Can you carbonate a non aqueous liquid? We have a home carbonator for makeing seltzer. It blows $\ce{CO2}$ under high pressure into water. Could I use it to carbonate corn oil? Is $\ce{CO2}$ soluble in a no aqueous solution?

$\ce{CO2}$, being non-polar (due to the 180 degree bond angles),  will probably be more soluble in solvents that are non-polar as well, although the quadrupole moment might cause higher solubility in polar solvents as well (thanks to @georg for this addition). Also temperature will play a significant role, the higher the temperature the lower the solubility.

These qualitative remarks however only take you so far. If you really want to know the solubility in a specific liquid you will have to look it up or, if that is not possible, test it yourself. In this paper they test the solubility of $\ce{CO2}$ in different types of crude oil and make predictions based on the structure of the oil.

I cannot find anything specific for corn oil, although the reverse process of dissolving palm oil in supercritical $\ce{CO2}$ might set you on the right track to find what you want to know.

Just a fun addition: don't ever try to carbonate milk as these guys do - it will get messy! (it is in Dutch, but you can still watch what happens). Click here an English video with similar footage

$\ce{CO2}$ is also soluble in various organic liquids, but your mileage may vary

From an excellent comment by Greg E. that he didn't opt to turn into an answer:

  CO2 is a non-polar molecule, so its solubility will be partially a function of the polarity of the solvent in question. Furthermore, solubility of gases is very sensitive to temperature. Solvation of a gas is entropically disfavored (lower degrees of freedom for the gas molecules), so at some point a further increase in temperature will take primacy over the heat of solution (enthalpy change due to solvation) and any further increase in temperature will actually result in more CO2 coming out of solution. Additionally, higher temperature will increase the vapor pressure of solution.


Problems regarding Kc,Kp and rate of reaction In our chemistry textbook, the equilibrium constant $K_c$ is derived this way:

$$\ce{aA + bB <=> cC + dD}$$

in this reaction, rate of forward reaction is $k_\mathrm f= k_1[A]^a[B]^b$

rate of backward reaction is $k_\mathrm b = k_2[C]^c[D]^d$

in equilibrium, equalizing the two rates we get
$$ K_c = \frac{k_1}{k_2}=\frac{[A]^a[B]^b}{[C]^c[D]^d}$$

Now, there is a problem... we know the $K_c$ depends on how we equalize the reaction... like we can write
$$\ce{H2 + 0.5O2 -> H2O}$$ or
$$\ce{2H2 + O2 -> 2H2O}$$, etc ways. But how the rate of reaction depends on the concentration? to what power?

Well, there are two cases:

$a + b = c + d$; and
$a + b \neq c + d$

For case 2, it's fairly simple to tell how the equation was balanced because your $K_C$ will come in a different unit, but for case 1, because $K_C$ is dimensionless we depend on the convention that Chemical Reactions are balanced on the lowest whole number. So in the case of the example given, we would balance it as:

$$\ce{2H2 + O2 <=> H2O}$$

and $K_C$ would be defined on that basis.


Acidic strength order in gaseous phase I got this question in my exam:

Arrange the following in increasing order of acidic strength in gaseous phase
(i) $\ce{CH3CH2COOH}$ (ii) $\ce{CH3COOH}$

I felt that increasing order must be (i) $<$ (ii). But answer given is (ii) $<$ (i). Why will the order change in gaseous phase?

Acidity or acid strength can be expressed either in terms of the proton affinity (PA) of the conjugate base, e.g., PA($\ce{A-}$), or in terms of the acid dissociation constant, e.g., $\mathrm{p}K_{\mathrm{a}(\ce{HA})}$. The former expression is often favored among gas-phase chemists, whereas the latter expression is most familiar to solution chemists.

In aqueous solution, $\mathrm{p}K_{\mathrm{a}(\ce{CH3CO2H})} = 4.76$ while  $\mathrm{p}K_{\mathrm{a}(\ce{CH3CH2CO2H})} = 4.87$, thus acidity of acetic acid is greater than that of propanoic acid. Keep in mind that solvent plays a huge role in $\mathrm{p}K_{\mathrm{a}}$ measurements. For example, compare $\mathrm{p}K_{\mathrm{a}}$ measurements in $\ce{H2O}$ and in $\ce{DMSO}$:

$$
\begin{array}{c|cc}
\hline
\ce{HA} & \mathrm{p}K_{\mathrm{a}} \text{ in } \ce{H2O} & \mathrm{p}K_{\mathrm{a}} \text{ in } \ce{DMSO}  \\
\hline
\ce{CH3CO2H} & 4.76 & 12.3  \\
\ce{H2O} & 15.7 & 31.2 \\
\ce{CH3OH} & 15.5 & 27.9 \\
\ce{C6H5OH} & 9.95 & 18.0 \\
\hline
\end{array}
$$

However, the proton affinity (PA) measurements of the conjugate bases of these acids in gas phase tell a different story. Proton affinity is a thermodynamic measurement conducted in the gas phase. These measurements give a pure view of base/acid strength, without complications from solvent effects such as salvation, thus take the solvent out of the equation.

Cardwell, et al. (Ref.1) has measured the proton affinities of various $\ce{R-CO2H}$ acids, including ethanoic and propanoic acids ($\ce{R}=\ce{CH3}$ and $\ce{CH3CH2}$, respectively). They conclude that:

  The acidity increment (with the size of $\ce{R}$-group) is relatively small and can be attributed to the stabilizing effect of the polarizability of the alkyl group on the negative ion.

References:

Gary Caldwell, Richard Renneboog, Paul Kebarle, “Gas phase acidities of aliphatic carboxylic acids, based on measurements of proton transfer equilibria,” Canadian Journal of Chemistry 1989, 67(9), 611-618 (https://doi.org/10.1139/v89-092).
Also read: Diethard K. Bohme, Edward Lee-Ruff, L. Brewster Young, “Acidity order of selected Brønsted acids in the gas phase of $\pu{300  K}$,” Journal of American Chemical Society 1972, 94(15), 5153-5159 (https://doi.org/10.1021/ja00770a002) and references there in.

One possible explanation for this is hyperconjugation due to the attached alkyl chain, as shown below:

since the acetate ion has one more $\alpha$-H atom than the propionate ion, the acetate ion is more stable than the propionate ion in the gaseous state and has more tendency to donate the hydrogen atom, making it more acidic.


Why are condensation reactions endergonic? My lecture notes state that

  Condensation reactions, like all biosynthetic reactions, are endergonic.

I am unsure as to why this would be the case. I thought bond formation was an exothermic reaction, so you would get a negative enthalpy change from the bond being formed. I appreciate that you wold have a decrease in entropy when going from two small molecules to one larger molecule, but would this really counter-balance the negative enthalpy change every time? Perhaps the small molecule produced in the biosynthetic reaction, e.g. $\ce{H2O}$ for condensation reactions, is important here?

In a nutshell, I am just confused as to what ensures that all biosynthetic reactions (and condensation reactions) are always endergonic?

One of the most simple biosynthetic reactions that generates ‘oligomers’ from ‘monomers’ (between inverted commas because that description is not fully correct here) is the ester formation used to synthesise lipids. It can be broken down to:

$$\ce{R-O-H + H'-O'-C(=O)-R' -> R-O-C(=O)-R' + H'-O'-H}$$

With each important bond highlighted and the oxygen and hydrogen atoms made distinguishable by primes. A simplistic way of deriving standard reaction enthalpies is to use Hess’ law and add up the bond dissociation enthalpies of bonds that are broken and subtract that from the bond dissociation enthalpies of bonds that are formed. Doing that in this case gives:

Broken:

$\ce{C-O'}$
$\ce{O-H}$

Formed:

$\ce{C-O}$
$\ce{O'-H}$

Since the bonds being formed and those being broken are identical, at first approximation $\Delta_\mathrm{r} H^0 \approx 0$, meaning the reaction depends entirely on the entropic term.

On the entropic side of things, you are creating one huge, dissolved entity out of two not-as-huge dissolved entities. Making an exact call here is difficult, as it depends on a number of things, most notably how large the hydration sphere is for all three of these compounds. If each of them has a hydration sphere of approximately the same diameter, the product side will have more entropy since less water molecules are needed for an identical hydration sphere (akin to the hydrophobic effect). The opposite effect is also possible, though.

I would like to challenge the ‘all condensations are …’ It seems to simplistic to me. Many condensation reactions are probably endergonic but likewise I expect a few to be overall exergonic. The cell just takes no chances and uses an $\ce{ATP}$-dependent pathway in all cases to ensure the outcome it desires.

That all condensation reactions are endergonic is certainly not true. And as can be seen by the Helmholtz equation $\Delta G = \Delta H - T\cdot \Delta S$, this also depends on the employed temperature. One example is cupper(II)sulfate penta hydrate that spontaneously releases water upon heating, so there you have an exergonic condensation reaction. Forcing T higher will always favor a condensation reaction as typically $\Delta S_\text{condensation} > 0$ due to release of water.

That all (or most, anyway) biological condensation reactions are endergonic is hopefully the case, as otherwise biology would be unable to control the conversion by providing the energy to enable it. If there is an enzyme that converts A to B in an exergonic fashion, all of A will be converted to B eventually (one could argue that the activation barrier is high so that A to B conversion is prevented, but even when the activation barrier is too high for the process being spontaneous the cell would not be able to withdraw the enzyme so the enzyme still catalyzes the reaction in an uncontrolled fashion). But typically the cell will employ pyrophosphates (e.g. ATP) hydrolysis to drive and control those reactions. These will generally be endergonic, so that the cell can convert the endergonic process into an exergonic one by ATP hydrolysis and thereby control the process.

So, indeed my guess would be that most enzymes catalyze endergonic processes. However, there are also exergonic processes like the hydrogen peroxide hydrolysis catalyzed by the catalase enzyme. In that case, there is no need of the cell for A, so it gets converted to B as quickly as possible (and with equilibrium far on the side of B).

So my bet would be that most condensation reactions in the cell are ATP driven, and are thus endergonic.


Why do alkyl bromides and iodides develop colour when exposed to light? My book says alkyl halides are colourless when pure. However, bromides and iodides develop colour when exposed to light. Why do they develop colour when exposed to light? It is important to note that alkyl fluorides, and alkyl chlorides are not said to develop colour.

This is because of spontaneous breakdown in case of iodides and bromides which have a lower bond-dissociation enthalpy as compared to fluorides and chlorides. This low enthalpy means that even at normal temperature or on exposure to sunlight, the halides (iodides and bromides) decompose to a certain extent forming free iodine and bromine, both of which are coloured and hence impart colour to the original halides.

This is because of the fact that in presence of light iodine and bromine absorbs light and excitation of the electron from the ground state to the excited state takes place which on returning back to the ground state radiates energy. This radiating energy comes under the visible region. Therefore they are co lured.


Why exactly does atomic radius increase down a group? I realize this should be a fairly basic question, but I'm still not quite satisfied with what I've been told from numerous sources. The general explanation seems to state that as we go down a group, there are more electron shells present to contribute to a "shielding" phenomenon, where inner electrons cancel out part of the inward force from the nucleus with a repulsive force. Thus with more energy levels present and more shielding, valence electrons begin to reside further and further away from the nucleus.

But at the same time, textbooks make a big deal about the concept of effective nuclear charge, which states that the "effective" force a given valence electron feels is a function of the net charge of the nucleus and non-valence electrons combined—the effective nuclear charge. I find this contradictory, since as we go down a group, the effective nuclear charge remains constant (equal numbers of protons and non-valence electrons are added). So what am I missing here? Why does shielding matter? Is the force on a single valence electron a function of the net nuclear charge or do we have to look at it from a more piecewise level, considering how other electrons are in between it and the nucleus and how it interacts with those individual electrons?

The description of $Z_\mathrm{eff}$ you gave is a bit too simplistic. As you correctly said, that would predict the same value of $Z_\mathrm{eff}$ for all elements in a group, which is not true.

In general we have $Z_\mathrm{eff} = Z - \sigma$ where $Z$ is the nuclear charge (that solely depends on the number of protons) and $\sigma$ is the shielding constant, which reflects the electron-electron repulsions but is not a simple function of the number of core electrons.

One of the earliest models to determine $\sigma$ was described by Slater in 1930.1 These are easily found online as "Slater's rules". These are quite simplistic, and so people tried to find better ways to calculate $\sigma$. Nowadays, one very popular source for values of $Z_\mathrm{eff}$ are the Clementi values.2 That's still quite long ago, so you can imagine that since then, we have come up with even more complicated ways to calculate it.

A peculiarity is that the values of $Z_\mathrm{eff}$ actually increase down the group. At the very least, that should dispel the myth that $Z_\mathrm{eff}$ only depends on the number of core electrons.

(Source: Keeler & Wothers, Chemical Structure & Reactivity: An Integrated Approach, p 264)

"Surely that should mean that the atomic radii decrease down the group!" you say.

I don't fault you for thinking that at all. However, that's not what we see: the atomic radii clearly increase going down the group. The second factor is the principal quantum number, $n$.

In the hydrogen atom, an electron in the 1s orbital has a much smaller radius3 than an electron in the 2s orbital. That's something that comes right out of the quantum mechanical description of the hydrogen atom.4 The same can be said for other atoms: a 2s orbital in lithium is smaller than a 3s orbital in lithium.

Note that I didn't compare a 2s orbital in lithium with a 3s orbital in sodium. Why? That's because sodium has a larger effective nuclear charge, and this serves to contract all the orbitals in sodium relative to those in lithium. Essentially, the orbitals are pulled inwards by the larger $Z_\mathrm{eff}$.

So, as you can see, going from lithium to sodium, there are two competing factors:

Larger $n$ suggests that the outermost electron should be further away from the nucleus.
Larger $Z_\mathrm{eff}$ suggests that the outermost electron should be nearer to the nucleus.

Unsurprisingly, it turns out that the variation of atomic radius is dependent on a combination of the two. We know that the atomic radius increases going down the group, so that must mean that the increase in $n$ (going from 2 to 3) outweighs the increase in $Z_\mathrm{eff}$ (going from 1.28 to 2.51), although it's sadly difficult to find an exact mathematical formulation in any textbook. It's likely that the exact dependency is very complicated and nowadays, such atomic properties are mostly calculated computationally anyway, which removes the need for such a mathematical formula.

Notes and references

1 Phys. Rev. 1930, 36 (1), 57

2 J. Chem. Phys. 1963, 38, 2686

3 I'm talking about $\langle r \rangle$, not the radius of a Bohr orbit, although that does also increase going from 1s to 2s.

4 I'm avoiding the circular logic that "higher energy means further away". It is further away because that is just the form of the radial wavefunction obtained by solving the Schrodinger equation - period.

Down a group means that the valence electrons are on a higher energy level (they are further away from the nucleus). You would also expect that going down a group, nuclear charge increases (more protons), however, the electrostatic strength down a group is relatively the same. Why? well because there are more shielding due to more energy levels so overall nuclear attraction is similar.

Z eff = Z(no. of p) - S(shielding e) <-- finds effective/overall nuclear charge. Let's take oxygen and sulfur for example. Z eff for oxygen = 8-2=6 and Z eff for sulfur = 16 - 10 =6. This means that the overall electrostatic attraction between the nucleus and the valence electrons are similar, so the major factor must be that the electrons are on a higher energy level, so they are further away from the nucleus, that results in larger radius. So going down a group always mean larger radius.


Mechanism for conversion of imide to ester with NaH What is the mechanism of this reaction? My proposed mechanism is shown below, would it work?

NaH is a strong base but a bad nucleophile. This is one possible mechanism.


Which double bond is more electrophilic : carbon-carbon or carbon-oxygen? Question

Attempt

It seems to me that the carbon-oxygen bond would be more electrophilic due to the higher electronegativity of oxygen. Hence I believe $\ce{CH3-}$ should attack the carbonyl site. But this doesn't seem to be the case.

Is it because that the intermediate carbanion can undergo resonance with oxygen, had $\ce{CH3-}$ attacked carbon- carbon double bond?

What can be a valid reason?

step-1: cleavage of C=O bond:

the C=O bond undergoes heterolytic cleavage to give positive charge on carbon and negative charge on oxygen just like almost all other organometallic reagents react with a ketone, for example a grignard reagent (not exactly, but similar to it).

step-2:(the real step which caused deviation from your answer)-resonance of intermediate:

the positive charge on the carbon of the erstwhile C=O group now undergoes resonance with the double bond at the allylic position to give positive charge at carbon no. 3 instead of carbon no. 1
in this sort of reaction is 1,4 addition occurs instead of 1,2 addition. For more information, you can visit this link: https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/1%2C4-Addition

step 3:(as usual):-attack of CH3 group:

now on as usual, the CH3 group from our organo copper reagent ( called as gilmann reagent, used in Corey-house synthesis), just like a grignard reagent, attacks the new carbocation formed at carbon no.3 and the water molecule in the second step donates it's proton to the remnant oxygen of our old C=O group and it becomes an enol. tautomerisation takes place and we bet 3-methyl cyclohexanone as a result.


Pale blue species forming during electrolysis of NaHCO3 I have a solution of 3500 ml $\ce{H2O}$ + 454 g $\ce{NaHCO3}$ in electrolysis at a potential of 12.5V using a lead anode and copper cathode. The surface areas aren't measured but they are enormous. I made the electrodes myself from lead and copper foil.

The idea was to make NaOH from $\ce{NaHCO3}$ (an inexpensive source  $\ce{NaHCO3}$ $$0.86/454g  and NaOH $7/500g) meanwhile producing the PbO2 layer on the lead anode.

Everything seemed to be going as expected until I see there's a pale blue species being produced. What could it be? I'm thinking H2O2, water having the potential to completely become peroxide, so I used the 5V output instead and the blue species vanished.

I've read electrolysis of salts in basic solution usually result in the salt reaching its highest state of oxidation at the anode.

Eg.:

$\ce{NaCl + H2O ->Cl- + OH- -> HClO + OH- ->ClO2- + H2}$
eventually becoming $\ce{ClO4-}$

(correct me if I'm wrong but I believe that's the way that reaction carries out),

or

$\ce{(NH4)2SO4 -> (NH4)2S2O8}$

EDIT
It's just copper. Apparently in a solution of NaOH, there is some oxidation of a Cu cathode. I've seen it occur many times afterward. Hydrogen peroxide cannot be made without special laboratory equipment.

The blue color was caused by copper being stripped from the wire. btw, when your hair turns orange, it is because it is being stripped of color, among other other things (chemical bonds), which will cause it to be brittle and also absorb any hair dye to the extreme that it feels waxy...I'd suggest you cut off a piece of hair for a test before dunking your whole head in bleach (chlorine will turn it greenish).

P.S. Use nitric acid to dissolve Cu.


What are the magnetic quantum numbers for specific d orbitals? I mean I know they’re $\ce{d_{-2}}$, $\ce{d_{-1}}$, $\ce{d_{0}}$, $\ce{d_{1}}$, $\ce{d_{2}}$ but how do these numbers relate to each one of $\ce{d}$ orbitals $\ce{d_{z_2}}$, $\ce{d_{xz}}$, $\ce{d_{yz}}$, $\ce{d_{xy}}$, $\ce{d_{x_2 - y_2}}$)?
I couldn't find a definitive answer anywhere.

On p. 238 of my copy of McQuarrie (1983; his Table 6-6), he lists the following for the hydrogenlike atomic wave functions (expressed as real functions), describing these as "commonly used":
\begin{align*}
n=3, l = 2, m=0:     & \ \psi_{3d_{z^2}} \ ... \\
n=3, l = 2, m=\pm 1: & \ \psi_{3d_{xz }} \ ... \\
       : & \ \psi_{3d_{yz }} \ ... \\
n=3, l = 2, m=\pm 2: & \ \psi_{3d_{x^2-y^2 }} \ ... \\
       : & \ \psi_{3d_{xy }} \ ... \\
\end{align*}


Why is trichloromethane more acidic than trifluoromethane? In Concise Inorganic Chemistry by JD Lee (4th edition; adapted by Sudarshan Guha), page 78 under the topic "Back Bonding" it is given:

  $\ce{CHCl3}$ is more acidic than $\ce{CHF3}$ which is explained as follows:

  $$
\begin{align}
\ce{CHCl3 &<=> H+ + ^-CCl3}\\
\ce{CHF3 &<=> H+ + ^-CF3}
\end{align}
$$

  The lone pair on the $\ce{C}$ atom gets delocalised through 2pπ–3dπ bonding in $\ce{^-CCl3}$ which is not possible in the case of $\ce{^-CF3}.$

My doubt is, fluorine being the most electronegative element exerts a greater negative inductive effect (compared to that of chlorine) which stabilises the conjugate base formed. So I feel trifluoromethane must be more acidic than trichloromethane. But the reverse order is observed.

Please explain why is this so. Should we want to prefer the conclusion from back bonding over inductive effect?

We know that fluorine ($\ce{F}$) is more electronegative than chlorine ($\ce{Cl}$). Therefore, -I effect (Inductive Effect) of $\ce{F}$ would be more and trifluoromethane ($\ce{CHF3}$) would be more acidic.

But this is not the case. As -M (Mesomeric Effect) effect also plays an important role in determining the acidity. We also know that -M is considered over -I (Inductive Effect).

Now, If we compare the conjugate base of both the given molecules, $\ce{Cl3C-}$ and $\ce{F3C-}$, Chlorine has vacant $\mathrm{3d}$ orbitals whereas $\ce{F}$ does not have any vacant $\mathrm{d}$ orbitals (in fact, it does not even have $\mathrm{d}$-orbitals). Therefore, $\ce{Cl}$ can exert -M effect due to availability of vacant $\mathrm{d}$ orbital but $\ce{F}$ cannot exert the -M effect. Generally, -M effect stabilises the negative charge and increases acidic strength. Therefore, trichloromethane ($\ce{CHCl3}$) is more acidic than trifluoromethane ($\ce{CHF3}$).


Oxidation - Reduction Ionic Half Reaction Trying to figure out how to balance the following molecular equation:

$$\ce{KMnO4 + Na2C2O4 + H2SO4 -> MnSO4 + CO2 + Na2SO4 + H2O}$$

I have compiled all of their oxidation states

$$\ce{KMnO4 + Na2C2O4 + H2SO4 -> MnSO4 + CO2 + Na2SO4 + H2O}$$

$$\ce{(+1)(+7)(-2 * 4) (+1 * 2)(+3 * 2)(-2 * 4) -> (+2)(-6)(-2 * 4) + (+4)(-2 * 2) + (+1 * 2)(+6)(-2 * 4) + (+1 * 2)(-2)}$$

Now I know that theoretically Mn changes from +7 to +2, so it gains 5 electrons. I also know Carbon in this situation goes from a state of +3 to +4, so it loses one electron.

However, after this I am drawing blanks on how to setup my ionic half reactions with these polyatomic ions, and I am suspicious that I may have set this up incorrectly. Any help at all would be appreciated, I can not find much to explain this type of equation with this particular molecular formula.

The idea of Redox reactions make sense when talking about individual atoms or more clearly depicted changes, but I am having trouble figuring out what is even changing into what in this example.

Credits to my awesome teacher for this method.
(I had to add a potassium ion to the right hand side, there was no other way)


Is it possible to protect molten magnesium with Butane I am currently experimenting with simple methods of metal casting at home. It is conventional wisdom that magnesium is better left to the professionals due to the unfortunate arrangement of its combustion and melting temperatures. I dont have access to any inert gasses or vacuum furnaces; but like a challenge.

The first thing that occurred to me was to use CO2 cylinders you can buy at the supermarket to protect my casting area. Luckily I didnt try that experimentally, since it is a known bad idea. Same applies to water vapor. Molten magnesium will happily steal the oxygen from either, leaving carbon resp. hydrogen gas behind. And a lot of heat of course.

Which brought me to the following counter-intuitive idea: if magnesium cares so little for carbon or hydrogen, could you protect your magnesium from burning with hydrocarbon gas? If I douse my casting mould in butane, then pour my molten magnesium in, I am likely to set the butane ablaze, sure. But I can deal with that; as long as my magnesium stays put.

Is this the dumbest idea ever? Or might it actually work?

To make my question more concrete: what is the known chemistry of magnesium at 700c and pure hydrocarbons? Note that there will also be a lot of silicon dioxide present, which can form a thermite with magnesium, but normally shouldn't under these circumstances. But perhaps the hydrocarbons could catalyze that reaction?

if it might pull the oxygen out of sand , try using magnesium oxide, surely it won't pull the oxygen from its own self. Or using burnt lime, surely it won't pull oxygen away from calcium.
It might react with butane to form magnesium hydride and carbide ....which would be fun to play with . Also magnesium easily reacts with nitrogen to make the nitride...which is an excellent way to make ammonia


Why is the bond order of CO+ 3.5? According to my books the bond order of $\ce{CO+}$ is  $3.5$. But shouldn't it be $2.5$? On googling this, I found the following answer that is on Stack Exchange but its only talks about the bond length.

I am unable to understand why it is $3.5$ as I am in class 11.

For a long time it was taught in school and universities that the HOMO of carbon monoxide is anti-bonding. Without more context it was also often taught that the bond order in CO is three, since there are eight electrons in bonding orbitals and two in anti-bonding orbitals. $$\text{Bond order} = \frac12(\text{bonding} - \text{anti-bonding})$$
By assuming that the HOMO is anti-bonding (it is not!) and taking away one electron, the  bond order has to increase to 3.5. This is wrong.

When we have a look at the MO diagram, a calculated version can be found here, we know that the HOMO, i.e. 3σ, is a bonding orbital, while the anti-bonding orbital is the 2σ. Upon ionisation, we would indeed remove one bonding electron and therefore the bond order has to decrease to 2.5 as you suggested.
However, it is not that easy. Strictly speaking the below MO scheme is, as well as MO theory itself, an approximation, and only one possible configuration. While we do not have to use resonance structures with MO theory, we have to consider other configurations (analogous to excited states). So naturally the bond order of CO is not strictly 3. And removing an electron does not mean we are removing it from only one orbital, rather than decreasing the electron density. Therefore we cannot accurately predict the bond order with these simple considerations.
Experimental observations and theoretical calculations suggest that the bond indeed becomes stronger when removing an electron. See the linked question and Philipp's answer within for more detail. (Don't look at the other answers, they are as wrong as they could be.)

In short: The bond order of $\ce{CO}$ is not exactly 3 and removing an electron will not increase the bond order to 3.5. In both cases, the observed bond order is probably closer to 2.5, while experiments suggest that the bond is stronger in $\ce{CO+}$.

An orbital with bonding character has no node perpendicular to the bond axis; an orbital with anti-bonding character has at least one node perpendicular to the bond axis (electron density is zero). Strictly speaking there are no non-bonding orbitals.

It is $3.5$ due to synergic bond in $\ce{CO}$ which release a lot of energy hence which upgrade the energy of antibonding $2\sigma$, and when we change $\ce{CO}$ to $\ce{CO+}$ electrone is released from $2\sigma$ antibonding, resulting in bond order of $3.5$.


Le Châtelier’s Principle and Addition of Water With regards to gas volume, Le Châtelier’s principle predicts that an increase in the volume of the container causes the system to favor the side of the reaction that produces more moles of gas. However, would this principle apply equally well to an aqueous system, at equilibrium, that has now had water added to it? That would be analogous to expanding the volume of a gaseous system, wouldn’t it? Would adding water to an aqueous system at equilibrium affect the position of equilibrium at all?

Yes adding more water will affect the position of the equilibrium simply because you are adding more volume to the solution by adding more water. So adding volume is inversely proportional to pressure where the system needs to reduce the amount of water. Then the equilibrium will shift to the side with increased amount of pressure.


Why isn't dry CaCl₂ used to dry HCl gas instead of concentrated sulfuric acid? In our chemistry books, it is said that concentrated sulphuric acid is used to dry $\ce{HCl}$ gas.

But, concentrated sulphuric acid is a strong oxidising agent so, why isn't $\ce{HCl}$ gas oxidised to $\ce{H2O}$ and chlorine gas?

Why dry $\ce{CaCl2}$ (Calcium Chloride) isn't used as drying agent?

[Note: $\ce{CaCl2}$ have no reaction with $\ce{HCl}$ as they both have chloride ion in common]

There are several reasons:

Efficiency
When a stream of wet hydrogen chloride is run through a tube filled with coarse calcium chloride "rocks", interaction with the drying agent takes place at at its surface.
When the same wet $\ce{HCl}$ ist passed through sulfuric acid, the contact area is given by the surface area of the bubbles. Moreover, bubbles mean mixing. The trapped water is distributed over the drying agent.
Safety
Wet calcium chloride will clog, which might result in a pressure buildup in the drying tube.

This is because calcium is higher than hydrogen in the activity series of metals  due to which, the objective of the reaction will be destroyed. This is why concentrated sulphuric acid is used.


Adiabatic mixing of liquids One mole of benzene at $\pu{20^\circ C}$ is adiabatically mixed with 2 moles of toluene at $\pu{40^\circ C}$. The heat capacity of benzene is $\pu{84 J K-1 mol-1}$ and the heat capacity of toluene is $\pu{96 J K-1 mol-1}$. If benzene and toluene form an ideal solution, compute the entropy change for the process.

I tried to use

$$\Delta S = n C \ln{(T_\mathrm f/T_\mathrm i)}$$

where $C$ is the heat capacity and $T_\mathrm f$ and $T_\mathrm i$ are the final and initial temperatures of the liquids, respectively. However, I am unsure of what the final equilibrium temperature of the mixture comes to.

Let's say the final temperature is $T$. Also we assume the system is isolated and the process is adiabatic, as you said. Also all energy that is released by toluene or gained by benzene is heat.

The absolute amount of heat then released by toluene equals to the absolute amount of heat gained by benzene:

$|Q_\mathrm t| = |Q_\mathrm b|$

Considering signs, it will look like this:

$Q_\mathrm t = -Q_\mathrm b$

By definition, $Q = nc\Delta T$. Thus,

$Q_\mathrm t = 2\cdot96\cdot(40-T)$, and

$Q_\mathrm b = 1\cdot84\cdot(20-T)$

Plug in the latter two equations into the second one, and here's what you get:

$2\cdot96\cdot(40-T) = -1\cdot84\cdot(20-T)$

Now just solve it for $T$. And plug that in your equation as $T_\mathrm f$.

Note that it's not really important to do the temperature in kelvin, since we are dealing with changes in temperature. And yeap, your $T$ should be between $20\ \mathrm{^\circ C}$ and $40\ \mathrm{^\circ C}$, because there are such things as thermodynamics laws. So if it's not, you know for sure you did something not really correct.


What is meant by work done "expanding against atmosphere"? In the book, Chemistry : A molecular approach, I came across this line:

Over here what does the author mean by work done by expanding against atmosphere?

Imagine someone was pushing you against a wall. If you spend some energy overcoming that push, wouldn't you say that you did some work (expenditure of energy) against that pressure?

With a similar line of reasoning, the atmosphere is exerting a pressure of one atmosphere trying to compress your gas down. That's a huge pressure (nearly $10^5$ pascals!) and the gas in question has expanded against it. The wording makes sense.

In general, whenever gases are produced as a result of some reaction (in this case combustion), they are initially at a higher pressure than atmospheric pressure. This isn't an equilibrium situation. So, they expand, and lose their pressure until it finally becomes equal to that of the atmosphere.

It just means that when you burn natural gas, the reaction increases the volume of the matter involved in the reaction, so the system expands.  The author's wording is odd when they said it expands "against the atmosphere"; that just means the atmosphere is always there, exerting pressure, but the system expands anyway.


Why does the size of hybrid orbitals vary as sp³>sp²>sp? In my text book it said that

  Size of hybrid orbitals vary as sp³>sp²>sp

Does this size variation in hybid orbitals means that S orbitals are smaller than P orbitals?

But we know that larger the quantum number "n" is, the larger the orbital. I guess it has nothing to do with type of orbital.

What could be the reason for this variation in sizes?

Looking at the shapes of atomic orbitals shown in many textbooks can be deceptive. The probability density of the radial part of the wavefunction ($r^2|R_{n,l}(r)|^2$)  shows that the maximum probability for 2s orbitals is noticeably greater than for 2p, and for d orbitals the same applies with p greater than d.  (As the principle quantum number increases so does the position of the maximum probability density of an orbital, i.e. increasing $n$ increases the size of the orbital.)

The hybrid orbitals form from  the atomic orbitals and are added together in different proportions. So the position of maximum probability is a type of average value of the s and p. The sp hybrid has $2s \pm 2p$, the $sp^2$ in the z direction $2s+\sqrt{2}.2p$ and the $sp^3$ also in the z direction just as for the sp orbital. It seems to be that the maximum position of probability taken along, say, the z direction is the same in each case.

Simply as p character increases length of the hybridised orbital increases. As value of wavefunction of p orbitals is root 3 times of the wavefunction  of s orbital. What else you expect??


is water a non-metallic oxide? and hence acidic in nature? I just read about non-metallic oxides in my textbook which says that non-metallic oxides are a compound formed of an oxide ion and a non-metal. It also said that they are acidic in nature. It gave examples of carbon dioxide and sulfur dioxide.

But then wouldn't hydrogen oxide (water) also be a non-metallic oxide and hence, acidic in nature. But most textbooks say that water is neutral. So how's that possible?

When textbooks say that water is "neutral," textbooks mean that the water meets this condition:

$\ce{[H_3O^+]} = [\ce{HO^-}]$.

That's all.

Water can be neutral, in that its hydronium ion concentration equals its hydroxide ion concentration. Water can also be acidic if there is carbon dioxide dissolved in the water. Carbon dioxide can dissolve in water simply by letting the water come into contact with the atmosphere. Water can be basic too if you put a basic solute in water.

Water itself can also act as a weak Brønsted acid by donating a proton. This however is not a favorable process.

Water can also act as a Lewis acid because the oxygen in water withdraws electron density from the hydrogens, giving each hydrogen a partial positive charge. This in turns makes the hydrogens electrophilic - fitting the definition of Lewis acidity.

I think what your textbook is getting at is Lewis acidity. Consider carbon dioxide. This doesn't fit the Brønsted definition of acidity because it has no hydrogen protons to donate. So consider the partial charges on the carbon dioxide molecule. There is a positive partial charge on the carbon because the oxygens are more electronegative than carbon and have withdrawn electron density from the carbon. This makes the carbon electrophilic. And hence carbon dioxide as a whole can be considered a Lewis acid (but do understand that it's the carbon specifically in carbon dioxide acting as a Lewis acid).

So in general, your textbook says that non-metallic oxides tend to be acidic (in the Lewis sense) because the oxygens withdraw electron density from the non-metal (remember that oxygen is one of the most electronegative elements). This in turns creates partial positive charge(s) and makes the centers of these partial positive charges electrophilic.

Also, you could say that non-metallic oxides are precursors to Brønsted acids; consider this Lewis acid/base reaction. Remember that the central conceit of Lewis acid/base chemistry (coordination chemistry) is electron flow; nucleophiles attack electrophiles (not the other way around; that would be the fatal conceit).

So consider this reaction:

$\ce{SO_3 + H_2O -> H_2SO_4}$

We have sulfur with an electronegativity (EN) of 2.6. Oxygen has an electronegativity of 3.5. The difference is 0.9. The bond is clearly polar covalent. In addition, we have not one but three oxygens isolating electron density from the sulfur. Therefore, the sulfur most definitely has a partial positive charge and the oxygens partial negative charges. We say the oxygens are nucleophilic. And back to water, we have already concluded the hydrogen atoms in water are electrophilic due to their positive partial chage. The sulfur trioxide will attack the hydrogens and form the sulfuric acid molecule - a strong Brønsted acid (its first ionization is complete in water solution).

Hydroxonium(H3O+)=hydroxide(OH-) one acts as a strong acid and the other a strong base- the instant they come to be, we have water, which results in a neutral PH in isolation; the PH is altered, in solution, by changing relative concentrations of these ions caused by exposure to molecules such as atmospheric carbon dioxide(distilled water can absorb more CO2 and hence tends to be slightly acidic) which creates carbonic acid in solution with water. This equilibrium is also part of the fundamental nature of pure water.  H2O (the molecule) is an oxide of hydrogen mathematically. Water is bound H2O plus a relatively tiny amount ofhydrogen; hydroxide ions(oxides are  binary compounds of oxygen with another element).   Additionally "water", under various circumstances can have oxygen dissolved at varying levels(from very low dissolved oxygen (fish breathe) in the open underwater areas of the sea to relatively high oxygen in babbling brooks and rivers) and carbon dioxide (added by contact with air)- so there are many forms of "water" but "pure water in isolation" is without these common additions. Water is in a category of it's own. There is much yet to be learned about this wonderful substance, with all of it's unique properties, including long held beliefs regarding bonds and bond strength in my opinion, but that is another discussion. The bound molecule, H2O, has been categorized as a neutral, non-metal oxide albeit with significant unique aspects as water which have not been fully explored.   The molecule of H2O is existant, in our reality, in the form  of water.  Water is not H2O; H2O is not water.  H2O, should it be isolated, is not water and does not have the properties of water. Additionally, all H2O molecules are not the same, as a small percentage of H2O molecules, in water, are deuterium, as opposed to the aneutronic hydrogen(protium), which is far more common(with the definition that "pure water" is not "heavy water"(water with a high amount of deuterium)). Both protium (most H2O) and deuterium are stable non-radioactive isotopes of hydrogen.

Water Ph, in solution, changes up or down depending upon the relative concentrations of the two ions, hydroxonium (acid) and hydroxide(base). When the acid ions outnumber the base there is an acid and when the base ions outnumber the acid there is base. That is, when there are more hydroxonium ions than hydroxide ions, we have an acid and when there is more hydroxide we have a base- pure water has an equal number of these 2 ions and hence is neutral, by default. The confusing part of the process is the value of "neutral PH" which changes as the temperature of water changes, matching the value of the PH of water at a given temperature.  Water, on it's own, always maintains a perfectly neutral PH value, even upon change in temperature, albeit it water can react amphoterically- as an acid or a base when in solution with other substances.  The equilibrium point of water shifts when the temperature is raised (an effort, at reattaining balance, by effectively moving against change). Thus, water responds with an opposite reaction to the change in temperature. The counterbalance, to heating, is an effort to move in the direction of cooling. Heating pure water also causes both of water's ions to increase in numbers(equally); this causes a seeming decrease in the numerical value of the PH of water. However, this doesn't cause the water to become any more acidic as temperature increases. Water, in solution, is only acidic with an excess of hydrogen ions, as i have already said above. Pure H2O, is PH neutral, even if it's PH value changes due to the shifting of it's equilibrium point by increasing temperature because the new PH value instantly becomes the neutral PH value for water at that particular temperature.  As another way of looking at it, the neutral PH value of water at room temperature, PH 7, would actually be alkaline for water tested at higher temperatures. Effectively, the neutral PH value, of water, lowers, as water raises in temperature. The "real PH" effectively remains unchanged within the instantly stable equalibria of water at a given temperature. The same result is seen, but in the other direction, as the temperature of water is lowered from room temperature. The PH value of water rises as the temperature of water falls from room temperature but so to does the neutral PH value of water at the new temperature. The unaffected PH state of pure water, at any temperature, is a neutral PH- meaning that water is not acidic and not base in addition to meaning an equal number of hydroxonium and hydroxide ions.


Mechanism of acetylene formation from iodoform and silver My book mentions a synthetic route for turning iodoform into propyne.

That conversion was somewhat as follows:

  $$\ce{CHI3->[Ag] HC#CH ->[NaNH2] NaC#CH ->[CH3Br] CH3C#CH}$$

The first step is the conversion of iodoform into acetylene with silver. I am unsure how that reaction works. Could someone provide a mechanism?

What happens is that the C-I bond being weak, the iodide group is attacked by the silver atom (similar to what Na does in Wurtz reaction). As, there are 3 I-atoms and 2 molecules of iodoform usually react, 6 silver atoms are used up in the reaction and they precipitate out 6 molecules of AgI. The 2 carbon atoms form a triple bond to form acetylene.

EDIT- $$\ce{2CHI3 + 6Ag-> C2H2 + 6AgI}$$
$$\ce{C-I + 2Ag +I-C->I\bond{...}Ag + Ag\bond{...}I + C\bond{...}C}$$
As there are 3 C-I bonds, 6 Ag atoms form bonds with I-atoms, Hence, triple bond is formed between the 2 C-atoms. I hope that explains it.

It might be undergoing the same mechanism as the Wurtz reaction. Which is through free radical mechanism (Some say that Wurtz reaction also takes place through the ionic reactions).

Firstly the $\ce{C-I}$ bond breaks from both the iodoforms to form $\ce{CHI2}$ free radical
which causes them to fuse forming $\ce{I2HC-CHI2}$. The same step repeats causing further breakng of $\ce{C-I}$ bond leading to formation of alkenes and alkynes later.


What defines an element's taste? A useful post by @Martin indicated that probably the naming of Sweetwater town is because of the sweet tasting lead compounds in it's water.
Then my question arose. I know that the taste of any material is dependent on which tongue buds it provokes. Now, my question is, which chemical property of an element defines or is directly related or is causing the chemical property of taste? (I deliberately avoid asking this for all chemical species to prevent very broad answers, if you think you can give not-too-long answers, my question is about all chemical species.)
Note that by a responsible chemical property I mean things related specifically to an atom, like e.config.

TL;DR: Don't taste pure elements. They either taste of nothing, taste foul, or kill you. Or all of the above.

Edit: Also, for clarification, elements don't have a taste. Taste is a biological reaction to chemical interactions happening with an element, not a property of the element itself. So, like everything else in chemistry, the taste is decided by the electron configuration of the element.

Taste as we know it happens in chemical reactions between our taste buds and the chemical species in question. Most of the taste is actually governed by the nose, where very specific receptors interact with very specific chemicals to send very specific signals to the brain, the chemistry of which is, while not trivial, also not interesting. It's your basic, run of the mill presumptive test.

In the taste buds, however, the receptors look for certain specific characteristics of the species being tasted. As an example, acids are sour, indicating that our sour taste buds are actually pH meters (although there might be more to it than this).

Similarly, most sugars have similar characteristics, that can be detected by a relatively simple chemical detector.

If, however, you go around tasting elements, you'll find two very interesting things.

First, you'll find that you're dying from all kinds of poisonings. At a rough count, all of the periodic table is poisonous in its pure form, with only a few exceptions (carbon, oxygen, hydrogen, nitrogen, silicon, titanium, noble gases, gold and platinum, possibly a few others).

Second, you'll find that the pure elements don't taste much at all, if anything. Taste is a completely biological concept, and these elements simply aren't found in nature, meaning that the tongue and nose have no frame of reference.

Most metals will taste metallic, due to tarnishes forming on their surface, mostly oxides, which all have the familiar, metallic taste. The non-metals are mostly fairly reactive, and taste absolutely foul as they react to the air in your moist mouth.

Gases like chlorine and fluorine may react and form chloride and fluoride, which taste sour, and then dissolve your mouth and nose, and then kill you.

The more radioactive metals will seem to taste of blood, but that's actually just acute radiation poisoning setting in, actually filling your mouth with blood. If they didn't kill you, they'd probably taste metallic like the rest of the metals, but they do, so they don't.

Finally, the alkali metals like lithium and its friends will taste bitter, also kill you.

"But wait!", I hear you scream, "What of, say, lithium in pharmaceuticals?"

When elements like lithium are ingested, they are done so in a rather more controlled fashion. You don't get a brick of pure, elemental lithium and told to go crazy when you go crazy, but rather, delivered in pills, in conjunction with a delivery agent, in this case, lithium citrate.

Most elements have some biological role or other, but if they are in the kind of quantity you could taste, they are most likely poisonous, and in many cases, fatally so.

I don't know how many variables determine the taste but we recently found out that materials can have completely different properties just by the amount. We know that with different structures of atoms materials get completely different like diamond and graphit but, we know at least since the discovery of graphene that the periodic table might be much more complicated than we had imagined.

Every element gets completely different properties with just the quantity of atoms. If you take for example a really small amount of an element with a spoon it glows yellow, then you take with an other spoon a bigger amount and it glow red, then you take a bigger amount it glows blue. Currently it doesn't make any sense why this happens since you don't change the atomic structure of the material.

In the future the periodic table will get a 3rd dimension with every element becoming different with just the amount. This is not about the structure of the atoms or something, just the amount.

For the colors of matrials we have the wave lenght of the light as a reference but for the taste there are to much variables since we taste with our nose and our tongue.

http://periodictable.com/Posters/PlaceMatBackSide3000.JPG

http://periodictable.com/Posters/LetterSizeBackSide1500.JPG


What is the difference between configuration and conformation in stereochemistry? I understand conformation to be any spacial arrangement of atoms that can be achieved through rotation of a single bond. I understand configuration to be the disposition of atoms/groups in space. However, would a conformational isomer also be a configurational isomer since the groups are arranged differently in space?

I recommend you this webpage: http://worldofbiochemistry.blogspot.com/2012/09/conformation-vs-configuration.html.
Configuration and conformation are two concepts are often used interchangeably, but represent very different things. The conformation regards the relative spatial orientation of a portion of a molecule relative to another. Thus, it is an aspect that is not directly related to the covalent bonds that are established within the molecule, but with their possible rotation.

When we talk about rotation around covalent bonds, we are only referring to the single bonds, as they are the only ones that can suffer rotation. Basically this concept is easily understood if we think that the bounds work as an axis...

It should be noted that when we speak of different conformations, it does not necessarily involve all the covalent bonds of a molecule, it can account only for one or few of them.
 Taken together, it is possible to convert one conformation to another without cleaving or forming chemical bonds, simply by rotating some simple covalent bonds. Thermal energy at room temperature is sufficient to rotate some simple covalent bonds.

The configuration is a concept that is related to the order by which different substituents linked to the same central atom establish covalent bonds. That means, in this case it is clearly an aspect that is a direct consequence of the covalent backbone of molecules.

To change the configuration, you must always cleave and form new covalent bonds...This costs a lot of energy!!

In conclusion, the concept conformation encompasses portions of a molecule which are not directly linked to the same atom and do not involve the covalent backbone of the molecules, while the configuration comprehends parts of the molecule which are bound to the same atom, which means that there is a direct involvement of the covalent bounds of the molecule.

Configuration differs from conformation by a fact that conformations are only due to the orientation of the atoms in molecule around the covalent bond considering the bond as an axis.  And changing the configuration causes the cleavage and formation of new chemical bonds.


Measuring Colloidal copper PPM & how to ignore the salt PPM in soluton I made some colloidal copper with 9V battery. To make this water I had to add small amount of salt. However, the salt itself shows up in PPM count of my PPM reader. So how can I discount the salt PPM and come up with correct copper PPM in distilled water.

First method is to look for Faraday's law of electrolysis and put in the equation your data.

Second method is to add to TDS reading the initial reading of the salt solution before the electrolysis.


Equation of water at triple point I was thinking about what exactly happens at the triple point of water, and what could be the effect of adding solutes/solvents. I feel we can represent the triple point using the following equilibrium reaction:
$$\ce{H2O (s) <=> H2O (l) <=> H2O (g)}$$

Is this correct? How would this change if I add solutes?

Note: I know that adding solutes would increase the concentration of the solution, but how exactly will it shift the equilibrium?

At the triple point the water system is assumed to be adiabatic. The system has 3 phases, ice from pure water, liquid water, and pure water vapor. The various equilibria  are shown below.

Note that the ice floats on the water. So the ice is in contact with both the liquid phase and the gas phase. Furthermore The ice can't cover the whole surface like a frozen over pond. There must be some liquid water in contact with the gas phase too.
A subtlety here... You can't make a solid phase of ice that is homogeneously say 1 molar NaCl. If you take a 1 molar NaCl and try to freeze it you get relatively pure water first, then saltier and saltier water freezing as the solid phase gets colder and colder. Since the impure ice has doesn't have a homogeneous composition, it doesn't have a specific melting point.
So for the system you proposed there is pure water ice,  a 1 molar NaCl solution, and a gas phase of water vapor. Unfortunately humans can't make a perfectly adiabatic container, so as the ice melts, the 1 molar NaCl solution becomes more dilute. Thus there is no specific triple point.
In other words... if you just throw some ice into the 1 molar NaCl solution, the final temperature of the theoretical system will depend on the relative amounts of the three phases - How much 1 molar NaCl solution, how much ice, and the volume of the gas phase.

The triple point of water is an example of Thermodynamic Equilibrium, which is explained by Wikipedia as:

  simultaneously in mutual thermal, mechanical, chemical, and radiative equilibria

Your equation only covers the chemical facet of this equilibrium and hence is not the correct way of representing total thermodynamic equilibrium. I am also not sure of the correct way of representing these kinds of equilibria which deal with both energy and physical states.

If this equilibrium is disturbed chemically i.e. by adding more water, according to Le Chatelier's principle, the equilibrium will shift towards the solid and gaseous states. As the enthalpy of vapourization is more than enthalpy of fusion, more ice will be formed as compared to the amount of steam evolved in order to balance the energies involved. Hence, it would tend to shift to the left.


What causes incomplete combustion? During the combustion of hydrocarbons, there is a difference between the amounts of carbon or hydrogen that results in incomplete or complete combustion of the material.

My question is, besides from an insufficient amount of oxygen, what can cause incomplete combustion?

What are the physical or chemical properties of the hydrocarbon responsible for incomplete combustion, such as the hydrogen to carbon ratio (mass, volume, number of atoms, etc.) or the saturation.

Please keep the reply suitable for a student studying high school to early university chemistry.

According to the Iowa State University page Carbon Monoxide Poisoning: Checking for Complete Combustion (AEN-175) suggest the following mechanisms that cause incomplete combustion:

  Insufficient mixing of air and fuel.
  Insufficient air supply to the flame.
  Insufficient time to burn.
  Cooling of the flame temperature before combustion is complete.

The main notion is the lack of air to cause complete combustion (either through lack of supply, mixing or time for combustion).

However, it should be noted that, practically speaking, according to the ChemWiki page Burning Alkanes, the size of the hydrocarbon has an affect in 'normal' conditions:

  Provided the combustion is complete, all the hydrocarbons will burn with a blue flame. However, combustion tends to be less complete as the number of carbon atoms in the molecules rises. That means that the bigger the hydrocarbon, the more likely you are to get a yellow, smoky flame.

The reason being (according to ChemWiki) is:

  If the liquid is not very volatile, only those molecules on the surface can react with the oxygen. Bigger molecules have greater Van der Waals attractions which makes it more difficult for them to break away from their neighbors and turn to a gas.

So, the larger the hydrocarbon, the less likely it would vaporise sufficiently, thus less chance of the optimal mixing of air and fuel, resulting in incomplete combustion.

However, provide the right conditions with sufficient supply of air and optimal hydrocarbon-air mixing ratio, it is still possible to cause complete combustion, in the general formula (for many hydrocarbons):

$$\ce{C_{x}H_{y} +O2 ->H2O + CO2}$$

A quite comprehensive resource explaining each of the factors is the University of Tulsa chapter Fuels and Combustion.

a particular flame temperature is recommend for each types of sample, if your sample required higher temperature to vaporize and given temperature is not the required one, incomplete combustion may occur


Why does ice water get colder when salt is added? It is well known that when you add salt to ice, the ice not only melts but will actually get colder. From chemistry books, I've learned that salt will lower the freezing point of water. But I’m a little confused as to why it results in a drop in temperature instead of just ending up with water at 0 °C.

What is occurring when salt melts the ice to make the temperature lower?

When you add salt to an ice cube, you end up with an ice cube whose temperature is above its melting point.

This ice cube will do what any ice cube above its melting point will do: it will melt. As it melts, it cools down, since energy is being used to break bonds in the solid state.

(Note that the above point can be confusing if you're new to thinking about phase transitions. An ice cube melting will take up energy, while an ice cube freezing will give off energy. I like to think of it in terms of Le Chatelier's principle: if you need to lower the temperature to freeze an ice cube, this means that the water gives off heat as it freezes.)

The cooling you get, therefore, comes from the fact that some of the bonds in the ice are broken to form water, taking energy with them. The loss of energy from the ice cube is what causes it to cool.

No. Unfortunately we seem to all be forgetting a fundamental property of ice - I kicked myself when I realized it. Think of a large chunk of ice taken out of your home freezer, sitting in a bucket, floating in its own melt-water. All the liquid water surrounding that ice is at $\pu{0 ^\circ C}$. The extreme surface of that ice, that is exposed to the meltwater, (the layer that is about to transition to liquid) is also at $\pu{0 ^\circ C}$.  BUT the solid (non-transitioning) interior of the ice chunk is still at about the temperature of the freezer that it came from, about $\pu{-18 ^\circ C}$! So we shouldn’t think of all the water in the bucket (liquid & solid) as being at $\pu{0 ^\circ C}$. Instead, most of the frozen water is much colder than $\pu{0 ^\circ C}$. So now we see that within the chunk of ice there is a significant temperature gradient. The exact center is coldest (at perhaps $\pu{-18 ^\circ C}$). Moving out from there the temperature increases until you reach the surface where the ice has warmed to a temperature of $\pu{0 ^\circ C}$ and has begun melting. Now we can readily see why adding salt to the liquid surrounding the ice chunk would cause a lowering of the temperature of that salty melt-water. By adding salt you have lowered the melting point. The surface layer of the ice that was $\pu{0 ^\circ C}$ would rapidly melt because the salt is in close contact with it and acts on it such that the ice surface is now much too warm to maintain the solid state. Underneath that quickly melting ice layer is another layer at say $\pu{-1 ^\circ C}$. The salt would be able to act on that colder ice in turn and cause it also to melt while maintaining that negative temperature creating a small amount of $\pu{-1 ^\circ C}$ briny water. This below-zero melting action caused by the briny liquid would continue toward a new temperature equilibrium (between the solid surface and the liquid) somewhere below $\pu{0 ^\circ C}$ in accordance with the salinity level of the melt-water.


Relation between activation energy and rate constant Consider two types of reaction. One has more activation than other. If we increasing temperature by same amount in both, rate constant of which reaction will increase more?

My attempt: as we know that $k = A\mathrm e^{(-E/(RT))},$ if temperature is same in both, then the one having more activation energy will have lesser rate constant. Am I correct?

This can be answered both conceptually and mathematically.

Conceptually: Let's call the two reactions 1 and 2 with reaction 1 having the larger activation energy. Here is a plot of the arbitrary reactions.

It is clear from this graph that it is "easier" to get over the potential barrier (activation energy) for reaction 2. Thus if we increase temperature, the reaction would get faster for reaction 2 before it gets faster for reaction 1. Therefore, when increasing the temperature by the same amount, the rate constant for the reaction with the lower activation energy will increase more.

Mathematically: You can understand how the rate constant changes with temperature by simply graphing. Here is a very quick and very rough graph on how the rate constant changes with temperature (plotting the equation $k=Ae^{-E/RT}$).

Notice how increasing the temperature causes a quicker increase in the reaction with the lower activation energy $(E_{a2})$. However, this is only an a priori expectation.

You mentioned in a comment above about differentiating. You can use that method as well. Taking the derivative of the equation for $k$ with respect to temperature while assuming $A$ and $E_a$ do not depend on temperature yields

$$\frac{dk}{dT} = \frac{AE_a}{RT^2}e^{-E_a/RT}$$

Edit: As porphyrin pointed out in the comments below, this equation elucidates the true behavior of the temperature dependence. Plotting this function for the two energies, we see that the derivative for the reaction with a smaller activation energy increases faster for lower temperatures. But for higher temperatures, there is a shift in the change of the rate constant. Therefore, the process is temperature dependent. At low temperatures, the rate constant for the reaction with low activation energy changes is more sensitive to temperature change while at high temperatures, the reaction with larger activation energy is more sensitive to temperature change.

As a side note: at very large (infinite) temperatures, both functions approach the same value and there will be an equivalent distribution among the reactions' products.

From comments below: Prove that if we increase temperature in an endothermic reaction, equilibrium proceeds in forward direction.

There are a couple ways to do this. As this is now an equilibrium question, I will keep the regime to equilibrium instead of rate constants. First, let's use fundamental chemistry to get an expected result then use equations to show what the result would be quantitatively.

Fundamentals: Say you have a simple reversible reaction
$$A\rightleftharpoons B$$
Given this reaction is endothermic, energy is required to drive the reaction. Because of this, we can write energy as a "reactant"

$$A + Energy \rightleftharpoons B$$

According to Le'Chatelier, if you increase temperature then the equilibrium must shift towards the products, thereby answering the question.

The Maths: Now, let's apply some math as overkill. The equilibrium constant is dependent on the standard free energy of the reaction given by the following equation

$$\Delta G^o = -RT\ln(K)$$

If we want to understand how the equilibrium constant will change we can rearrange this equation to solve for $K$ and plot as a function of temperature. Doing this and making the substitution for the free energy $\Delta G^o = \Delta H^o - T\Delta S^o$ gives

$$K = e^{-\frac{\Delta H^o}{RT} + \frac{\Delta S^o}{R}}$$

Since the reaction is endothermic, $\Delta H^o > 0$. This gives the same monotonically increasing profile that we see for the Arrhenius equation (plotted above). Therefore $K$ must always increase, and therefore the reaction proceeds in the forward direction, for increasing temperature. Of course, this is making the assumption that $\Delta H^o$ does not change with temperature.

The activation energy is the kinetic energy required from the reactants to form the products.Now lets go to your second part.Since the first reaction has a bigger activation energy than the second reaction , if we add energy to the system the first reaction has better chances of happening than the second  reaction because more molecules from the first reaction will be converted in products than in the second case.To actually understand it you have to take quantum mechanics of every molecule in both cases and you will  see it.


Is melting/boiling point of ionically bonded substance higher than of covalently bound? Is the melting and boiling point of ionic bond usually higher than covalent bond?

I know that compounds with ionic bonds are usually solid at room temperature, so I want other answers than this. (and this question can also be about why compounds with ionic bonds are usually solid in room temperature.)

The answer relates to the strength of the interactions between the component units that make up a crystal or a solid.

The reason why anything is a solid at a given temperature is, crudely, that the interactions between the units that make up the solid (atoms, ions or molecules) are stronger than the amount of thermal energy available at that temperature.

Diamond, for example, is a solid because each unit carbon is bonded to its neighbours by a strong carbon-carbon bond that takes a great deal of energy to break. Non-polar molecular compounds, like candle wax (in reality a mixture but imagine a pure long chain hydrocarbon to keep it simple) are held together by the intermolecular forces between the molecules (often called dispersion or Van der Waals forces). These are much weaker than chemical bonds and depend on the surface area of the molecules. So small non-polar or not-very polar molecules tend to have low melting points. Hexane is a volatile liquid. The shape and symmetry of the molecule matters a bit to these interactions so toluene is a liquid down to -95°C but benzene, which is very similar to toluene but far more symmetric and slightly flatter only melts at about 6°C as the molecules pack together better encouraging stronger interactions. Polar molecules have stronger interactions because the molecular dipoles interact and this produces stronger forces than the non-polar dispersion forces. So phenol (very similar in size to toluene but far more polar) melts at about 41°C.

Simple ionic solid are made from charged ions. Table salt crystals consist of Na+ and Cl- ions. The key interaction between these is electrostatic and this is a strong force that takes a lot of energy to break. The melting point is about 800°C. Permanent electrostatic interactions are strong especially if the components can get close (remember, it's an inverse square law). This explains why not every ionic compound is a solid. There are ionic compounds that are liquid at room temperature. These tend to consist of molecular ions that are large and sometime irregular so forcing structures where the ionic force has to act over much larger distances and, therefore, be much weaker. An example is [BMIM]PF6 which has a melting point well below room temperature:

These ionic liquids are an interesting area of research in chemistry as they sometimes enable much less wasteful chemical reactions in industry (see wikipedia).

In the case of ionic compounds, there is a complete transfer of electrons from one atom of the element to another atom of the another element. So, there exists a strong electrostatic force.

In the case of covalent compounds, electrons are not completely dragged, the atoms of different elements are attracted by a weaker electrostatic force. Thus, ionic compounds have atoms bound by stronger electrostatic force than in case of covalent compounds.

So, in general it is easier to break covalent bonds than ionic bonds. Thus, the melting and boiling points of ionic compounds are greater than covalent compounds.


How to find the number of valence electrons? I need to understand the following:

Considering an element Sulfur - S which has 16 electrons.

How do we calculate the number of valence electrons of S?

Please correct me if I am wrong:2+2+6+2+4=16. So, the valence electron = 4+2=6

Is it that always valence electron is the addition of the last two shells? In that case for Na, sodium, if we take 11 and arrange it as 2+2+6+1, then the valence electron is 1 or 6+1?

Now, considering the fact valence electron is 1, if we go by, up down up down rule, then as it is 1, hence the spin is up?

For 6: is it up(1) down(2) up(3) down(4) up(5) down(6) as it is both up/down hence it would be +-1/2 spin?

For Phosphorous 2,8,5, what would be the valence electron? 5 or 13?

If 5 then up down up down up, will it have an up spin?

Is it that for even numbers like 4,6 the spin will be both up/down i.e. +-1/2?

'p' and 'd'.

There is one type of 's' orbital. It is spherical. So an s orbital can hold 2 electrons only.
There are three types of 'p' orbitals. They are the same shape (like a dumbell) but are oriented in different directions - along the x, y and z axes.  So a 'p' orbital can hold $2$x$3$ = 6 electrons.
There are five types of 'd' orbitals (10 electrons) and seven types of 'f' orbitals (14 electrons).

Fill up the orbitals in the following order (following the red arrows), which is increasing order of energy. Lowest energy fills up first.

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p

So let's take the example of sulfur.
The total number of electrons is 16. We'll distribute them in the following manner.

$1s^2, 2s^2, 2p^6, 3s^2, 3p^4$, where the superscripts of each orbital represent the number of electrons in it. Now, we have a total of 6 electrons in the last (3rd) orbital. Therefore the valence number is 6.

It is not simply the last 2 numbers while writing the electronic configuration.

For Phosphorous, we have:
$1s^2, 2s^2, 2p^6, 3s^2, 3p^3$, which gives us 5 electrons in outermost (3rd) orbit.

For Lithium (3 electrons), we have:
$1s^2, 2s^1$, which gives us a total of 3. Outermost orbit is the 2nd orbit and the total number of electrons in it is just one. Therefore valency of Lithium is one.

However, it is much simpler to do it the way I've shown above as with Sodium. Don't need to bring in all the orbitals

For the spin, just remember that you have to fill up the entire orbital with up spin electrons before you start putting the down spin electrons. Don't just alternate up and down.

Let's talk about the 'p' orbital. Each 'type' of the p-orbital must have up-spin electrons before adding the down-spin. So there will be 3 up spin electrons before you start adding the down-spin ones. Also, you can not move on to the next orbital without completely filling the previous one with an EQUAL number of up and down spins.

For Phosphorous; let's say we've filled the 1s, 2s, and 2p orbitals. Now we add one up spin to the 3s, followed by a down-spin. Which leaves us with 3 more electrons to be added to the valence orbit (3rd). It is a 'p'-orbital, which needs to have 3 up spin electrons before any downspin ones can be accommodated. But we have only 3 remaining! So all will be up-spin electrons in the valence orbit. So we'll have a total spin of +3/2.

Hope I didn't complicate too much, but I've actually tried to make it as short as possible. I've still left out quite a bit.

It is not so difficult I could not understand your method of finding valence electrons, easiest way I could explain is try to find the atomic number of the element. For example chlorine it's atomic number is 17 now find it's electronic configuration which is 2,8,7. Now we know that there are 3 shells in it hence third shell would be it's last shell and valence electrons are the electrons of valence shell therefore valence electrons would be 7.


What is Resonance, exactly? I know the that it(Resonance) is the delocalisation of pi electrons but how and why are those electrons delocalised? What is the driving force that causes the effect?

Resonance is a conceptual tool to help make sense of the real observed structures of molecules
Resonance is not a real physical phenomenon. That is the first important thing to realise. It is merely a tool to bridge the simple view where all bonds consist of integral numbers of electrons and the real structures of molecules.
The simple view of bonds often fails to represent the "real" observed structure of molecules. The archetypal example is benzene. A simple view of the bonding of benzene has the molecule consisting of alternating single and double bonds (which might be called cyclohexatriene). This theoretical molecule has the bonds neatly grouped into simple bonds that use up all the available electrons into either single bonds (one pairs of electrons) and double bonds (two pairs of electrons). But this would imply that the bond lengths alternate. But we know the structure of benzene and the bond lengths are all the same. We can explain this with molecular orbital theory when we recognise that some of the bonds are delocalised, but that gets fairly sophisticated fairly quickly and chemists prefer easier ways to keep track of the electron counts when writing structures on paper and explaining reaction mechanisms.
One halfway house that saves the ability to write structures where simple electron counts are possible is the idea of "resonance". Instead of needing to invoke molecular orbital theory all the time we can write the structure of benzene as a combination of two structures where the double and single bonds alternate. We pretend that the real structure is a "resonance hybrid" of the two hypothetical structures. This is like the mathematical average of the two competing structures.
More complex molecules can also be represented as combinations of (more complex) resonance structures. We often say that the structure consists of resonance hybrids. But resonance is not a real physical phenomenon. The different structures are not real independent structures that rapidly interchange: they are merely a convenient model so chemists can talk about electron counts and draw structures without having to invoke much more complex bonding theories.
So there is no such thing as "resonance" as a physical phenomenon. What drives the structures of molecules is the way orbitals in atoms combine together in molecules and that is more complicated than simple views of bonding can represent. The underlying driving force is quantum mechanics and energy minimisation of the possible ways atomic orbitals interact in molecules. And that is sometimes far more complicated than simple views of bonding can easily represent.

Why do electrons delocalise?
Delocalising is a stabilizing force in a way such that when the electron has a greater number of orbitals to move into the energy spreads over a larger area than in a single orbital as the distribution of energy in multiple orbitals brings extra stability in the system.
There is also a possibility that after energy gets distributed there is a lowering down of the total internal energy of the molecule had there been a single orbital holding the energy ie. if there were no delocalisation.
Have a look at this one.
Why does delocalization of π electrons bring stability?


Why lone pair occupies more space than bond pair? I'm not a chemistry expert and merely a high school student. In my book there is a slight introduction to VSEPR theory and therein they mention that lone pair-lone pair repulsions are maximum, followed by lone pair-bond pair and then bond pair-bond pair repulsions.

Now the reasoning they provide to this is that as lone pairs are localized on the central atom, and each bond pair is shared, the lone pair occupies more space and hence the repulsions are greater.

Now I don't understand how localisation of lone pair leads to its large size and secondly what is meant by"size"here, because it's hard to imagine size of a lone pair for me.

It's a request not to mention about advanced theories like molecular orbital theory, because I'm not familiar with aaj that right now.

EDIT
I would say that the other is also in a way similar but I could not find an answer of my question there. That's why I posted it.

Any of your help is appreciated.

Thanks.

The basic premise of this statement is that the repulsion exerted by an electron system depends upon the area it occupies. More the area, more is the repulsion.

As the bond pair is formed from the sharing of electrons, it experiences attraction from not one, but two nucleii. On the other hand, the lone pair is attracted by only one nucleus.

So, the bond pair is more localised while the lone pair is spread out and occupies a larger area.

The 'size' here refers not to the size of electrons but to the size and nature of the orbitals these electrons occupy.


Fehling's solution turning yellow and then black? While testing for formaldehyde (an aldehyde), I added Fehling's solution to the sample and let the test tube simmer in a water bath for some time. I returned ~15 minutes later to see a black/dark grey precipitate awaiting me, which looked very metallic (Almost like Tollen's test, but darker)

I try again and this time heat the tube for only five minutes: I obtain a buff yellow coloured precipitate which, on standing, gives me the same metallic precipitate I observed earlier.

I expected that Fehling's solution will give a red precipitate of $\ce{Cu2O}$ but it looks like the formaldehyde seems to be oxidizing it all the way down to metallic copper. Also, what is the reason for a yellow precipitate to form rather than a red one?

I could not take any pictures: We're not allowed to carry phones/cameras into the lab.

In the reaction with aldehydes, Fehling's solution produces $\ce{Cu2O}$ as an orange or yellow precipitate. I don't know why the color is not always the same. The oxidation of an aldehyde produces an acid. And usually an acid cannot be more oxidized. There is one exception : methanal $\ce{HCHO}$. Because it can be oxidized in formic acid $\ce{HCOOH}$, and then into $\ce{CO2}$ . That is why Fehling's reagent produce metallic copper $\ce{Cu}$, $\ce{CO2}$ and no acid when reacting with $\ce{HCHO}$.


Do I have to systematically show the protons in my reaction equation to find the Pourbaix diagram's line equations? I was trying to get the line equation for the Pourbaix diagram for the couple $\ce{Ni(OH)2}/\ce{Ni}.$ I can write two chemical equations for basic and acidic conditions, respectively:
$$
\begin{align}
\ce{Ni(OH)2 + 2 e- &-> Ni + 2 OH-}\tag{R1}\\
\ce{Ni(OH)2 + 2 H+ + 2e- &-> Ni + 2 H2O}\tag{R2}
\end{align}
$$
Now I can determine the line equation for both chemical equations using Nernst equation:
$$
\begin{align}
E^\circ + \frac{0.06}{2}\log\left([\ce{H+}]^2\right) &= E^\circ - 0.06\,\mathrm{pH};\tag{1}\\
E^\circ - \frac{0.06}{2}\log\left([\ce{OH-}]^2\right) &= E^\circ + 0.06\,\mathrm{pOH}.\tag{2}
\end{align}
$$
I need to make the $\mathrm{pH}$ appear in the previous equation. To do so, I use the relation
$$\mathrm{pH} + \mathrm{pOH} = 14 \quad\Leftrightarrow\quad \mathrm{pOH} = 14 - \mathrm{pH},\tag{3}$$
which gives
$$E^\circ + 0.06(14 - \mathrm{pH}) = E^\circ + 0.84 - 0.06\,\mathrm{pH}.\tag{4}$$
I've found two slightly different expressions depending on whether I choose basic or acidic conditions. I've spend quite a few hours trying to figured out what is wrong here, and I've found a PDF Diagrammes E-pH = Diagrammes de Pourbaix (courtesy of Académie de Clermont-Ferrand) stating on page 4 that even if the species only exists in basic conditions I need to make the $\ce{H+}$ appear in the half-reaction.
Is there a convention when writing chemical equation prior to the determination of the line's equation for Pourbaix diagram? Or did I make a mistake somewhere? If yes, could help me please?

the standard potentials are given in the literature generally in an acidic medium at pH = 0. In this case, we must therefore write the equations in an acidic medium with $ H ^ + $ and not with $ HO ^ - $


Hybridisation of Mn in potassium permanganate I'm clear with the concepts of crystal field theory. But I can't figure out the exact reason why the hybridisation of manganese in potassium permanganate ($\ce{KMnO4}$) is $\mathrm{d^3s}$. Can anyone explain please?

This is formally a manganese(VII) compound and hence there are no 3d electrons.

The four $\ce{O^2-}$ ions are considered to be donating two electrons each to the atomic orbitals. Tetrahedral "hybridisation" can be achieved by using the $\mathrm{4s},~\mathrm{3d}_{xy},~\mathrm{3d}_{yz}$, and $\mathrm{3d}_{xz}$ AOs. Since in Mn(VII) the $\mathrm{3d}$ AOs are lower in energy than the $\mathrm{4s}$ and $\mathrm{4p}$ AOs, the "hybridisation" is best considered as $\mathrm{d^3s}$.

I have put "hybridisation" in quotes because it is not used in advanced chemistry (it is an artifact!) but it is correct in saying that the bonding MOs contain mostly $\mathrm{4s},~\mathrm{3d}_{xy},~\mathrm{3d}_{yz}$ character but they'll also probably mix some $\mathrm{4p}$ character in there for good measure!

yes because there is d3s hybridisation in this complex
and in d3s hybridisation only dxy, dyz and dzx atomic orbitals can only participates in this hybridisation so only three AO's out of five are utilised in this
and according to CFT energy difference between t2g and next s orbital is less so hybridisation can be easily occurs


Un-denaturing industrial alcohol I have a question regarding the denaturation of industrial ethanol with methanol to discourage its consumption.

The boiling point of methanol is 64.7 °C, while the boiling point of ethanol is 78.37 °C. Wouldn't it be trivial then, to separate the methanol by simply distilling industrial ethanol at 64.7 °C ?

Is there some technical reason as to why this wouldn't work? Because if it does work, then why isn't it a widespread practice among criminal elements and such.

(I'm no chemist, but here's my take on it)

Let's see: I’m an alcoholic, and I really hate having to dole out large amounts of cash to purchase your usual "drinking"-ethanol.

So I get this amazing idea to separate the ethanol by distillation (methanol has a lower boiling point than methanol, so what I'll really be doing is distilling out the methanol, leaving behind ethanol and water). Pretty smart, right (for someone who's still inebriated)? Thus invigorated, I sally forth... and return with a dozen gallons  of industrial (denatured) ethanol. However, as I slowly get sober, I begin to see two major obstacles to my plan:

It’s going to be very, very difficult to maintain a constant temperature of exactly 64.7 °C (boiling point of methanol). Now, even though ethanol boils at around 78 °C, trying to maintain the denatured alcohol at under (say) 70 or 75°C isn’t really an option either. Remember, what I’m trying to do here is criminal (where I'm from at least), plus I don’t really have a Nurdrage-worthy laboratory. If say, I were to simply buy a large metal vat with a distillation apparatus clamped on the top and light a fire below, at such a large scale (a dozen gallons in my case) and with the non-professional equipment I use, it’ll be impossible to ensure that the vessel is uniformly heated at a specific temperature. So what’ll happen is, I’ll end up distilling out a mixture of ethanol and methanol, and the ethanol that remains will also have significant amounts of methanol in it (10ml of methanol's enough to make you go blind… just to give you an idea about what I mean by “significant”). Regardless, I wouldn't want to engage in something that involves producing methanol vapors (be it in a proper laboratory or otherwise)

Say I did manage to get hold of the required laboratory apparatus (I, uh, happen to know this cancer-stricken, bald, high-school Chemistry teacher who was willing to lend me them), and I were to distill the alcohol in small batches to ensure I evenly heat the solution at the desired temperature, but the whole process would be so time-consuming and labor intensive that it would probably just be better if I just went and bought "drinking"-ethanol in the first place.
(Should have mentioned this first, but since you were talking about methanol, ah well…) To begin with, I probably wouldn't know what the manufacturers actually used to denature the alcohol. They could have used methanol, pyridine, or some other foul-tasting, foul-smelling additive for all I know, or maybe even mixed all of them together. So trying to get ethanol out of that mess, is going to be a HUGE pain in the neck.

Also you might want to read up on denatured alcohol here.

I wasn't able to find anything on ethanol-methanol azeotropes in this Wiki page, so if anyone knows where I can find something on it, gimme a shout! :-)

Distilling your own spirits is fine and legal as long as you make no more than five gallons of finished ethanol. Anyways, you just need to throw out the first ounce or two of distillate if small scale, up two 4oz if large scale. That will be mostly methanol, other than that just be careful and God bless chemistry!


is bleach residue (after evaporation) safe? I have been preparing bleach disinfectant lately in a spray bottle and has been spraying on my couch, door mat, door knots, etc...
I don't rinse it... So I assume there is bleach residue left after evaporation?
Is it safe to leave the residue?

I felt some very mild burn like feeling on my arms when putting my arms on my couch lately; I don't know if this is some kind of placebo effect or if it is real.

Bleach, especially employed in laundry applications, may have, in addition to the active ingredient sodium hypochlorite (NaOCl), an equal concentration of NaCl (as it is prepared from the action of Cl2 on NaOH) and at times, an alkaline, like lye (NaOH) or Washing Soda (sodium carbonate, Na2CO3). The latter can cause skin rashes and irritation (and will be present on drying).

The rationale for the latter is to keep the pH elevated in aqueous sodium hypochlorite as upon exposure to CO2, there is the possible introduction of hypochlorous acid. While HOCl is more powerful (and reportedly nearly a hundred times more potent disinfectant than the associated hypochlorite) is far less stable (a short shelf-life). Further, the decomposition product is O2 and also HCl (and not NaCl).

The created hydrochloric acid can further react with the sodium hypochlorite or hypochlorous acid as follows:

$\ce{HCl + NaOCl -> NaCl + HOCl}$

$\ce{HCl + HOCl <=> Cl2 + H2O}$

leading to a chain decomposition of the bleach, hence the advantage of adding NaOH or Na2CO3. To quote the cited source:

  The pH has a significant effect on the stability of sodium hypochlorite solutions. Below pH 11 the decomposition of sodium hypochlorite is significant due to the shift in the equilibrium in favor of the more reactive hypochlorous acid. A pH between 12 and 13 gives the most stable solution. This equates to 0.4 to 4.0 grams per liter (gpl) excess NaOH. Greater concentrations will not improve the stability. Excessively high improve the stability. Excessively high alkalinity will damage textiles and retard the bleaching and disinfecting actions of the hypochlorite.

I would suggest that you can even dilute your current NaOCl dosing and add some lemon juice, just prior to use, to create (per the source reference noted above) a much more powerful mix (and a more potent bleaching agent, so be careful where you select to employ it) albeit, with a much shorter shelf-life.

To rapidly remove bleach smell (likely Cl2O from the dehydration of HOCl), one can apply a spray of dilute H2O2, which rapidly converts NaOCl to NaCl and oxygen. Caution: this further promotes bleaching action via intermediate radical formation (see 'Hydroxyl radical from the reaction between hypochlorite and hydrogen peroxide'), so again be careful where you apply it.

Bleach is usually a mixed solution of sodium hypochlorite NaClO and a lot of sodium chloride NaCl. NaClO is decomposed when dry. So your couch is probably soaked in salt (NaCl). It may produce an uncomfortable feeling.


Acetic acid reaction sequence Can anyone tell me where I am doing it wrong for the following reaction?

In the first step, Br should be attached to the alpha carbon.

In second step, the bromide is replaced by cyanide.

I am able to get the third product that is final compound.

Step 1: $\alpha$-bromination
$$\ce{CH3-COOH + Br2/P -> Br-CH2-COOH}$$

Step 2: Nucleophilic Substitution by Sodium Cyanide
$$\ce{Br-CH2-COOH + NaCN -> NaBr + NC-CH2-COOH}$$

Step 3: Acid-catalyzed Conversion of Nitrile to Amide or Acid

(1) First hydration... $$\ce{NC-CH2-COOH + H2O -> H2N-CO-CH2-COOH}$$
(2) Further heating... $$\ce{H2N-CO-CH2-COOH -> HOOC-CH2-COOH + NH4+}$$

Special Notes

Be VERY CAREFUL with Step 2 as you are potentially forming $\ce{HCN}$ which could get everyone in the lab killed if you're not careful.

There are two possibilities. You will have a two step reaction

NC-CH2COOH -> HOOC-CH2-COOH This is malonic acid. It can be further decarboxylated to

HOOC-CH2-COOH + H$^+$ + heat -> CO2 + CH3COOH

the final product is acetic acid.


Will the solubility of a solute increase over the time a solute dissolve? would like to ask about a question that I came across in one of my assessment books. Appreciate the help that I can get.

Qns: In an investigation, two substances X and Y of the same mass are dissolved in the same volume of water. It is found that X dissolves faster than Y. Which of the following graphs correctly shows the solubility of the substances against time in this investigation?

The answer given is A.

I would like to ask why is that so. Intuitively, I would think that the solubility would be a constant since solubility is defined as "the maximum amount of solute that will dissolve in a given volume of solvent at a given temperature and pressure", therefore it will not change with time as the solute is dissolving.

Even if we're talking about rate of dissolving here, shouldn't the rate be decreasing, as when a higher concentration of particles are dissociated, there would be more of the dissociated particles in the solvent which will shift the equilibrium to the left?

Thanks for all your help!

Ivan's comment really nailed the problem, but I'll throw in a bit more.

In the question the term solubility would be ok in an English class, but the term is incorrectly used within chemistry. The more correct tense would be solvation. Likewise the diagrams should have solvation rather than solubility.

  In an investigation, two substances X and Y of the same mass are dissolved in the same volume of water. It is found that X dissolves faster than Y. Which of the following graphs correctly shows the solubility solvation of the substances against time in this investigation?

It is 'A', the graph is showing the solubility of a substance over time , the answer 'A' shows that 'X' completely dissolves in less time than 'Y'.

Different substances have different solubility values in a solvent, so the same mass of two different substances in the same volume of the same solvent can have different solubility values.

A more soluble substance will dissolve at a higher rate.


In my homemade electrolysis setup, only the negative end bubbles? I've created an electrolysis setup by connecting a $6~\mathrm{V}$ battery to a cup filled with saline water via pencils; I am confused as to why only the negative pencil bubbles though. After running the setup for longer than five minutes, the water turns brown but so far that information has not been useful. Why is the negative pole the only one bubbling and if possible, why is the water turning brown?

Your setup, using table salt ($\ce{NaCl}$ plus additives) will initially electrolyse according to the following two reactions, separated by cathodic reduction and aniodic oxidation:

$$\ce{2 H+ + 2 e- -> H2 ^}\tag{CatRed1}$$

$$\ce{2 Cl- -> Cl2 + 2 e-}\tag{AnOx1}$$

These reactions assume no counterions. However, you don’t have dissolved protons in your solution because you aren’t electrolysing hydrochloric acid. Instead, the proton is taken from water molecules and if we include them (and sodium counterions) we get the following improved cathodic reduction equation:

$$\ce{2 Na+ + 2 H2O + 2 e- -> H2 ^ + 2 NaOH}\tag{CatRed2}$$

This still displays the hydrogen gas bubbling away at the cathode but chlorine gas at the anode should also bubble after some time (or at least evolve noticeably). However, with the increase of $\mathrm{pH}$ due to the generated hydroxide, the chlorine produced at the anode will react further:

$$\ce{Cl2 + 2 OH- -> Cl- + OCl- + H2O}\tag{AnOx2}$$

And both of these are soluble ions. The overall reaction of your electrolysis setup is:

$$\ce{NaCl + H2O -> H2 ^ + NaOCl}\tag{RedOx}$$

There will be hydrogen evolution at the cathode but no gas evolution at the anode. The browning will likely be due to a number of side reactions induced by the oxidant hypochlorite anion.

In case you’re wondering why water’s oxygen is not oxidised, this is due to its higher standard electrode potential.

Try collecting the gas at the negative electrode and testing it. Hold a match to the gas and my guess is that it will go pop. (Positive test for hydrogen)
If there is a group 1 metal salt in your solution like $\ce{NaCl}$ the ions would be Na+ and $\ce{Cl^-}$.
When the Na+ is reduced to Na by accepting the electron from the cathode (negative terminal) it becomes the metal atom.
$$\ce{Na+ + 1e^- \rightarrow Na}$$
This then reacts with the water in the solution to release hydrogen bubbles at the negative electrode.
$$\ce{2Na + 2H2O \rightarrow 2NaOH + H2}$$

This is probably why you have bubbles at the negative electrode.
If the cup vessel you are doing the experiment in has any imperfections in the ceramics or has any metals in the surface then there is probably a displacement reaction from the Na ions or the Halide ions floating around.
There could also be displacement of other ions from inside the saline (saline is a mix of different salts)

That's my best argument for your observations


Treatment of ambidentate ligands in geometrical and optical isomerism Are ambidentate ligands considered different entities when they donate through different atoms? For example will a hypothetical square planar complex like [Pt(ONO)2(NO2)2] show geometrical isomerism?

To specify [Pt(NO2)4]2+ would mean the ligand donates electrons through the N-atom, and not the O-atom. If it were to donate through O, it would be written as (ONO) and not (NO2).

Thus, this compound would not show geometrical isomerism. Also note that, if NO2 ligands are replaced by ONO, then that is a form of structural isomerism in complexes, called linkage isomerism.


Why is fluorine more reactive than chlorine? Why is fluorine more reactive than chlorine despite chlorine having a higher electron affinity?

It depends on what reaction you are talking about.

In the context of organic SN1 reactions, the reactivity is $\ce{R-F}$ (least) $\ce{< R-Cl < R-Br < R-I}$ (most). [1] This reflects the leaving group ability, with larger atoms better able to accommodate the extra electron.
In the context of radical homolysis of $\ce{X-X}$ the order of reactivity is $\ce{F2 < Cl2 < Br2 < I2}$ as the sigma bonds become weaker as you move down the period.[2]
In polar aprotic solvents (DMSO) nucleophilicity follows $\ce{F- > Cl- > Br- > I-}$. This order reverses in polar protic solvents due to solvation: J. Chem. Educ., 1970, 47 (6), p 473
Transition metal $\ce{MX_n}$ bond enthalpies have been measured empirically and calculated: J. Chem. Phys., 1957, 26 (6), 1644–1647.. It would be inappropriate to over-generalize, as any discussion here needs to also account for metal oxidation states, HSAB theory, electronic configurations, and ligand-field theory

References:

Clayden, Greeves and Warren Organic Chemistry, p. 348.
Clayden, Greeves and Warren Organic Chemistry, p. 971.

Fluorine molecule has a small atomic radius with 5 electron in 2p orbital. This means it has a high charge density on its atom. When fluorine molecule breaks up into F- ions it remains highly unstable and just wants to bond with other element. Due to this charge density fluorine has high polarizability which makes it more reactive.


Aromaticity of different compounds I'm new here and I'm still learning, I understand that there four rules for aromaticity:

cyclic
conjugated double bonding (every atom in the ring should be sp2)
huckle number should be 2, 6, 10, 14, 18, ..18
planar ring

The 2nd question should be 4 electrons and is antiaromatic, but how about the first question?

My answer would be to refuse to answer this case.
The 4n+2 rule applies only if the electrons are conjugated in a cycle -- nothing more, nothing less.  If you have pendant pi bonds in the conjugated system, like the carbonyl group in cytosine and many similarly structured compounds, you don't have the right conditions to apply the 4n+2 rule at all.
Interior conjugated atoms, as in pyrene, also mean you don't have specifically conjugated cycle.

Cytosine is aromatic, there are 6 electrons in the ring. Two pairs of pi bonds(thus 4 electrons) and a lone pair on Nitrogen(the one which has H written below it). The other Nitrogen (which is forming a pi bond) in the ring, does have a lone pair, however it is "localised". That lone pair cannot participate in resonance, and hence it is not counted when considering aromaticity.