to the Valence Bond Theory, Hybridization and VSEPR.

There are successes as well as shortcomings of both the Molecular Orbital Theory, and the other three.

In general, the VSEPR theory is used to predict shapes of molecules and ions, not the Molecular Orbital Theory. It can be extended to a large number of Covalent compounds, but it has a few limitations as far as ionic compounds and multi-centered bonds are concerned. The MOT tells us that when atoms combine, they form molecules orbitals, which are lower in energy than the atomic orbitals. It is generalizable, and can be extended to even semiconductors and superconductors. It is very useful when it comes to finding the actual energies associated with every single orbital.

The Valence Bond Theory, along with Hybridization and VSEPR, is the one that is generally used to find the shapes of molecules. The reason is simple: The Molecular Orbital Theory tells us the arrangements of delocalized electrons in Molecular Orbitals, whereas, the other theories tell us the possible positions of the combined atoms after considering that the electrons are localized and repel each other.

However, there is the case of multi-centered $\pi$-bonds, also known as banana bonds, and the 4c-3e bonds, which can be explained only by the Molecular Orbital Theory. A famous example would be Di-Borane, or $\ce{B2H6}$.

Thus, both the Valence Bond approach as well as the Molecular Orbital approach are often required for an accurate prediction of the shape of a molecules.

I quote an article from Chemistry LibreTexts:

  Both the MO and VB theories are used to help determine the structure of a molecule. Unlike the VB theory, which is largely based off of valence electrons, the MO theory describes structure more in depth by taking into consideration, for example, the overlap and energies of the bonding and anti-bonding electrons residing in a particular molecular orbital. While MO theory is more involved and difficult, it results in a more complete picture of the structure of a chosen molecule. Despite various shortcomings, complete disregard of one theory and not the other would hinder our ability to describe the bonding in molecules.

Thus, although the VB, VSEPR, and Hybridization theories are outdated and have many limitations, they, along with the Molecular Orbital Theory provide us with a toolbox for almost any molecule possible.

(If there is any place where I have erred, please correct me)


Identifying Nucleophiles and Electrophiles Are all polar molecules both nucleophilic and electrophilic, depending on which atom you are talking about? Take water for example. Oxygen bears the partial negative charge. Oxygen is nucleophilic. Because oxygen isolates electron density from the hydrogens, the hydrogens bear partial positive charges. So would the hydrogens be electrophilic?

Nucleophilicity and Electrophilicity are no absolute measurable quantities - they are based on an per entry basis and usually in reference to some kind of reaction. So basically you are right in assuming that polar molecules have both properties. In the special case of water these properties also stem from $$\ce{H2O <=> {}^{-}OH + H3+O}$$

But let us have a look at some molecules:

$\ce{CO}$ pretty good nucleophile especially towards transition metals. It has quite a small dipole towards the carbon, but a massive overload of electron density due to the $\sigma$-type HOMO
$\ce{CO2}$ also quite polar bonds, but no (permanent) dipole - still quite a potent electrophile given the right nucleophile
$\ce{SO3}$ has no permanent dipole (but a transitional one) although quite polar bonds. However, it is only a strong electrophile.
$\ce{NH3}$ quite polar, hence a reasonably good nucleophile. But it is also almost always way too strong a base to act as an electrophile
$\ce{NH(Et)2}$ not so different from ammonia, quite polar, but yet a terrible nucleophile, because it is too big
$\ce{BH3}$ not very polar (no permanent dipole) but quite a good electrophile
$\ce{I-}$ a monopol, very polarisable, good nucleophile, good leaving group.
...

The consideration of the strength of nucleophiles and electrophiles lies in many variables. It might be polarity, local electron density, molecular orbitals, polarisabilities. It is often highly dependent on the solvent and the other reactants. If you add $\ce{I-}$ to a acyl chloride you will certainly have a substitution to some extend:
$$\ce{R-COCl + I- <=> R-COI + Cl-}$$
If you add hydroxide ions, then you will certainly end up with the acid:
$$\ce{R-COCl + I- + {}^{-}OH <=> R-COOH + I- + Cl-}$$
Look at it nucleophilicity or electrophilicity from a case to case basis.