Steric Inhibition of Resonance in ortho-substituted Benzoic Acids It is given that due to Steric Inhibition of Resonance (SIR), the acidity of ortho-substituted benzoic acid increases with respect to benzoic acid. However, I'm unable to understand the reason behind this fact.

We know that resonance stabilises a molecule and the absence of which decreases the stability. In case of ortho substituted benzoic acids, resonance is inhibited due to the loss of planarity due to the steric interactions from the ortho substituent.

If this is the case then the conjugate base of the acid will not be stabilized by resonance with the benzene ring. So it seems the ortho-substituted benzoic acid will not readily donate H+ and is less acidic than benzoic acid, contrary to what is being given in various sources.

Then kindly explain why is ortho-substituted benzoic acid is more acidic than benzoic acid.

Important information before understanding the reason for steric inhibition of resonance and the ortho effect

In general, resonance stabilizes a molecule. But in the case of carboxylic acids, the conjugate base (anion after losing $\ce{H+}$) there are two equivalent resonance structures.
Take the example of acetic acid, $\ce{CH3COOH}$, and thus acetate ion, $\ce{CH3COO-}$:

The hybrid would look something like this:

It could be easily seen that equivalent resonance structures mostly stabilise the hybrid greatly.
Therefore, equivalent resonance tends to stabilize a molecule/compound more than normal resonance. (in most cases)

Now let's take the case of o-toluic acid.

There are two ways by which resonance can take place:

In (I), it takes place through $\ce{-COOH}$. Its ion is expected to be more stable due to the above result. Also, here $\ce{OH}$ is highly polar and it will tend to lose $\ce{H+}$ to become uncharged and thus be more stable (as compared to $\ce{O-}$).
In (II), it takes place through the ring. Its ion is less stable due to more formal charge and less aromatic character (due to + on the ring). Also, $\ce{OH}$ is less polar than $\ce{H+}$ due to the above point.

Therefore, it will be better to prevent the taking place of resonance through ring.

That is done by $\ce{-CH3}$ group which provides bulky group repulsion and takes the ring out of the plane of $\ce{-COOH}$ and thus providing condition for equivalent resonance in $\ce{-COOH}$.


Derivation of work done by real gas According to the Wikipedia on real gases, it is stated that the expansion work done by a real gas differs from the ideal gas by $\int\left(V-\frac{RT}p\right)\,\mathrm dp$.

Firstly, does this mean $W_\text{real} = W_\text{ideal} + \Delta W$ or $W_\text{real} = W_\text{ideal} - \Delta W$?

Secondly, how is this formula derived?

For a reversible process in a closed system (no mass entering or leaving), the general equation for the expansion work done by a real gas on the surroundings is the same for an ideal gas, namely
$$W = \int p\,\mathrm{d}V$$
However, for a real gas, we use the equation of state for that gas $p=p(n,V,T)$ rather than $p=\frac{nRT}{V}$, the equation of state for the ideal gas.


How are ions formed? I know that $\ce{H2O}$ is neutral.
So how does adding one more hydrogen atom to it make it positive ion ($\ce{H3O+}$)?

The flaw in the question is exactly what Gaurang Tandon said: $\ce{H3O^+}$ is formed by a water molecule and a proton ($\ce{H^+}$), not a hydrogen atom.

Compare the number of electrons in both sides of the following two equations:

$$\ce{H2O + H^+ -> H3O^+}$$

$$\ce{H2O + H^. -> H3O^+}$$

The last equation is wrong: there's an extra electron in the left hand side.


Why ZnS can be seen as the solution is made alkaline? Explain with reactions There is an extract in my book which i am not able to get. It says:
Place strip of metallic zinc in aqueous solution of copper nitrate for about one hour. You may notice that the strip becomes coated with reddish metallic copper and the blue colour of the solution disappears. Formation of Zn2+ ions among the products can easily be judged when the blue colour of the solution due to Cu2+ has disappeared. If H2S gas is passed through the colorless solution containing Zn2+ ions, appearance of white zinc sulphide, ZnS can be seen on making solution alkaline with ammonia.

The final solution contains $\ce{Zn^{2+}}$ ions, which reacts with $\ce{H_2S}$ to produce white $\ce{ZnS}$. But you may not avoid that maybe some of the initial $\ce{Cu^{2+}}$ ions still remains in the solution. And these rare $\ce{Cu^{2+}}$ ions do react with $\ce{H_2S}$ to produce Copper sulphide $\ce{CuS}$ which is dark black.
$$\ce{Cu^{2+}  +  H_2S  ->  CuS  +  2 H^+}$$
So the obtained final precipitate is probably a mixture $\ce{ZnS}$ and some $\ce{CuS}$ ; it is probably not perfectly white, but grey or even black. To be able to see the white color of $\ce{ZnS}$, you add some drops of ammonia $\ce{NH3}$ in the dark mixture containing $\ce{ZnS}$. This will dissolve the copper sulphide, producing a solution containing $\ce{[Cu(NH3)4]^{2+}}$.
$$\ce{CuS  +  4 NH3 -> [Cu(NH_3)_4]^{2+}  + S^{2-}}$$
So the black component of the precipitate will disappear, and you will be able to see the white color of insoluble $\ce{ZnS}$.


Reaction of propylene oxide with ammonia or hydrazoic acid I am supposed to identify the products A and B in the following scheme:

The three options I have been given are:

I think that since $\ce{HN3}$ is a stronger nucleophile than $\ce{NH3}$, B should form via an SN2 mechanism. So, cleavage should take place in such a way that $\ce{OH}$ group remains with the carbon attached to the $\ce{CH3}$ group. Hence I think option 1 is the answer, but I am not sure about it.

In fact, these are at most chemoselective reactions, so there will probably be a mixture of all possible products.

Ignoring stereochemistry and assuming that we want to know the 'most probable' products, as far as I can tell the main difference is that $\ce{NH3}$ is a rather nucleophilic basic compound, whereas $\ce{HN3}$ is an acid and not very nucleophilic. We are not given the solvent, so we cannot know what side reactions can happen, or what solvation effects we have. So we may assume an 'inert' solvent is used.

Within these hypotheses, I'd say that $\ce{NH3}$ will react by $\mathrm{S_N}$2, attacking mostly on the least substituted side ($\ce{CH2}$), whereas $\ce{HN3}$ will protonate the oxygen first, and the opening by the azide anion will happen by $\mathrm{S_N}$1, i.e. attacking the most stable carbocation (the one with the positive charge on the most substituted carbon. So for me, the 'correct' answer would be 2.


Good measures ot atomic size I was looking up the Lennard-Jones parameter $\sigma$ (which, to my understanding, is a measure of molecular size) when I was surprised to discover that the value for the $\ce{N2}$ molecule (3.667 Å) was slightly lower than the value for the $\ce{CH4}$ molecule (3.78 Å). It's my understanding that $\ce{N2}$ molecules are significantly larger than $\ce{CH4}$ molecules, and this is exploited in, for instance, molecular sieves.

My question: is $\sigma$ a reasonable measure of mean atomic radius? And if it isn't, what is a good measure of molecular size, with which I can compare molecules?

Atoms and molecules, as you surely know, are not tiny hard pebbles that you may measure using a slide gauge, unambiguously and with great precision. Instead, they are clouds of electron density with rather fuzzy edges, like things on an out-of-focus photograph. Their apparent size depends on how you define what "edge" is, or how hard you squeeze them.
This is one of the reasons why we don't have a single definitive list of atomic radii, but a whole variety thereof.

As to the point of your question, yes, $\sigma$ is a good measure of molecular size. But there are also others, equally good and all subtly different. Which one to use depends on your purpose, and may be a hard problem by itself.

Also, remember that (1) the Lennard-Jones potential is defined for spherically symmetrical objects, which your molecules are not, and (2) even for these objects (atoms) it is but an approximation.


How hot can Thermal Grease get? Thermal Grease is used to remove the air gaps between CPU and heatsink. However, computing aside, what would determine the max temp that the grease can handle before it burns or breaks down?

Since most thermal greases are ceramic and/or metal suspended in a silicone or organic oil, the temperature the grease remains stable at will likely primarily be determined by the properties of that oil -- either its ease of evaporation, or decomposition temperature (or, I suppose, its flashpoint, but that seems unlikely).

As Ben Norris says in the comment above, your best bet is to contact the manufacturer or check their data sheets: for common CPU uses, most manufacturers will give you stable and extreme temperature ranges which are likely to be fairly conservative (for example, here), and they may be able to help with more esoteric data if you're using it for other purposes.


Cooling down and heating of solution unexpectedly during cation identification test While conducting qualitative tests for identification of cations in the given salt, I encountered this unexpected temperature change. Following are the steps of tests:

Step 1: We made an aqueous solution of the salt and added dilute hydrochloric acid. There was no precipitate, which showed that Group 1 cations ($\ce{Pb^2+}$) are absent.

Step 2: We then divided the solution into two parts, and through one part, passed $\ce{H2S}$ gas. There was no precipitate, which showed that Group 2 cations ($\ce{Cu^2+}$) were absent.

Here, the temperature of the solution through which we passed the gas, rose above the room temperature. It could be easily felt while holding the test tube.

Step 3: To the other part in the test tube, we added $\ce{NH4Cl}$ to make the solution completely alkaline, and then added $\ce{NH4OH}$ till the solution was ammoniacal. We did not get any precipitate, which showed that Group 3 cations (viz. $\ce{Al^3+}$, $\ce{Fe^2+}$ and $\ce{Fe^3+}$) were absent.

It is here that we encountered the temperature change. Before adding $\ce{NH4Cl}$, the solution was at room temperature. However, as we kept on adding $\ce{NH4Cl}$ to make the solution saturated and alkaline, the temperature of the solution went down considerably. After some time, it was so cold that water molecules in the atmosphere had started to condense on the outer side of the test tube. The temperature still went down when we added $\ce{NH4OH}$.

Step 4: Now we divided the solution into two parts and through one part, passed $\ce{H2S}$. We got a white precipitate, indicating the presence of Group 4 cation ($\ce{Zn^2+}$).

It must be noted here that as soon as we passed the $\ce{H2S}$ gas, the temperature of the solution again increased to room temperature.

Step 5: To the other part of the solution, we added $\ce{(NH4)2CO3}$. We got a white precipitate, which indicated that Group 5 cations (viz. $\ce{Ba^2+}$, $\ce{Sr^2+}$ or $\ce{Ca^2+}$) was present.

It must be noted that the temperature remained almost same now.

Later, by confirmatory tests, we determined the salts to be $\ce{Ca(CH3COO)2}$ and $\ce{ZnSO4}$.

Can anyone explain what chemical reactions had occurred which were exothermic and endothermic, thus resulting in a change in temperature of the solution?

After the help by @paracetamol, I found that the process of dissolution of $\ce{NH4Cl}$ is an endothermic reaction, which results in the decrease in the temperature of the solution.

An experimental demonstration can be found here.

Greater information may be found here.

Important parts of the second reference:

  When discussing this question, one needs to think about all the processes involved (or at least those that account for most of what happens). In general, any process will occur spontaneously, if $\Delta G$ is negative. To determine $\Delta G$, equation (1) is used.

  $$\Delta G = \Delta H - T \Delta S\tag{1}$$

  There is an enthalpic term related to the heat a reaction produces or requires and an entropic term that is temperature-dependent. We cannot say a lot about the entropic term a priori, but we can about the enthalpic term — which is luckily what the question is about. For that, we first need to write a reaction equation (2):

  $$\ce{NH4Cl (s) + n H2O <=>> NH4+ (aq) + Cl- (aq) + m H2O}\tag{2}$$

  Dissolving of the ammonium chloride can also be thought as two separate processes, see equation $(2')$.

  $$\ce{NH4Cl (s) -> [NH4+ (g)] + [Cl- (g)] -> NH4+ (aq) + Cl- (aq)}\tag{2'}$$

  Meaning that first we break up the ammonium chloride crystal structure and second we dissolve the bare ions. If we want to write that in single enthalpy terms, we can do that as shown in equation (3).

  $$\Delta H_\mathrm{tot} = -\Delta H_\mathrm{lattice}(\ce{NH4Cl}) + \Delta H_\mathrm{solv}(\ce{NH4+}) + \Delta H_\mathrm{solv}(\ce{Cl-})\tag{3}$$

  All of these values can be looked up. I have given those as calculated by Jenkins and Morris in table.

  $$\textbf{Table 1:}\text{ Values of enthalpies used in this answer as}\\\text{quoted from Jenkins and Morris (reference 1).}\\\begin{array}{cccc}\hline
\text{compound} & \Delta H_\mathrm{lattice} [\mathrm{kJ/mol}] & \Delta H_\mathrm{solv} [\mathrm{kJ/mol}] & \Delta H_\mathrm{tot} [\mathrm{kJ/mol}] \\\hline
\ce{NH4Cl} & -709.1 & - 694.7 & 14.4 \\ \hline\end{array}$$

  So we need energy (a lot of it!) to break up the crystal lattice of $\ce{NH4Cl}$. We then re-gain energy by creating the hydrated, i.e. dissolved, ions in solution. If energy is required, it is typically (excluding photochemical reactions — not the case here) heat energy which is simply drawn from the surroundings. If energy is released, it is typically (same caveat) released as heat into the surroundings.

  Thus:

  The heat is absorbed by the solid ammonium chloride. It is used to break up the salt crystal according to equation $(2'.1)$. Part of it is re-released (unnoticed) by forming solvent–ion hydrogen bonds (the solvation mechanism of $\ce{NH4Cl}$ — equation $(2'.2)$).
  This heat is drained from the surroundings, which in this case is primarily the water in which you wanted to dissolve the ammonium chloride. (Of course, the solution will then further exchange heat with whatever is around it, meaning that either the air around a bottle/flask or an outer flask as in your described experimental setup will lose heat at the benefit of the $\ce{NH4Cl}$-solution.)


Symmetries of the v=0 and v=1 quantum harmonic oscillator wave functions and those of the Cartesian Coordinates (References) I'm busy with a writing project for my third year Chemistry course and in the project one of the things we need to consider and answer is the symmetries of the v=0 and v=1 quantum harmonic oscillator wave functions and those of the Cartesian Coordinates. This question falls under the suh-heading (in the project) 'Determination of IR-activity of normal modes and the prediction of spectra'.

Now I'm not expecting anyone to give me the answer to this question, but I would greatly appreciate it if you could let me know of websites and/or accessible textbooks that I can use to find this answer. I still have two months to complete the writing project and am essentially done with it, but its just this question that I have been really struggling to answer since I can't seem to find it online or in the textbooks by Atkins and Engel & Reid respectively.

Any help would be appreciated, Thanks in advance.

In general you need to understand point group tables.

These always have the same form, which is to say that symmetry operations are on the top line, characters (1,-1, 0 etc) in the main body. The most left hand column the Mulliken symmetry labels (also called symmetry species) for each irreproducible representation (one row of characters) .

The right hand two columns, often without header labels contain the operators for different type of transitions. x,y, z are used to dipole transitions because a dipole transforms as x, y or z. (The z direction is usually that of the principal axis). Rx, Ry, Rz are used, for example, spin orbit operators (These are coupling between electron spin and orbital motion, Not rotations).

The last column contains product operators $x^2$, $x^2-y^2$ etc and are generally used to describe Raman transitions since the Raman transition operator depends on a molecules polarisability (a volume) which in projection is proportional to an area hence squared terms.

If you want more complicated operators say $x(x^2-y^3)$ then generally these can be made from simpler operations. When making more complex functions consider the x as a $p_x$ orbital with two lobes of different 'colour'. The subtraction as in $x^2-y^2$ then looks like the $d_{x^2-y^2}$ orbital with opposite lobes coloured in the same way but each pair of a different colour. You can the use the symmetry of this object to determine which irreproducible representation it belongs to. In $C_{2v}$ (below) it is $A_1$.

The C$_{2v}$ point group shows these features. The operators are E, $C_2$ etc,. The Mulliken symbols for the symmetry species, in your case vibrational normal mode symmetries, are $A_1$, $A_2$. The characters are 1, -1 etc.

A totally symmetric normal mode $A_1$ has characters 1,1,1,1, is active to transitions with a dipole on the z direction and is also Raman active, as it has at least one squared term, x$^2$ for example. The $A_2$ symmetry species in not ir active, but is Raman active, due to the xy operator. You can follow the rest.

If you want to work out what the symmetries of a wavefunction are in a given point group, then the easiest way is to sketch the wavefunction and apply the symmetry operations from that point group. Choose a set of axes, z vertical and x horizontal and y out of the page, perpendicular to the other two. Count 1 if the symmetry operation is present and -1 if not.

For example the $v$=0 wavefunction for the harmonic oscillator has a gaussian or bell shape. By applying in turn the symmetry operations in C$_{2v}$ this wavefunction belongs to the $A_1$ symmetry species. The v=1 wavefunction has a node at the origin with an inversion in the values between negative and positive x. In this case C$_2$ is -1, $\sigma _{xy} = 1$ and $\sigma _{yz} = -1$ which is the $B_1$ symmetry species.


Microfiltration for saturated solutions of highly hygroscopic compounds? Is it possible to pass saturated solutions of compounds with very high solubility through filters with small pores (e.g. passing a saturated aqueous ZnCl2 solution at 90C through a 2mcm glassfiber filter with the help of partial vacuum)?
Will it work well? Are there preferred methods for this? If not, is there any (reasonable) solution to filter out microparticles from such solutions?

Filtration is a mechanical operation.  Thus, if you have a solution of $\ce{ZnCl2}$, which (assuming you use water as solvent) dissociated into ions of $\ce{Zn^{2+}}$ and $\ce{Cl^-}$, a typical filter will not retain these ions.  Instead of using some vacuum (potential danger to evaporate some solvent, thus altering the concentration of he solution), you may consider filtration under pressure (e.g., across a Schlenk fritt for larger scale, or a syringe filter at smaller scale [e.g., testing a spin coating]).

(credit)
If the solvent capacity for $\ce{ZnCl2}$ is not (this much) temperature dependent, you equally could perform the filtration near room temperature (e.g., $\pu{432.0 g}$ $\ce{ZnCl2}$ per $\pu{100 g}$ of water at $\pu{25 °C}$ listed here).


Atomic radius and nuclear effective charge in a group Why even though the effective nuclear charge increases in gruop from top to bottom , the atomic radius increases too? Shouldn’t it decrease with more effective charge on valance electrons and eventually more force affecting them?

Just so that we are on the same page before I attempt to answer this question, the definition of effective nuclear charge we are considering Zeff=Z−σ, where Z is the nuclear charge (number of protons) and σ is the shielding factor of the inner shell electrons. The overall effective nuclear charge experienced by the valence electron in question is Zeff.

You are right in thinking that the effective nuclear charge increases down as a group as the increasing nuclear charge has a greater effect on Zeff than the shielding effect of more inner electron shells. This, as you quite rightly mention, should suggest that because Zeff increases down a group, a greater force of attraction would be experienced between the outer shell valence electrons and the nucleus, thus decreasing the atomic radius. However, what we have failed to consider is the size of the electron orbitals in question. The principal quantum number, n, of electron orbitals affects their size. The quantum mechanical description shows that the 1s orbital of a hydrogen atom is a lot smaller in radius than the 2s orbital of a lithium atom for example. This comparison can be further extended where the 3s orbitals are larger in radius than 2s and these are larger than 1s. Thus we have a quantum mechanical picture whereby the principle quantum number of electron orbitals affects their radius.

Essentially we have two competing factors that affect the trend of atomic radii in question:

The effective nuclear charge, Zeff, increases down a group which draws electrons closer towards the nucleus, decreasing atomic radius.
The principal quantum number, n, of electron orbitals that increases down a group and due to the quantum mechanical nature of electrons, the radius of these electron orbitals increases with increasing n, thus increasing in size as you go down the group.

Simply put, the factor of increasing n down a group is greater than the effect caused by the increasing Zeff, thus causing the atomic radius of atoms to increase down a group in the periodic table.


How can you inhibit crystallisation of hot superconcentrated magnesium sulphate I’m trying to create a gel like solution of primarily magnesium sulphate heptahydrate or Epsom salt and water.  It’s simple to create a hot super concentrated solution but when it cools it’s obviously begins to crystallise. Continuous stirring as it cools and cooling quickly creates a slush of tiny granules which eventually settle to the bottom of the solution like sand in water if too thin.

How can a thick concentrated smooth liquid be created?

I’ve looked into ways of inhibiting crystal formation but no good examples.  Maybe an impurity if some kind and then possibly a thickener if the mix is still watery.

First, your question involves inorganic chemistry, not organic, unless you're talking about the unmentioned gelling agent.

Second: Impurities serve as a source of nucleation, and stirring increases crystallization. If you leave the supersaturated solution in a clean container and allow it to cool slowly, e.g. in a Styrofoam cooler chest, it may take a long time to form crystals. If you use some thickener, such as PEG or gelatin, it might take longer yet. Left undisturbed, though, supersaturated $\ce{MgSO4}$ can persist for years. There are videos of it crystallizing after a seed is added.


What Causes Atomic Orbitals to Hybridise? I'm finding it difficult to visualise the process of hybridisation. Taking the example of formation of $\ce{CH4}$, can it viewed as the collapsing of the p orbitals of the Carbon atoms as they get "heavier" on excitation even before the introduction of Hydrogen atoms or is it the collision with Hydrogen atoms that causes the fusion?

Hybridization is a mathematical concept developed to cope up the flaws of Valence Bond theory. It hinges solely on the linear combination of wavefunctions of atomic orbitals having more or less same energy. Hybridized state is a purely theoretical concept & cannot be detected even spectroscopically.

There is nothing like "collapse." The wavefunction of a hybridized orbital is "mixing" of the atomic orbitals only. The number of hybrid orbitals are same as the number of atomic orbitals participating in the "mixing up". The wavefunctions interfere constructively or destructively to give rise to equivalent orbitals known as hybrid orbitals.

The Methane Case:

Carbon has valence configuration: $2s^2 2p_x ^1  2p_y^1$. So, VB theory predicts that it can form two bonds. This is one of the main defect of VB theory. Then promotion comes to the rescue! One electron from $2s$ gets promoted to unoccupied $2p_z$ orbital, so as to make the configuration: $2s^12p_x^12p_y^12p_z^1$ . So, $\ce{C}$ can form four bonds. But still there is a problem: experiments show that the four bonds in methane are exactly equivalent which is not possible by this configuration since $s$ & $p$ orbitals are not  same. To deal with the problem, concept of hybridization was developed. Hybridization gives rise to four equivalent orbitals but having different orientations (four corners of the tetrahedron.) The non-normalized wavefunctions of four orbitals are as follows: $$\psi_1 = \psi_s + \psi_{p_x} + \psi_{p_y} + \psi_{p_z}\\\\\\ \psi_2 =\psi_s - \psi_{p_x} + \psi_{p_y} - \psi_{p_z}\\\\\\ \psi_3=\psi_s - \psi_{p_x} - \psi_{p_y} + \psi_{p_z}\\\\\\ \psi_4 = \psi_s + \psi_{p_x} - \psi_{p_y} - \psi_{p_z}$$


Why does water favour nucleophilic substitution over elimination? The two broad ways a halogen can leave a haloalkane are nucleophilic substitution and elimination reactions.

I was told that elimination reactions (that form a double bond as a $H^+$ is also removed) do not occur when the haloalkane is dissolved in water; here nucleophilic substitution (anion replaces halogen only) is favoured.

Why is this? I guess it's not just water - what are the general conditions (including type of solvent) that prefer nucleophilic substitution over elimination, and vice-versa?

I understand there are different variants of both elimination and substitution ($S_N1, S_N2, E1,E2$); are any of them more prone to occur than others of the same type in water?

The nature of the substrate (in particular the degree of substitution of the α-carbon) is of principal importance in determining which mechanism predominates. Generally, $\mathrm{S_N1}$ and $\mathrm{E1}$ are both less common than $\mathrm{S_N2}$ and $\mathrm{E2}$, partially as a consequence of their first-order kinetics and the stability constraints imposed by the charged intermediate species the mechanisms generate. $\mathrm{S_N1}$ and $\mathrm{E1}$ both require tertiary, allylic, or benzylic α-carbons in order to be viable (although secondary carbons can occasionally react by these pathways as well, but slowly, and only if conditions disfavor competing reactions), mainly because the carbocation intermediate that forms during the rate-determining step can be stabilized by the neighboring electron-donating alkyl groups via hyperconjugation (or via resonance in the case of carbons in allylic or benzylic positions). Tertiary carbons also disfavor $\mathrm{S_N2}$ because the approach of the nucleophile is sterically hindered by the β-alkyl groups. Under those conditions, $\mathrm{E1}$ and $\mathrm{S_N1}$ reactions would be in competition. $\mathrm{E1}$ would be favored if there were no strong nucleophiles present, as well as by the application of heat. Obviously, $\mathrm{S_N2}$ reactions are favored by the presence of strong nucleophiles (ideally, ones that don't also act as strong bases) and a primary α-carbon substrate. $\mathrm{E2}$ reactions are favored over $\mathrm{S_N2}$ if strong bases, especially non-nucleophilic, are present. $\mathrm{E2}$ also prevails when either the substrate and/or the base is sterically hindered, since abstraction of a proton is vastly less impacted by steric factors than is the approach of a nucleophile attacking a carbon with simultaneous loss of the leaving group. Finally, $\mathrm{E2}$ reactions impose an additional restriction in that the β-hydrogen and the leaving group must be anti-periplanar to one another, in order for simultaneous loss of the leaving group and formation of the π-bond to occur in a coordinated fashion. If the molecule can't adopt such a conformation (for example, it may be rotationally locked due to the presence of π-bonds, or lack the kinetic energy necessary for a ring flip, etc.), then $\mathrm{E2}$ is categorically ruled out.

(Parenthetically, I should add that elimination reactions, in general, require higher temperatures than substitution reactions. The naive explanation for this is based on the equation for the free energy change of a reaction: $\Delta G = \Delta H\ -\ T\Delta S$. Elimination reactions, at the most simplified level, involve two reactant molecules [namely, a base and the substrate], and yield three products [the protonated base, the substrate molecule that has undergone elimination, and the substrate's leaving group]. Consequently, elimination reactions are entropically favorable [i.e., the $T\Delta S$ term is positive]. However, elimination reactions are also endothermic [i.e., $\Delta H$ is positive], because heat energy is required to break the two σ-bonds of the abstracted proton and the leaving group, while the resulting products form only one σ- and one π-bond, the latter of which is less stable. The implication of this is that, in order for the overall change in free energy, $\Delta G$, to be negative [a requisite condition for spontaneity of the reaction], $T\Delta S$ must be greater than $\Delta H$, so the temperature needs to be sufficiently high if the reaction is to take place.)

The solvent is of only slightly less importance. Polar protic solvents (water, alcohols, weak carboxylic acids, etc.) have partial-positive hydrogens, which serve to stabilize negatively charged species. Consequently, $\mathrm{E2}$ and $\mathrm{S_N2}$ mechanisms, which are aided by having negatively charged bases or nucleophiles in high concentration, are somewhat dampened by protic solvents, allowing otherwise disfavored $\mathrm{E1}$ and $\mathrm{S_N1}$ pathways to compete. Polar aprotic solvents (THF, DMF, DMSO, chloroform, ethers, ketones, etc.) cannot stabilize negative charges as effectively, making them the preferred choice when $\mathrm{E2}$ or $\mathrm{S_N2}$ mechanisms are desired.

Finally, to your specific question about haloalkanes. I've explained above why various combinations of conditions might favor one class of reaction over another. In water, specifically, a few distinct factors, beyond the general principles outlined above, might be relevant:

Due to the leveling effect, no base stronger than hydroxide can exist in water at any appreciable concentration (that is, any base much stronger than hydroxide will simply deprotonate the water molecules, forming hydroxide and the base's conjugate acid). However, strong, non-basic nucleophiles are not subject to this effect, and though they may be solvated and weakened by hydrogen bonding, this effect would be less drastic than the protonation that strong bases would undergo.
Hydroxide is actually a much stronger base in many other solvents than it is in aqueous solution, which is chiefly the result of it being very thoroughly solvated and stabilized by hydrogen bonding. Hence, competing nucleophiles become stronger by comparison, while stronger bases simply end up being consumed, producing more hydroxide.
Because elimination reactions require heat, substitution reactions will be heavily favored and immediately take place unless the reaction vessel is heated prior to mixing of the reagents. The problem is compounded by the fact that, with hydroxide as the nucleophile, an alcohol would form, which would be stabilized by hydrogen bonding with water molecules. Further, because haloalkanes are poorly soluble in water, and immiscible liquids in the same container will boil at lower temperatures than they would unmixed, it could become very difficult in practice to achieve the temperatures necessary to favor elimination reactions while also maintaining the agitation necessary to keep the haloalkane and the aqueous phase from completely separating.


Why is latent heat of vaporization not exactly proportional to boiling point? Latent heat of vaporization and boiling point are both chemical properties related to the bond strength of the molecule. In general, the higher the boiling point is, the higher the heat of vaporization will be. However, there are some molecules that break this trend. For example, ammonia has a higher heat of vaporization than you’d expect based on its boiling point relative to other molecules with similar boiling points. See table here: https://en.m.wikipedia.org/wiki/Enthalpy_of_vaporization. Why does ammonia (and other substances) break this trend?

Hm, the proportionality is not so very good. (data from the wikipedia link above)

The line is a linear regression through all ten data points.

The answer to your question is given on the same wp article. The quotient of enthalpy of vaporisation and temperature is the difference in entropy between the liquid and gaseous phase.

According to Trouton´s rule, it is about 85 to 88 J/molK (ca. $10.5 R$) for most liquids. https://en.wikipedia.org/wiki/Trouton%27s_rule explains the reasons for deviations, like hydrogen bonding, small rotational excitation, dimerisation in the gas phase etc.


Does HF react with ethanol? HF is needed to be dissolved in ethanol in many cases, e.g. in electrochemical etching for fabrication of porous material or in sol-gel synthesis of transition metal oxides being doped with F. Does HF only dissolve in ethanol or can it also react with it to form some fluorocarbon compounds under ambient conditions? If it reacts with ethanol, it will introduce complications in my sol-gel synthesis (ethanol used as solvent and HF as dopant), which I need to take care of.

I don't think HF will react with ethanol under normal conditions. Nucleophilic substitution could be a possibility, but $\ce{F-}$ is worse nucleophile than $\ce{OH-}$, so this won't happen.


Why is triplet carbene sp2-hybridised? Triplet carbene has a carbon with 6 electrons, of which 2 are unpaired. I would expect the carbon to be $\ce{sp}$ or $\ce{sp^3}$ hybridised and the two singly occupied orbitals to be degenerate. However, Clayden et al. (p. 1011) say that, since the triplet carbenes are almost always bent, they are best considered to be $\ce{sp^2}$ hybridised, with the unpaired electrons occupying a $\ce{p}$ and an $\ce{sp^2}$ hybrids .

The very basic explanation I see why the two SOMOs are not degenerate is simply that $5E_{\ce{sp^2}} + E_{\ce{p}} + 2E_{\ce{sp^2},\ \text{pairing}}$ is less than $4E_{\ce{sp}} + 2E_{\ce{p}} + 2E_{\ce{sp},\ \text{pairing}}$ or $6E_{\ce{sp^3}} + 2E_{\ce{sp^3},\ \text{pairing}}$. However, I wonder if there's a better explanation.

That's actually a good question. I was lucky enough that Dietrich Gudat, who wrote a long chapter on this type of chemistry in the Riedel - Moderne Anorganische Chemie was my professor in inorganic molecular chemistry and we discussed these things in detail back then but it's been a while.

I think it's due to how you want your final structure to look like or VSEPR. You have bascially three ligands, two rests and the lone-pair on the carbon. And by having it sp² you get quite some distance with the 120° angles. This splits your possible energy states in one which is called a1, that is basically the sp² and an b1 (the p-orbital). As the a1 has more s-character it is lower in energy and thus you can, depending on how big the gap is, either get a triplett or a singulett carbene.

This will change however if you move to the isolobal and isoelectronic nitrenes. Here we have no sp² but an sp which is automatically filled with two electrons. The other two electrons occupy the degenerate sets of px and py and thus you always end up with a triplett.


How do I calculate the color of a compound? I know this is complicated, but I'll ask. How do you calculate the color of a compound with pen and paper? Everyone knows that iron(II) sulfate is yellowish. How do I know without looking at it?

Since this is quite a broad question, I will also only be able to provide an answer in very broad terms.

The short answer is: You cannot. At least not with (a real) pen and (real) paper.

The slightly longer answer is:
With a simple pen and paper approach this is virtually impossible. Even the use of more elaborate methods like MO theory can lead you to completely wrong results. Electronic transitions involve excited states, which can be incredibly hard to calculate. In most cases you can get an approximate spectrum by using time-dependent density functional theory. This can be right for the wrong reasons - it is still somewhat a crude theory. The best results you will get with multi-reference methods or big automated computations, which may take a long time on a supercomputer.

In other words: There is not easy, general, rigorous way to determine the colour of a compound.


What does 'organic/non-organic molecule' means exactly? I am new to the organic chemistry world.
I was delving for the exact answer to my question, and I've found some results:

In wikipedia (french), is was mentioned that an organic molecule contains Hydrogen, so that $\ce{CO2}$ is a non-organic molecule, but wait, what about $\ce{H2O}$? It's almost said a non-organic molecule and it does contain Hydrogen.
And on other websites, I've read that an organic molecule is supposed to have 'Carbon' in its components, else, it will be called 'a non-organic molecule', the result that is an obvious contradictory to the first one.

What I've deducted from all my results is, that the organic molecule must contain Hydrogen or Carbon, and I'm not pretty sure of my deduction.

There are many different answers to the question, "What is an organic compound".  Here are a few examples:

Pudue - "compounds that contain both carbon and hydrogen"
Britanica - compounds in which one or more atoms of carbon are covalently linked to atoms of other elements, most commonly hydrogen, oxygen, or nitrogen. The few carbon-containing compounds not classified as organic include carbides, carbonates, and cyanides
Wikipedia - any member of a large class of gaseous, liquid, or solid chemical compounds whose molecules contain carbon. For historical reasons discussed below, a few types of carbon-containing compounds such as carbides, carbonates, simple oxides of carbon (such as CO and CO2), and cyanides are considered inorganic
UC Davis- Any compound containing carbon atom(s) is classified as an organic compound

Trying to pull the best from each of these while still keeping it simple, perhaps a good definition of an organic compound would be - any covalent compound containing, at least, carbon and hydrogen.  Everything else would be inorganic.


Would an amalgam filling under or next to a gold crown cause any electrochemical reaction? Would a mercury amalgam filling deteriorate faster if it was in contact with a gold crown?

The answer is basically yes, if the dental procedures are done improperly.  Here is a good answer from a DDS:

  Galvanic current is a term that has been used in dentistry for over
  100 years. It is a condition created by the presence of dissimilar
  metals in the oral cavity of the teeth and gums, with saliva serving
  as the electrolyte. Have you ever felt a “shock” to your teeth caused
  by a piece of tin foil or a spoon that touches a sliver or mercury
  filling in your mouth? If you answer yes to this question, you’ve
  experienced a galvanic event.

  There are several different types of galvanism: 1) A silver/mercury
  filling is placed in opposition or adjacent to a tooth restored with
  gold. These dissimilar metals in conjunction with saliva and body
  fluids constitute an electric cell. When brought into contact, the
  circuit is shorted, the flow of electrical current passes through the
  pulp, and the patient experiences pain. 2) Dissimilar metals coming
  into contact when the upper and lower teeth come together and touch
  each other. 3) Two adjacent teeth are restored with dissimilar metals.
  The current flows from metal to metal through the dentine, bone and
  tissue fluids of both teeth resulting in discomfort and tooth
  sensitivity.

Additionally, from the same site:

  Metals will not necessarily always cause the galvanic effect in the
  mouth. It depends on the specific metal and alloy being used along
  with quantity and placement in the mouth.

In other words the amalgam filling and the gold cap may form a weak galvanic cell.  The proximity of the amalgam to tooth nerves creates a situation in which only a very weak cell current is required in order for the resulting sensation be perceived.  The rate of degradation could thus be very small, or if the filling and capping procedures are done correctly there will be no galvanic affect at all.


Is Lightning Basically a Cathode Ray? I've heard the term cathode ray pretty much every year in school since they introduced chemistry, and I know that it's a stream(?) of electrons. I'm just having troubles relating it to the real world. I think lightning would be a prime example, but I'm not sure; looking for confirmation.

As you are already aware, cathode rays are electron beams. You take evacuated glass tube, and  add two electrodes on either end. Once you apply a sufficiently high voltage, a stream of electrons flows towards the "cathode"; this is what you call a cathode ray. Inside the tube there exists near perfect vacuum, so there are barely any other gas molecules, ions etc.

Lightning is similar in the sense that you have a large potential difference that leads to electric discharge. The key difference is that this takes place in air, and not in vacuum. Since air is "insulating", it has to be ionised to for the electric discharge to take place. Indeed this is what happens, when the potential difference is sufficiently high. However, what this creates is a plasma, not a stream of electrons.

A plasma is the "fourth" of matter, and is created by ripping apart some electrons from gas molecules resulting in a collection of electrons and ions.


Regarding the probability density of finding an electron in between a bond I'd found the image attached on a website explaining the molecular orbital theory.

My question is, shouldn't the graph have a maximum at the middle of the two hydrogen atoms? since my intuition tells me that the electrons have a great chance of being in the middle rather than anywhere else. hmm, this train of thought has led me to another question, what exactly does this graph even represent? I mean, which electrons probababilty does it even show? or does it show us the probability of finding any electron participating in the constructive bonding.
another question that I have is this, if say, we were to excite one of the electrons into the higher antibonding orbital, what would exactly happen to this orbital, the shape I mean, would it simply cease to exist? since we're literally making the bond cease to exist (bond order=0) or would it assume the shape of this?

I still haven't completely grasped the true nature of such orbitals, I think it's evident from all these questions.

Bonds are in general not maxima of the electron probability distribution. Local maxima are typically centered at the nuclei. This is due to the Coloumb potential. The electron can lower its potential energy by getting closer to the nuclei so they will always try to maximize their probability around the nuclei, where all the positively charged protons are bundled. This is a bit oversimplified but in essence the reason why you see the maxima at the H position or in general at the position of nuclei.
To see bonds, you should look at the difference of the probability density of two lonely  hydrogens and a H2 molecule. I.e. calculate the electronic wavefunction for a single Hydrogen Atom and then add the plots of two such single hydrogen calculations placed at the positions as in H2. Compare this with the plot coming from the H2 calculation and you should see that the H2 distribution has a higher density in between than the two unbonded hydrogens simply taken together.
Bonds as we draw them in simple chemistry are not rigorously defined mathematically and more or less a useful approximation to rationalize reactions without doing difficult calculations. There are ways to obtain a mathematical definition of bonds but these definitions are not neccessarily agreeing with the way that we draw bonds in general chemistry. One such model is Atoms in Molecules which evaluates the electron density and its topology to define bonds. But there are other models, for example Electron localization functions and i am sure that there are more, unknown to me.
Overall it is important to realize that bonds are a useful concept but they have no unambiguous clear cut underlying physical definition. Physics/Quantum mechanics yields no equation that unambiguously defines a chemical bond.


What material should I mix with graphite to make a brick? I want to make flake graphite into a brick but I don't know what material will work best. Would mixing it with clay work?

Clay would work, yes. It is, after all, dirt. Adding graphite to it would only be changing the composition of that dirt.

Typically, you'd only need any sort of binding agent: If you're willing to make a weak brick, you can add some cornstarch and literally press it into a brick. You could also try things like epoxy and acrylic compounds, or things like plaster.

If you wanted to keep the composition of the brick to be mostly carbon, you could consider using chemicals like bitumen and tar, although these bricks would be more flexible (depending on the graphite to whatever ratio).


Molecular antibonding orbitals energy When we heat hydrogen the energy of the molecular orbital of H2(1s) is increased .What happens to the energy of the antibonding molecular energy of hydrogen(1s*)?

Energy is not stored in the energy levels themselves but in the states which occupy them.
At normal room temperature the bonding orbitals are occupied according to Maxwell-Boltzmann's statistics $P_i =e^{\epsilon_i /kT}/Z$, and store that chemical energy. At higher temperatures the antibonding orbital will become occupied releasing the chemical energy.


Why is my conclusion inconsistent with the van't Hoff equation? Let's say I hypothesize that a graph of $\ln K$ vs. $1/T$ has a slope of $-∆G^\circ/R$ and a $y$-intercept of $0$. I prove it simply:

$$∆G^\circ = -RT\ln K \quad\to\quad \ln K = -\frac{∆G^\circ}{RT}$$

This matches the linear form $y = mx + b$. Thus, plotting $\ln K$ vs. $1/T$ would have a slope $m = -∆G^\circ/R$ and a $y$-intercept $b = 0$.

However, I understand that a van't Hoff plot defines a graph of $\ln K$ vs. $1/T$ to have a slope of $-ΔH^\circ/R$ and a $y$-intercept of $∆S^\circ/R$. It is clear from the relation $∆G^\circ = ∆H^\circ - TΔS^\circ$ that my final equation is thermodynamically equivalent to the van't Hoff equation. I do not disagree that

$$\ln K = -\frac{∆H^\circ}{RT} + \frac{∆S^\circ}{R},$$

but if I were to experimentally measure temperature and calculate the equilibrium constant temperature, why should I expect the y-intercept to be $∆S^\circ/R$ as defined by van't Hoff rather than $0$ as I defined above? Why should I expect the slope to be $-ΔH^\circ/R$ instead of $-ΔG^\circ/R$? What makes the van't Hoff equation match experimentally determined values over the equation $\ln K = -∆G^\circ/(RT)$?

In the linear form $y = mx + b$, both $m$ and $b$ are constants, i.e. they don't depend on $x$. On the other hand, $\Delta G^\circ$ definitely depends on the temperature (and consequently on its inverse $1/T$). So if you plot a function $$f(x) = m x$$ where $m$ is not a constant but a function dependent on $x$, you might get something unexpected. In your case, $x$ is $1/T$ and $$m = -\frac{\Delta H}{R} + \frac{T \Delta S}{R}$$

The $y$-intercept corresponds to an infinitely high temperature where $-\frac{\Delta H}{R} \times \frac{1}{T}$ tends to zero and $\frac{T \Delta S}{R} \times \frac{1}{T}$ cancels to be just $\frac{\Delta S}{R}$.


Graphite: Thermal and electrical conductivity I recently encounter this question:

Compare, while providing an explanation, the thermal and electrical conductivity of graphite to that of diamond.

Provided answer:
Graphite conducts electricity(only across a layer) where as diamond does not.

Diamond is better at transferring heat than graphite.

Explanation:
Diamond is covalent network solid of carbon atoms which are linked to each other through sigma bonds.

Graphite has a layered structure where each layer is bonded to the other later through vanderwall forces only. Each layer consists of hexagonal rings which have a delocalized pi electron network just above and below the plane.

These delocalized pi electrons are the reason why graphite can conduct electricity along the planes, where as diamond cannot due to the lack of delocalized electrons.

Diamond is a better conductor of heat because the transfer of heat takes place through adjacent atoms transferring their vibrational energy and diamond has a very close knit network of atoms where as graphite does not.

My doubt:
In metals it is accepted that both electrical and thermal conductance is due the the free electrons present in the metal body.

Graphite too has delocalized free electrons so according the same logic graphite should be both a good conductor of heat and electricity.

So where am I or the answer provided going wrong?

P. S. I'm writing from a mobile device due to the lack of access computers so there might be some errors.

Edit:

J. D. Lee: Concise inorganic chemistry, fifth edition :

This would mean that the electrons are truley free since they can interact with other species nearby not preventing them from being bound to a system.

You are wrong to say that graphite has free electrons.  The electrons are bound to sheets of aromatic rings--delocalized but not free. In electrical conduction, the electron pairs are able to jump from bond to bond on the sheet and between sheets. It is worth noting that graphite has anisotropic conduction where between sheets conduction is lower than conduction along a sheet.

For thermal conduction, energizing an atom does not transmit the energy via free electrons like a metal would.  instead, you are limited to a predominately phonon driven heat transfer again, but now the structure is less dense and the carbons are less rigidly bonded.

Diamond as your correctly identified has a rigid structure that can easily conduct phonons but not electrons which gives good thermal, but not electrical conduction.


Can all compounds that plants and animals produce be synthesized in a lab? I guess a better way to ask the question would be, Are there compounds that exist in biology that humans don't have the capacity to produce in a lab?

I suspect, in its current form, this question is largely opinion based. Currently, as it stands in September 2015, there are definitely limitations. But never say never.

On the small molecule scale, Maitotoxin is a famous example of a molecule that has yet to been synthesised in the lab. But perhaps only for a matter of time (or money)

3D printing and automated synthesis is expanding the scope for small molecule synthesis, giving more flexibility to synthesising recalcitrant targets.

On the larger scale, human DNA contains up to about 300 million base pairs. Currently, the largest DNA synthesised in the laboratory is about 600,000 base pairs, so still a way off. However, with increasing capacity in areas like artificial gene synthesis, complete replication of entire DNA is not unforeseeable. One day.


Interaction radius of a molecule What is the most commonly used radius of an atom or molecule to describe the area that must touch another area of another molecule for the two to react. Is there one radius that always works or does it depend on the reaction? Of course the actual distance at which bonds can form or break varies but I was wondering if at some distance the atoms can somehow bring themselves together for bond 'swapping'

The short answer: no, there is no such a general distance.

The long answer: atoms and molecules are not just hanging is the vacuum and waiting someone to grab them. They actually have rather high energy in translation, rotation and vibration modes and particles actually collide during a reaction some of these energies can be involved. The science that study these collisions is called reaction dynamics, and you can read e.g. Polanyi's Nobel lecture.You should note that:

It is a quantum event, so technically speaking it is not deterministic.
The orientation of the collision and the geometry of the potential energy surface along the reaction can be favourable, and rotation, vibration etc can help the reaction, or can prevent it.
The details are also influenced by the medium (is it a reaction in gas or condensed phase, on surface?) and what kind of orbitals are involved in the reaction. About frontier orbitals and their roles you also can find lot of materials in textbooks, eg. any text that discusses the work of Fukui and Hoffmann.


How does an electrolyte react to voltages below its electrolysis potential? An electrolyte can be electrolysed by a voltage higher than the reverse reaction’s cell potential, while conducting a current in the process. But what happens at the electrodes when the applied voltage is below this cell potential, such as below 1.23 V for water or 4.07 V for NaCl? Does the reverse reaction occur simultaneously, or is the electrolyte unable to conduct a current with such a low voltage at all?

Electrolysing water is NOT required for it to conduct electricity. Water ionizes into $\ce{H+}$ and $\ce{OH-}$. Pure water will contain $10^{-7}\ \mathrm{M}$ of either ion at room temperature at equilibrium. This will result in $18\ \mathrm{M\Omega\ cm}$ of resistivity. If and when potentials above the electrolysis potential is applied, the local pH around the electrodes will change and more ions will be created.

$\ce{NaCl}$ dissociates as soon as it is dissolved in water. The $\ce{Na+}$ and $\ce{Cl-}$ ions created in the solution carry current and increase the conductivity.


What will "Efavirenz amino alcohol methyl carbamate" look like? I'm translating a text in which a possible derivative (produced by harsh conditions) of this drug is mentioned:

Its name is put as "Efavirenz amino alcohol methyl carbamate".

I wonder what it will look like.

The carbamate group, according to Wikipedia, will look like this:

And there will also be an alcohol (hydoxyl) group and an amino group somewhere. Will there be two nitrogens, or will the only nitrogen atom particiape both in the "amino group" described in the name and in the "methyl carbamate"?

The "efavirenz amino alcohol methyl carbamate" is (S)-methyl 4-chloro-2-(4-cyclopropyl-1,1,1-trifluoro-2-hydroxybut-3-yn-2-yl)phenyl carbamate (CAS 211563-41-6),

according to this reference.


What would oxygen or nitrogen snow look like? I was watching a video of what would happen if the earth got kicked out of the solar system and at one point it was mentioned that oxygen and nitrogen snow would form. So would such snow be any different physically to good ol' water snow? Would it be more fine or less dense? I'm asking this because I'm thinking of writing a story or something that uses that. Curiously, there seem to be almost no related results on the internet.

I don't know if there would actually be oxygen and nitrogen snow. I would think under those circumstances, there might be carbon dioxide  snow, but the rest of the atmosphere would probably liquify first, there may be oxygen and nitrogen rain with some hail or sleet, but snow? Not sure it would happen.
If it did, oxygen and nitrogen snow would be different from our normal water based snow because of the absence of hydrogen bonds.
Individual snowflakes hexagonal in shape would not form, as this is a function of hydrogen bonds and this would not happen with oxygen or nitrogen. Water ice is also less dense than liquid water because of the way hydrogen bonds cause "voids" in the solid structure. This would not be the case with oxygen and nitrogen. So you'd probably get something more like needles than snowflakes.
Also, solid oxygen is blue as noted by Ed V below, so blue snow if it happens at all.
Qualifier:
If the earth is out there (beyond the orbit of Pluto) long enough, enough of the atmosphere would eventually freeze and the pressure may drop below the triple point of nitrogen. If that happens nitrogen would sublimate, then fall as snow as temperatures fluctuate.
But even Pluto's surface temperature and pressure is above the triple point of oxygen most of the time (temperature does sometimes drop below though) so any oxygen in the atmosphere would probably fall as rain (blue rain) or possibly hail/sleet rather than snow.
Another thing to remember is that if the atmospheric pressure is that low, there won't be much drag/resistance either, so no "softly drifting" snow. Those needles will reach terminal velocity pretty damn fast and shatter on the ground.


Is there a difference between parchment paper for writing on and the type you use in cooking? I have tried searching this question many different ways but to no avail. My instinct and common sense say the processing is probably exactly the same but I would like to be sure.

Parchment paper for cooking is coated with silicone.  Therefore, most traditional writing implements would not even be able to leave a permanent mark on it.

In addition to the coating, at least one brand says it uses additional manufacturing steps to strengthen its cooking parchment paper:

  A spokesperson for Reynolds Consumer Products, says that the company’s
  parchment paper is pressed into a sheet, then dipped into an acid
  bath, washed, and “passed over a series of hot rotating drums that
  realign the fibers and give the paper its strength,” before the
  silicone coating is applied.


Why is the hypoiodite anion unstable in solution, whereas hypochlorite is relatively stable? $\ce{Cl2}$ and $\ce{Br2}$ can disproportionate in a cold, aqueous, basic environment to form $\ce{ClO-}$ and $\ce{BrO-}$ respectively, yet $\ce{I2}$ directly forms $\ce{IO3^-}$. Why isn't $\ce{IO-}$ isolatable in solution?

When iodine is added to water, the following reaction results:

$$\ce{I2(l) + H2O(l) -> OI^{-}(aq) + 2H+(aq) + I^{-}(aq)}$$

$\ce{I2}$ molecules and water molecules react to form hypoiodite ($\ce{OI-}$). The reaction can move both ways of the equilibrium, depending on the pH of the solution. Iodine may also occur as $\ce{I3-}$(aq), $\ce{HIO}$(aq), $\ce{HIO3}$(aq) . The compounds that are form behave differently when they come in contact with water. When iodine ends up in surface waters, it may escape as iodine gas.

$\ce{IO-}$ rapidly disproportionates to form $\ce{IO3-}$ and $\ce{I2}$
$$\ce{IO- -> IO3- + I2}$$.

See here for more details.

For the disproportionation reaction, consider the Latimer diagram for iodine:-

A species has a tendency to disproportionate into its two neighbor species if the potential on the right of species in a Latimer diagram is more positive than that on the left. In this case, right hand potential($\ce{+0.42}$) is greater than the left hand potential ($\ce{+0.15}$). So , it disproportionates.


Resonance in Cyclohexenone I don't think that cyclohexenone should have any resonance forms. Moving the double bond around would necessitate moving hydrogens around as well. For example, I know that benzene exhibits resonance (duh) and the second picture attached shows the pi bonding electrons shifting but doesn't show any hydrogens shifting positions.

That being said, is there a easier way to determine if a resonance form of a cyclic compound is valid? Drawing out each individual atom and bond is time-consuming! Should I use Huckel's rule? Do all cyclic compounds which exhibit resonance follow Huckel's rule?

The structure you've drawn on the left does have a resonance structure, but not the one you've drawn on the right.

Drawing resonance structures out by hand gets quicker over time.  You find that you don't need to actually put in a "C" for carbon, or label the hydrogen atoms.  Huckel's rule generally applies to planar, cyclic molecules with continuous conjugation, although it has also be applied to transition states where an S or P orbital on a shifting C or H moiety completes a loop of continuous conjugation (for example, the 1,5 hydrogen shift in cyclopentadiene).


Galvanic reaction between aluminum and carbon fiber composite I understand that there is a galvanic reaction between aluminum and carbon fiber composite. Structural aluminum parts in contact with carbon fiber composites should be insulated by something like fiberglass.

But what exactly is it that happens, chemically? I know that aluminum has a tendency to be oxidized; aluminum oxidizes readily by itself in air. I also know that carbon fiber itself is a conductor due to delocalized electrons.

So assuming that aluminum is oxidized then the carbon fiber composite must be reduced.

What exactly is reduced? The carbon fiber, or the epoxy? What is the role of the epoxy in all of this? Wouldn't the epoxy be electrically insulating? How exactly does this redox reaction take place across the epoxy?

The reason is that aluminium carbide, $\ce{Al4C3}$ is formed. The reaction occurs at around 500°C on the interface. The problem with this substance is that it is brittle, and it also reacts with water:

$\ce{Al4C3 + 12H2O -> 4Al(OH)3 + 3CH4}$.

More info at this and this wikipedia page.


What are D_b and D_a in the Generalized Fugacity Coefficients table? I've been referring the book: Chemical Process Principles: Part-II Thermodynamics. I found a table that gives me fugacity coefficients in a tabular form corresponding to various reduced pressures and reduced temperatures. This is it:

I dont know what the $D_b$ and $D_a$ in the table. In the bottom of the table, I see:

$$\left(\frac{f}{p}\right)^{'} = \left(\frac{f}{p}\right)(10^{D({z_c}-0.27)})$$

Is the $D$ in this equation related to the $D_b$ and $D_a$ in the table?

$D_b$ and $D_a$ are deviation terms which must be used in the place of $D$ to obtain the fugacity coefficients for fluids whose critical compressibility factor is not 0.27. Replace $D$ with $D_a$ if $z_c > 0.27$ and with $D_b$ if $z_c < 0.27$.


How to measure ozone output of an ozone generator using chemical methods I have the need to optimize an corona discharge based ozone generator. There are certain variables like frequency and voltage which I can change.

The easiest way would be to use an ozone analyzer – however these things are quite expensive and over my budget currently.

I have tried certain ways to get a hold on numbers but failed so far.

Example: I used activated charcoal to convert the $\ce{O3}$ to $\ce{CO2}$ and tried to measure the $\ce{CO2}$ ppm, however it turned out that the activated charcoal did absorb the $\ce{CO2}$, so this measurement did not work..

Is there a easy to do chemical way to react the $\ce{O3}$ with something, that can be measured/done by a hobby chemist? My aim is to find at least a setup with the maximum ozone output. If it is possible to calculate how much grams ozone were produced it would be even better.

You can do the quantitative estimation of ozone ($\ce{O_3}$) by two reactions. First one is the reaction between $\ce{O_3}$ and potassium iodide ($\ce{KI}$), and the second one is an iodometric titration.

First, ozone is reacted with excess $\ce{KI}$ in presence of borate buffer ($\mathrm{pH} =9.2$), and due to this reaction, iodine ($\ce{I_2}$) and oxygen ($\ce{O_2}$) are liberated as follows, $$\ce{O_3 + 2KI + H_2O -> I_2(^) + O_2(^) + 2KOH}$$

Now this mixture of gases is passed through potassium pyrogallete solution, which absorbs $\ce{O_2}$, and we are left with $\ce{I_2}$. Next, this liberated iodine is dissolved in a solution to make a particular volume, and then titrated with sodium thiosulfate ($\ce{Na_2S_2O_3)}$ of known strength in the presence of starch indicator.

$$\ce{I_2 +_ 2Na_2S_2O_3 -> Na_2S_4O_6 (sodium tetrathionate) + 2NaI}$$

At the end point of this titration, colour of the solution turns colourless from blue ($\ce{I_2}$ + starch gives blue colour complex  which is lost at the end of this reaction).

So, thiosulfate reacts with iodine in a quantitative manner, and from there you can determine equivalents of $\ce{I_2}$ = equivalents of $\ce{Na_2S_2O_3}$. Thus, you can calculate what equivalents of ozone had produced that much of $\ce{I_2}$, and also the grams of ozone present.


Why carbon monoxide reacts with hydrogen to give different products in the presence of different catalysts My books says how carbon monoxide will react with hydrogen to give different products with different catalysts, but how exactly do different catalysts yield different products. Do they provide different pathways for entirely different mechanisms, or is the formation of some products is better thermodynamically and for that reaction to occur some catalysts provide the necessary activation energy and the others cant.

The thermodynamics are always the same. If you go from carbon monoxide to methane, there is only one final enthalpy, and if you have the choice of going to methane or methanol the thermodynamically preferred product is always the same.

The difference indeed less within the different pathways and the activation barriers needed to get from one intermediate to the next. This isn’t really thermodynamics, it is mainly kinetics (although you can argue about thermodynamically preferred intermediates). The pathways are often very complex and entire papers revolve around how one reaction is catalysed on such a solid state catalyst, so please forgive me for not looking them up; but the general point still stands: one catalyst will lead to intermediates where the $\ce{C-O}$ bond is easily broken (and thus give methane), the other will not (and thus give methanol).


How to determine peroxy oxygen? Which of the following compound doesn't contain peroxy oxygen?

  $\ce{CrO5}$
  $\ce{Cr2O3}$
  $\ce{Na2O2}$
  $\ce{BaO2}$

I think peroxy oxygen is peroxide, am I right?. I have read in this answer that there is no direct way to recognise peroxide oxygen, and I think that's the reason no one has taught me that in the school.

I am familiar with many peroxides like $\ce{Na2O2}$, $\ce{BaO2}$, etc. But I not able to recognise between 1 and 2.

Is there any way for this type of question to recognise peroxide oxygen?  And if there is no way to determine then really I need to remember all, because we can't use X-ray crystallography in exam or while studying high school chemistry!

Answer given is 2. $\ce{Cr2O3}$

I have become familiar with many peroxide like Na2O2, BaO2, etc. But not able to recognise between 1 and 2. Is there any way for this type of question to recognise peroxide oxygen?

Yes. In the exam, if you do not know whether (1) or (2) contains peroxy, simply look at the charges of Chromium if all the oxygen is  in the -2 oxidation state.

For $\ce{CrO5}$, if all the oxygen have a charge of -2, then Cr would have to have a charge of +10, which is impossible, therefore you know that some of the oxygen must have a charge of -1(peroxy). On the other hand, for $\ce{Cr2O3}$, Cr has a charge of +3, which is possible (Cr(III) salts are plenty)


C−H bond length in organic compounds I am reading Fundamental concepts of organic chemistry by Dr. Anju Srivastava (12/04/2007).

It shows that carbon-hydrogen bond in ethane is 1.1 Å long, in ethene it's 1.08 Å,  and in ethyne it's 1.06 Å. According to the book, it's clear that carbon-hydrogen bond decreases with increasing multiple bonds on the carbon atom onto which the bond is!

According to my understanding, the shorter the covalent bond, the stronger it is, and it's contrary to the fact that the hydrogen atoms in ethylene are more acidic than those of ethene and so is ethane. The more acidic the hydrogen atom, the weaker it is held in a bond.

Help please where am wrong, and also guide about the book concept.

Bond strength (bond dissociation energy, BDE) refers to homolytic cleavage of a bond ($\ce{C-H}$ in this case). Heterolytic cleavage is measured by $\mathrm{p}K_\mathrm{a}$. The two processes are the numerical inverse of one another. In the case of acetylene with $\mathrm{sp}$-hybridization (50% $\mathrm{s}$-character) of carbon, the electron density of the $\ce{C-H}$ bond lies closer to carbon when compared to ethylene (33% $\mathrm{s}$-character) and ethane (25% $\mathrm{s}$-character). Thus in acetylene, it is more difficult to break the $\ce{C-H}$ bond homolytically because the carbon is "unwilling" to give up an electron but is happy to keep both electrons and give up a proton. Thus, acetylene has a high BDE ($\pu{133 kcal/mol}$) and lower $\mathrm{p}K_\mathrm{a}$ ($=25$) than ethylene ($\pu{110 kcal/mol}$, $\mathrm{p}K_\mathrm{a} = 44$) and, ultimately, ethane ($\pu{101 kcal/mol}$; $\mathrm{p}K_\mathrm{a} \approx 50$).


Increasing the length of a copper wire via electrolysis I'm currently experimenting with copper coating using electrolysis. I'm using copper wire from my home, like the one you use for grounding. I want to electrolysis with longer copper wire (say 5 times the height of the container) to find out if the process will be shortened. My question here is about how I should arrange the wires in the container and if one arrangement has different effect than the other.

Here is what I have in mind:

all horizontal wires connecting to one vertical wire(this will be 5 separate pieces same length connected to the vertical wire)
bent like snake up and down (in one piece)

Do you think these two arrangements would work the same or differently?

The two arrangements would produce identical results, because the copper wire conducts electricity so much better than the electrolyte solution.
However, the second arrangement is more practical because no joins are needed so it would be more reliable.
Generally, the surface area of the copper anode should be similar to the cathode which is to be coated to avoid concentrating the current unevenly, so increasing the anode area would probably help shorten the coating time somewhat, as it would avoid oxygen being evolved from the anode which can happen if the anode area is too small.


Diastereomers and H-NMR What effect does stereochemistry have on the number of signals in an H-NMR?
For example, if you consider the two structures below, would the number of signals and splitting differ due to the restricted rotation? If so, which hydrogens are being affected?

EDIT: This is what I think the splitting would look like.

The first thing you need to take into account is that cyclobutane rings are not planar, as they are drawn on paper, but puckered. Thus, a three-dimensional structure of the two isomers would look something like this:

I expect the structure on the left to be rather stable as it is, in the way I drew it, minimising steric interactions. Any other puckering would can either be traced back to that one by rotation or mirroring or put the two methyl groups into axial position which is sterically undesirable. Thus, we can expect only one significant species in solution.

This species should have $C_\mathrm{2v}$ symmetry if I didn’t miss anything. Using the principal axis — the one through the centre of the cyclobutane ring — we can identify four sets of homotopic hydrogens. We therefore expect four signals in the proton NMR spectrum.

Coupling constants, especially those along the ring, should be clearly visible and identifyable (if not confounded due to signal overlap).

The species on the right is less rigid. It can interconvert easily between different puckered states and in each one there will be one methyl group in axial position and one in equatorial position. Still, all structures can be traced back to the one I’ve drawn through mirroring and rotation unless I missed something. So here again, we have only one set of signals at all temperatures.

This species has a lower symmetry: $C_\mathrm s$. There is no rotation symmetry axis, so none of the ring protons are actually homotopic. Transforming one onto another is only possible by mirroring, hence they are enantiotopic. Since the methyl groups and the tertiary hydrogens are also non-identical, I expect a total of six signals at low temperature. However, repuckering would transform the axial into the equatorial methyl group and vice-versa. Therefore, I expect only a set of three signals to be visible at room temperature due to rapid interconversion.

These three signals would be:

the methyl group
the tertiary proton
the set of secondary protons

Since the protons rapidly exchange positions, I would expect the coupling constants at room temperature to be averaged. At very low temperatures, after freezing the rearrangements, these should become distinct.

So yes, rigidness of a ring due to stereochemistry does affect the number and type of signals. cis-1,3-dimethylcyclobutane is very rigid and unlikely to transform from one arrangement to another. This leads to a clean, full set of signals. trans-1,3-dimethylcyclobutane, on the other hand, interconverts rapidly between different orientations, mapping the otherwise non-identical methyl groups and protons onto each other. It therefore shows less signals than it should at low temperature.

A similar effect can be observed in cyclohexane and derivatives, where cooling to low temperatures also freezes the ring flip and causes the signals of axial and equatorial protons to be distinct.


UV-Visible Spectroscopy in the analysis of sodium chloride in potato chips Here is the question and answer out of an exam paper:

Firstly, I thought UV-Visible can also use radiation in the visible spectrum. Also when analyzing sodium chloride (a molecule), then UV-Visible would be more appropriate than AAS, because AAS would be used for just Na (Sodium). Would I be wrong to have said UV-Visible?

NaCl solution as you know is almost transparent in the VIS region, UV-VIS spectroscopy can be used to determinate concentration of soluble salt see J. Phys. Chem. A 2008, 112, 2242-2247 however UV part of the spectra is used instead of VIS part, so this does not fit the restrictions of your question.

Surely AAS has a greater sensibility and so is the best technique to use in this case.


Rearrangement of a bicyclic dienone under acidic conditions Find the major product of the following reaction.

Attempt:

First step would be the formation of oxonium ion. After that, there are two possibilities. Either the upper double bond or the lower bond will shift and create a carbocation (A and B)

In product A there are two possible migrations. The better migrating group is the carbon which is a part of the second ring. But it can't migrate because of formation of a bridged carbocation. So methyl group migrates.

If B carbocation is formed, then the better migrating group can attack the carbocation.

  My teacher told that the product A is wrong because a weaker migrating group is acting while in B, the stronger migrating group is acting. Is this correct? Because B involves ring contraction and more steps.

Your analysis of the possible products is excellent. I would consider both of the products to be major. Waring and coworkers studied this problem with a slightly different substrate (an extra methyl group at the 1-position, next to the carbonyl on the 'bottom' of the molecule). Under aqueous acidic conditions (sulfuric acid), the corresponding products were obtained in ~25% (mechanism A) and ~40% yield (product B). Only when run with sulfuric acid in acetic anhydride did mechanism B truly dominate (~90% vs 5% mechanism A).

There is actually a third pathway, where the methyl group shifts from ring juncture to ring juncture, and then goes through a mechanism similar to B. For the dienone in the question, the product is indistinguishable from mechanism B, but for the dienone studied in the publication, this gave another product in ~30% yield. I suspect that for the dienone in the question, the total yield of product B would be ~70%. That makes it appear as if mechanism B is dominant, but it is really two similar pathways operating that happen to give the same product.


Simulation of Miller-Urey experience The Miller-Urey experiment shows that it easy in a primordial atmosphere
similar to early earth to generate amino acids,my question: is it theoretically possible to simulate this experience on any computer ?

Theoretically you can simulate any reaction an a computer, provided it is big enough of course.

Practically however, this should be close to impossible. Here are some thoughts on how to approach that and their related problems.

First, one would have to model a complex reaction mixture including the educts $\ce{H2O}$, $\ce{CH4}$, $\ce{NH3}$, and $\ce{H2}$. Simulating those alone, without considering reactions, should be possible using molecular mechanics. This should scale up to a couple of thousands of these simple molecules.

Modelling all the possible reactions requires quantum mechanical approaches. But this does not scale anywhere close to macroscopic quantities.

One could model all the individual possible reactions one at a time. Then it is possible to compare how fast they are and how much energy is required. From that one could deduce how likely a certain reaction is, and feed this into the molecular mechanics calculation mentioned above.

First one would have to figure out all possible product of the reaction mixture, including all intermediates. This alone should yield a huge number of candidate compounds. But maybe one could try to focus just on a single product and one path there. Based on experimental results, pick a target compound of interest and a possible reaction path/mechanism. Experiments should give information on both, products and intermediates.

But even simulating a predefined reaction path (i.e. the educts, intermediates and final product) is a challenge. Different relative positions of the atoms in a reaction need to be considered (the energy landscape along the reaction coordinate). Ab initio molecular dynamics (AIMD) is an approach to do that. But there are still a number of approximations involved. This is costly and of limited accuracy.

We also need to consider the spark from the electrodes. This will lead to electronically excited state and ionic intermediates. This adds quite a lot of complexity to the already difficult task.

In summary, tackling the whole thing at once is impossible. One would have to break it down into many small steps, maybe then it becomes feasible to focus only on very specific aspects.


Creating 1L of 1000 ppb stock solution of Pb2+ I’m testing on samples that require a concentration of 1000 ppb (or 1 mg/L) of Pb 2+. My first thought was to simply dissolve 1 mg of Lead nitrate in 1 L of deionized distilled water to create 1 L of stock solution at the desired 1000 ppb (1 mg/L) concentration.

Is the process as simple as that, or is 1 mg proportionally too small a mass of solute to dissolve in 1 L of distilled water? Should I create a more concentrated stock solution and dilute it to 1000 ppb by adding requisite amounts of water? What concentration stock solution would be recommended in this case?

All the best!

Making the solution in a single step is inaccurate up to impossible.

Try to find a crystal of lead nitrate with the mass close enough to $\pu{1 mg}$
Try to weight up $\pu{1 mg}$ on scales with good enough accuracy.
Try to make such $\pu{1 ppm}$ solution stable enough.

The usual thing is to create a concentrated stock solution ( e.g $\pu{1000 ppm}$ and creating the working solution in 1 (1000 times) or 2 step ( 20x50 times) dilution whenever needed.

There is need to consider stabilizing by an acid, to prevent hydrolysis and adsorption. Very diluted solutions, like for AAS, are often stored in PE instead of laboratory glass to prevent ion Vs glass surface interactions.


When to use iso and sec while naming compounds in alkyl halides? I am having confusion on when to use iso- or sec- while naming the alkyl halides?

iso- is used when the all carbons form a straight chain except one methyl at "2nd Carbon", like 2-methyl-R (where R refers to straight chain alkane)

And, sec- is used when any functional group is linked to a secondary carbon.


Why is ATP, but not ADP or AMP, a sole source of energy? Why are three phosphate groups (such as in ATP) required for release of energy, but not two (ADP) or one (AMP) phosphates? How does breaking the phosphate bond release energy?

Let me add to iad22agp's excellent answer.

Often times in introductory biochemistry classes, the difference between phosphoanhydride and phosphate ester bonds is not adequately explained.

ATP (and ADP) contain phosphoanhydride bonds.  In these bonds, two phosphoric acid groups are condensed into one:

$\ce{R~-H2PO3 + R'~-H2PO3 -> R-HO2P-O-PO2H-R' + H2O}$

In contrast, AMP and many other phosphate-containing biomolecules such as glucose-6-phosphate glyceraldehyde-3-phosphate, contain phosphate ester bonds, where a phosphate group is condensed with an alcohol:

$\ce{R~-H2PO3 + R'OH -> R-HPO2-O-R' + H2O}$

Just as "regular" (i.e. carboxylate) esters like ethyl acetate are more stable to hydrolysis than carboxylic anhydrides (e.g. acetic anhydride), phosphate esters are much more stable and "lower energy" than phosphoanhydride bonds.  They don't have enough energy to power many of the key reactions of biosynthesis.  So that explains why AMP isn't a good energy source.

(By the way, mixed phosphate-carboxylate anhydrides are as high or higher in energy as phosphoanhydride bonds.  Examples from biochemistry include acetyl phosphate or phosphoenolpyruvate.)

So ATP has two phosphoanhydride bonds.  And as iad22agp mentioned, the electrophilic nature of the central P atom makes the outermost phosphoanhydride bond especially prone to hydrolysis.  However, ADP still contains a high-energy phosphoanhydride bond.  Consider this reaction:

$\ce{ATP + AMP <=> 2 ADP}$

The equilibrium constant for this reaction is about $2.82$, meaning the rightward "forward" direction is favored, but not by a whole lot.  So the "extra" energy of ATP's particularly high-energy phosphoanhydride bond is not really that much higher than that of ADP's regular run-of-the-mill phosphoanhydride bond.

There are somewhat rare examples where ADP powers key metabolic reactions but ATP does not.  That shows it is nearly as good an energy source as ATP.

The reason that ATP is so prevalent is probably because of evolutionary reasons.  Also, another trick that enzymes used with ATP is sometimes to hydrolyze the interior phosphoanhydride bond, leading to AMP and pyrophosphate.  Because pyrophosphatase enzymes quickly degrade pyrophosphate to two free ortho-phosphate  groups, this trick effectively makes what would have been a reversible reaction effectively irreversible.  That trick doesn't work with ADP because it has only one phosphoanhydride bond.


Can a strong electrolyte be sparingly soluble? A strong electrolyte is an electrolyte that completely, or almost completely, ionizes or dissociates in a solution.
My book has a question:

  In a saturated solution of sparingly soluble strong electrolyte $\ce{AgIO3}$ the equilibrium with sets in is $\ce{AgIO3→Ag+ + IO3-}$

Is the above statement wrong or is there anything as sparingly soluble strong electrolyte?

It means that by just adding it to water wont make it dissociate into ions(sparingly soluble) but if current is passed through its solution, it will easily dissociate and give ions. These cases look rare.

But Curt.F's reasoning looks better :

  In the low concentration regime where the salt is "soluble", it completely dissociates, but that the solubility limit is just low.


Why aren't all bond lengths equal in cyclobutadiene? even when there is a possibility of conjugation why aren't bond lengths equal?

Anti-aromatic compounds have no delocalization of electrons hence no resonance. To check aromatic character we use huckel's rule of (4n+2)π e- present in a conjugated closed loop. An anti aromatic compound would have only 4nπ electrons which is the case with above mentioned compound . Hence the bond lengths are different.


Entropy change during irreversible process Entropy change is defined as $\Delta S = \int \frac{\delta Q_{rev}}{T} $, where $Q_{rev}$ is heat transferred through a reversible process.  I’m a little confused how this works, so I have a couple of related questions which I hope will clarify what’s going on.

If a reversible process is defined as $\Delta S = 0$ then how can this integral be non-zero for a reversible process?  Or is it that just $\Delta S_{Univ} = 0 = \Delta S_{sys} + \Delta S_{surr}$ and therefore $\Delta S_{sys} = -\Delta S_{surr} \iff T_{sys} = T_{surr} $, in which case

How would one calculate it for an irreversible process, in particular, some heat, Q, transferred across a finite temperature difference?
I would try (considering isothermal transfer between infinite temperature reservoirs)
$$ \Delta S_{Univ} = Q \big( \frac{1}{T_{sys}} - \frac{1}{T_{surr}} \big) $$
But this is no longer reversible, so I need to do something else?

Since entropy is a state function, is the entropy change through any process that is irreversible, but between equilibrium states, equivalent to $\int \frac{\delta Q}{T}$ through a reversible process?

Is this equivalent to pretending I’m doing 2 reversible transfers to some fictitious bodies, the first taking Q reversibly from the system to the fictitious body, both at $T_{sys}$, and then magically dropping the fictitious body to $T_{surr}$  and transferring the heat Q to the surroundings?
If so, why use $\delta Q$ rather than $dQ$?  My understanding was that $\delta Q$ was for path-dependent processes, but there is only one fixed (adiabatic) path for reversible transfer?

Sorry for yet another entropy question, thanks in advance!

The answers to your first and third questions are "yes."  Here is a tutorial on how to calculate the change in entropy of the system or the surroundings for an irreversible process:  https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/. The reference also provides worked examples.  Note that, in considering entropy changes for both the system and the surroundings from an irreversible process, one needs to first separate the system from the surroundings and then subject each of them separately to a reversible path.  These separate reversible paths do not have to match.
It simplifies things a little to think about how entropy of a system can change.  There are only two ways:  entropy transfer across the boundary of the system by heat flow in and out at the boundary temperature, and entropy generation (due to irreversibility) within the system.  In a reversible process, only the first mechanism is present.  In both cases, the transfer across the boundary is $\int{dq/T_B}$ where $T_B$ is the boundary temperature.
Regarding questions 2 and 4 (involving ideal constant temperature reservoirs), these are somewhat limiting cases.  Here, if you imagine a very thin conductive slab between the two reservoirs (with high temperature at one face of the slab and low temperature at the other face), then for the irreversible process involving the transfer of heat between the reservoirs, all the entropy generation would take place within the intervening slab, while the reservoirs would each separately experience only entropy transfer with the slab.  In the limit where the slab becomes infinitesimally thin, we would obtain the system you have envisioned, but all the entropy would still have been generated at the interface, and each reservoir would only experience entropy exchange with the interface.


How can Ellingham diagrams be linked to Gibbs energies of formation? My textbook1 quotes the following as the limitation of Ellingham diagram:

  The interpretation of $\Delta G_\mathrm{f}^\circ$(standard Gibbs free energy of formation) is based on $K$ ($\Delta G_\mathrm{f}^\circ = -RT \ln K$).Thus it is presumed that the reactants and products are in equilibrium.

Definition of Standard Gibbs free energy of formation:

  The standard Gibbs free energy of formation of a compound is the change of Gibbs free energy that accompanies the formation of $\pu{1 mol}$ of a substance in its standard state from its constituent elements in their standard states (the most stable form of the element at $\pu{1 bar}$ of pressure and the specified temperature, usually $\pu{298.15 K}$ or $\pu{25 ^\circ C}$).

But in Ellingham diagram  reactants are normalized to consume one mole of oxygen having partial pressure of $\pu{1 atm}$. Thus products formed are not one mole. So how can it be a graph of Standard Gibbs free energy of formation and how this limitation even applies?

1. NCERT Chemistry Textbook for Class XII - Part I, Chapter 6, page 152

The Ellingham diagram doesn't actually use molar Gibbs energies of formation $\Delta G_\mathrm{f}^\circ$ per se; it is more accurate to say that it uses molar Gibbs energies of reaction $\Delta G_\mathrm{r}^\circ$. The difference is that the formation energy is only relevant to one specific chemical equation, for example:

$$\ce{Ca + 1/2O2 -> CaO} \qquad \qquad \Delta G_\mathrm{r}^\circ = \Delta G_\mathrm{f}^\circ(\ce{CaO})$$

in which the stoichiometric coefficient of CaO is equal to 1.** On the other hand, for any (balanced) equation with any stoichiometric coefficients, it is valid to define a Gibbs energy of reaction:

$$\ce{2Ca + O2 -> 2CaO} \qquad \qquad \Delta G_\mathrm{r}^\circ = 2\times \Delta G_\mathrm{f}^\circ(\ce{CaO})$$

which is related to the energy of formation, but is not the same thing, as evidenced by the factor of 2.

In the Ellingham diagram, every reaction has the same stoichiometric coefficient for $\ce{O2}$, which is typically 1. This is needed to make sure that different reactions are comparable. Let's say, for example, you want to see whether the reaction

$$\ce{C + 2CaO -> CO2 + 2Ca}$$

is feasible. This is done by checking the sign of $\Delta G_\mathrm{r}^\circ$: if it is negative, then the reaction is feasible, and vice versa. The point is that this $\Delta G_\mathrm{r}^\circ$ can be calculated by subtracting two reactions together:

$$\begin{align}
\ce{C + O2 &-> CO2} & \Delta G_\mathrm{r}^\circ &= c_1 = \Delta G_\mathrm{f}^\circ(\ce{CO2}) \\
\ce{2Ca + O2 &-> 2CaO} & \Delta G_\mathrm{r}^\circ &= c_2 = 2 \times \Delta G_\mathrm{f}^\circ(\ce{CaO}) \\ \hline
\ce{C + 2CaO &-> 2Ca + CO2} & \Delta G_\mathrm{r}^\circ &= c_1 - c_2 \\
\end{align}$$

but these two equations add up nicely only if the coefficients of $\ce{O2}$ in both equations are the same. What the Ellingham diagram does is to plot the Gibbs energies of reaction, $c_1$ and $c_2$: if $c_1 < c_2$, then the reaction is feasible. It doesn't plot the Gibbs energies of formation, because comparing those wouldn't tell us anything about the sign of $c_1 - c_2$.

As a final remark, note also that the equation

$$\Delta G_\mathrm{r}^\circ = -RT \ln K$$

holds true for any reaction, whether or not it actually corresponds to a formation reaction.

** Having a stoichiometric coefficient equal to $x$ does not mean the same thing as $x$ moles of the compound are produced in the reaction. The coefficient is purely a mathematical expression which tells us the stoichiometric relationship between different species in the reaction. It does not correspond to a real-life reaction, where a defined quantity of reactant is added to a defined quantity of product. To illustrate this, let's say you go to a lab and mix 0.4 mol of HCl to 0.4 mol of NaOH. You're asked to write a balanced equation for this. You can write

$$\ce{0.4 HCl + 0.4 NaOH -> 0.4 NaCl + 0.4 H2O,}$$

and that would be correct, but it is hardly the only correct possibility: the more conventional

$$\ce{HCl + NaOH -> NaCl + H2O}$$

is equally correct, even though the stoichiometric coefficients (1 in all cases) do not match up with the actual amount of substance used in the reaction (0.4 mol). Note also that the units are different: stoichiometric coefficients are dimensionless, but amount of substance is measured in moles. The difference is subtle, but one well worth pondering about, as mixing these two up can lead to a lot of misconceptions in thermodynamics.


What is the structure of [Co(NH3)5SO4]+? What is the structure of $\ce{[Co(NH3)5SO4]+}$? If sulfate is a bidentate ligand, then the coordination number of cobalt would become 7. How is this possible?

There's a certain amount of misinformation here:

  sulfate is a bidentate ligand

Sulfate is actually a flexidentate ligand, which is a term used to describe a polydentate ligand which is found to show different denticity in different complexes. Sulfate can act both as mono- or a bidentate ligand. Remember, just because it has two donor atoms doesn't make it a bidentate ligand.

The occurrence of complexes where sulfate acts as a bidentate ring would also be rare owing to the fact that four-membered rings are strained. This page lists $\ce{[Pt(SO4)(P(C6H5)3)2]}$ as one example of such an occurrence.

  coordination number of cobalt would become 7. How is this possible?

If this implies that coordination numbers of 7 aren't possible, it isn't true. Coordination numbers of 3, 5, 7, 8-12, 14 are possible, albeit such compounds are quite rare. See this Wikipedia page for some examples.

Thus, the structure of the complex $\ce{[Co(NH3)5SO4]+}$ should be octahedral with cobalt showing a coordination number of 6.


how to calculate mols Ag from AgNO3 solution Sorry for the very basic question but its been a very long time since I've taken chemistry and I'm having difficulty solving this problem.

If I titrate something with 1 molar AgNO3, adding 50 ml of AgNO3 solution... How would I calculate how many mols of Ag that equates to?

What I have so far is that 1 M means that there is 0.05 mol of AgNO3 in that 50 ml sample (M = mols/L). So then how many mols of Ag are there in 0.05 mol of AgNO3? Is there just 0.05 mols of Ag in AGNO3 or do I need to somehow take into consideration the proportion of the rest of the elements?

Thank you!!!

If I titrate something with 1 molar AgNO3, adding 50 ml of AgNO3 solution... How would I calculate how many mols of Ag that equates to?

A 1 molar solution of $\ce{AgNO3}$ is 1 mole per liter. 50 ml is 0.050 liters

$ 1 \dfrac{\text{mole }\ce{AgNO3}}{\text{liter}}\times0.050\text{ liters} = 0.050 \text{ moles }\ce{AgNO3}$

  So then how many mols of Ag are there in 0.05 mol of AgNO3? Is there just 0.05 mols of Ag in AGNO3 or do I need to somehow take into consideration the proportion of the rest of the elements?

$0.050 \text{ moles }\ce{AgNO3}$ is correct since there is only one atom of silver per "molecule" of $\ce{AgNO3}$. There would also only be 0.050 moles of N atoms, but 0.150 moles of O atoms in 0.050 moles of $\ce{AgNO3}$.