What functional groups can 900-2500nm IR spectrometers detect? 3700 cm-1 corresponds to about 2,700 nm
650 cm-1 corresponds to about 15,000 nm.
Many cheap IR spectrometers ($7000 or so like the ones at Stellarnet) has range of 900nm to 2400nm (or 11,111 cm-1 to 4166 cm-1). In the following table or list of the molecules functional groups, the range is between 2700nm to 15,000 nm (3700 cm-1 and 650 cm-1). Outside of the range of cheap IR spectrometers. So what functional group or stuff can you see with IR spectrometers that has range below 2400nm (above 4000 cm-1)?
https://www.sigmaaldrich.com/technical-documents/articles/biology/ir-spectrum-table.html

see the link above for complete list

You recall vibrations IR spectrometers detect in the range of $4000\dots\pu{800 cm^{-1}}$ typically are either fundamental vibrations and combinations.  Tuning for shorter wavelengths / higher wavenumbers however renders recording absorptions about overtones instead of fundamentals more likely.  On occasion, you may encounter this part of the electromagnetic spectrum accessed by the larger UV-Vis-NIR spectrometers (e.g., PerkinElmer Lambda series) with a spectral range from $175\dots\pu{3300 nm}$); here, NIR is the label about near infrared.
The typical overtone is less probable than the fundamental, thus -- in parlance of Beer-Lambert's law, $\mbox{Abs} = \varepsilon \cdot c \cdot d$ -- the molecular absorption coefficient $\varepsilon$ is smaller.  While this may appear on first sight as a drawback (at same analyte concentration $c$ and  pathlength $d$, this lowers $\mbox{Abs}$ recorded), it very well may become an advantage if you want to record absorption bands about matter in high concentration, e.g., to check the quality of grain and crops in general  where NIR spectroscopy is used as such.
IR spectroscopy has a long tradition of chemometrics, e.g. in the on-line process control of petroleum fractions where whole spectra are used to determine concentrations of components (e.g., keyword principal component analysis) allowing to map a sample for its chemical composition as tool in histology (a review).  Again, tuning the senors to record NIR instead of the range of $4000\dots\pu{800 cm^{-1}}$ allows to see some things literally in a different light (e.g., Landsat satellites, farming).

(credit)


Why can UV light initiate a reaction between hydrogen and chlorine gas? Can someone explain me how does UV light help combine chloride gas and hydrogen to produce hydrochloric acid?

$$\ce{Cl2(g) + H2(g) -> 2HCl(g)}$$

Before going into the mechanism of this reaction, I suggest you look up free radical mechanism, as this reaction takes place through that.

$\ce{Cl-Cl}$ bond in $\ce{Cl2}$ is weak enough to be broken by mere UV rays (present in sunlight), and hence they undergo homolytic cleavage (the resultant products are $\ce{Cl}$ atoms, not ions) to form two $\ce{Cl}$ free radicals. Now these $\ce{Cl}$ free radicals are extremely reactive (due to one free electron that it can share in a covalent bond) and hence attack hydrogen gas.

Now is the time you consider energy lowering. $\ce{H-Cl}$ bond is lot more stronger than $\ce{H2}$, and hence the formation of $\ce{H-Cl}$ bond formation will be favored instead of $\ce{H-H}$. Therefore, $\ce{Cl}$ free radical attacks $\ce{H2}$ gas to break the bond and form $\ce{HCl}$. The other $\ce{H}$ atom hence liberated forms a bond with the other $\ce{Cl}$ free radical.


Finding chair and boat transition states during ring-forming reactions Reactions that form 6 membered rings most commonly go via either boat or chair transition states but the products themselves do not necessarily adopt the same conformation as the TS.

I'm attempting to model a cyclisation in which an amine attacks a pi system to form a new ring, giving rise to two diasteromers depending on which face of the pi system is attacked.

The products have been explored (MM conformational search then QM optimisation of all low energy conformers) and are both chairs (the boat being significantly high in energy as to be essentially unpopulated).

In the reaction of interest there is precedence in the literature from computational studies that the reactions can go via either boats or chairs (depending on the exact structure) and so I would ideally like to find both possible transition states leading to each diastereomer.

My initial plan was to start with the product and work backwards (the transition state is thought to be late, meaning it should resemble the product) , elongating the bond by x angstroms (optimising at each step whilst constraining the distance) and then looking for a 'maximum' from which I would then optimise to a TS however this doesn't provide  a method for me to find the boat TS's.

How does one go about finding all possible transition states? Is it valid for me to find a high energy boat conformation of the product and apply the same bond lengthening exercise?

As you have figured out, this is a non-trivial problem.

It's okay to find a boat conformation that might be a bit higher in energy and see if it can map to a TS. Keep in mind that for 6-membered rings, there are many conformations beyond chair and boat. A half-chair, for example, might be a useful test.

In general, for 6-membered rings, using ring-puckering searches are useful. Calculate the plane of best-fit for the 6-membered ring and displace atoms up or down -- this is similar to sampling the out-of-plane vibrational modes of the ring.

So the ideal situation for the TS would be to consider a set of conformations, that might be higher in energy than the reactant and product. After all, it's a transition state. Then use a nudged elastic band or similar method (your bond-lengthening) to look for the real TS.

Good luck!


Using the quasi-steady assumption for reaction-diffusion equations I have a question regarding an equation for changes in chemical concentration. The chemical concentration we are looking at changes due to diffusion, decay and production of the chemical. Since the diffusion happens so much faster than the other two, it is given that we can you a so-called quasi-steady assumption. This then gives the equation:

$$D \nabla^2c-a \cdot c+b\cdot c=0$$

where $a$ and $b$ are rates of decay and production, respectively. $c$ is the chemical concentration and D is the diffusion coefficient.

Is it correct that the reason we only have a derivative expression for the diffusion because it happens so fast relative to the other two, and therefore we can assume the chemical concentration for the two other processes to remain constant (using the quasi steady assumption), or is that not a correct reasoning for why the equation has the simple expression for decay and production, while we need a more complex expression for diffusion?

This is an example of a reaction-diffusion system where the participating species both react as well as diffuse. Such systems can be described using partial differential equations.

The mathematical model describes how the system changes with time. So we have $$\frac{\mathrm dC}{\mathrm dt}=f(C)$$

where $C$ is a vector with each element denoting the concentration a species ($C_i$).

At quasi-steady state the rates of some of the species will remain unchanged over time.

$$\frac{\mathrm dC_i}{\mathrm dt}=0$$

The reaction part usually comes from mass action kinetic models (which gives you rates for zeroth/first/second order reactions).

The diffusion part comes from Fick's laws.

$$\frac{\partial C}{\partial t}=D\nabla^2C(x,y,z,t)$$

Here the $\nabla^2$ denotes spatial double derivatives $\large\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}$

which kind of describe how the different species would be distributed in space. The reaction-diffusion equation is a sum of the reaction and the diffusion parts. At quasi-steady state, the concentration of a given species at any given point in the space would remain unchanged over time. This only sets the time derivative to zero, as I previously mentioned.

The time derivative denotes the net accumulation rate of a species. In the absence of diffusion, it will simply be the sum of rates of all reactions leading to formation of the species minus the sum of rates of all reactions causing removal of the species (principle of mass balance).

In the absence of reactions the diffusion terms will describe how the spatial pattern of the species will change over time.

At (quasi) steady state, the time derivative goes to zero which means that all the reaction and diffusion terms balance each other out such that there is no change in the concentration of the corresponding species over time.


Why is p-toluidine more basic than m-toluidine? https://en.wikipedia.org/wiki/Toluidine

In m-toluidine the $\ce{CH3}$ group is closer to $\ce{N}$ so shouldn't that increase the electron density on $\ce{N}$ and make it a stronger base?

In p-toluidine you are not actually considering the hyper conjugating structures due to which the compound becomes more basic than m-toluidine which only has it's +I effect and I would like to suggest you that as far as I know resonance effect "usually" overpowers the other effect!


13C NMR peak location in butanone The above is the correct $ ^{13}C $ NMR spectrum of butanone.

In the $ ^{13}C $ NMR spectrum of butanone, I figure that the peak locations of the first and third carbons should be switched (27.3, 35.2). Shouldn't the third carbon be shifted upfield by the terminal methyl group and therefore have a "smaller ppm" than the first carbon?

No, substitution by an alkyl group produces a small downfield shift in both proton and carbon nmr. Therefore, in a typical hydrocarbon a secondary carbon will be downfield of a primary carbon.  The assignments in your figure are correct.

A possible explanation for the downfield shift of a carbon atom when a proton is replaced with another carbon atom is as follows.  Hydrogen is more electropositive than carbon, therefore, in a C-H bond a significant amount of the electron density in the bond resides around the carbon.  When we replace the electropositive hydrogen with a relatively more electronegative carbon, then there is less electron density around the central carbon atom.  This reduction in electron density deshields the carbon nucleus and shifts it downfield.  Every time another hydrogen is replaced by carbon a further downfield shift occurs.


Proof of Brillouin's theorem using the diagonality of the Fock matrix I'm working my way through Szabo and Ostlund's Modern Quantum Chemistry. I am having trouble understanding the proof and significance of Brillouin's theorem. According to the book:

  Now, by definition, solving the Hartree-Fock eigenvalue problem requires the off-diagonal elements to satisfy $\langle \chi_i|f|\chi_j\rangle = 0,  (i \neq j)$.[1]

However, isn't the Fock matrix only diagonal because we used "nice enough" spin orbitals that we constructed by unitary rotations? That is, for any other choice of spin orbitals, the Fock matrix is not diagonal but is still a perfectly valid matrix with a mostly-eigenvalue statement, right?

What am I missing here?

[1] A. Szabó, N. Ostlund, Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory, Dover Publ., 1989, p. 129.

Your question actually contains two linked inquiries. The proof itself depends on the nature of the Fock-operator to which I'll have a few words afterwards. For the proof itself it is only important, that the Fock matrix is diagonal. Let's restate Brillouin's theorem:

In case of the Hartree-Fock (HF) reference determinant $\Phi_0$, the Hamiltonian matrix element between it and any singly excited determinant $\Phi_a^r$ reduces to zero.
$%
\newcommand{\ll}{\left\langle}\newcommand{\rr}{\right\rangle}
\newcommand{\lb}{\left|}\newcommand{\rb}{\right|}
\newcommand{\op}[1]{\mathbf{#1}}$
$$\langle\Phi_0|\op{H}_\mathrm{el}|\Phi_a^r\rangle = 0$$

The proof of this then rather simple:
\begin{align}
&& E &= \ll \Phi_0 \rb \op{H}_\mathrm{el} \lb \Phi_a^r\rr\\
&& E &= \ll \phi_a \rb \op{H}^\mathrm{c} \lb \phi_r \rr +
     \sum_j^{N/2} \ll \phi_a\rb \left(2\op{J}_j -\op{K}_j \right) \lb \phi_r \rr\\
&& E &= \ll \phi_a \rb \op{F}_j \lb \phi_r\rr\\
\text{with}&& \op{F}_i\phi_i &= \varepsilon_i\phi_i\\
\text{and}&& \ll \phi_j \rb \op{F}_i \lb \phi_i \rr
          &= \varepsilon_i \ll \phi_j | \phi_i \rr = \varepsilon_i\delta_{ij} \\
&& E &=\varepsilon_j\delta_{ar} = 0
\end{align}

The direct consequence and therefore significance of this is that for any post-HF method the singly excited determinants only mix indirectly with the ground state. This simplifies a couple of the integral evaluations.

The remaining question to solve is why the Fock matrix needs to be diagonal. Even though the equation
$$\op{F}_i\phi_i = \varepsilon_i\phi_i
     \tag{1}\label{fock-pseudo}$$
suggests and eigenvalue problem, it is not. Remember the definition of the Fock operator and the operators contained
\begin{align}
  && \op{F}_i &= \op{H}^\mathrm{c} + \sum_j (\op{J}_j - \op{K}_j),\\
\text{with}&&
     \op{J}_j\lb \phi_i\rr &=
       \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
       \lb \phi_j(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr,\\
\text{and}&&
     \op{K}_j\lb \phi_i\rr &=
       \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
       \lb \phi_i(\op{x}_1) \rr \lb \phi_j(\op{x}_2) \rr.
\end{align}
As you can see, the "one-electron" Fock operator depends on the solution of all "one-electron" Fock operators (ref. Szabó-Ostlund p. 115). The only way to achieve self-consistency and therefore $\eqref{fock-pseudo}$ is to use unitary transformation and diagonalise the Lagrange multipliers from
$$\sum_j \lambda_{ij} \lb \varphi_j \rr =
  \op{F}_i \lb \varphi_i \rr.\tag{2}\label{fock-lagrange}$$

In other words, the Fock operator is not actually associated with the total energy, but only with the variation of it (due to the Lagrange minimisation). To know one orbital you have to know all orbitals, and to know one orbital energy, you have to know all orbital energies. The Hamilton operator is not the sum of all Fock operators, and the total HF energy is not the sum of all orbital energies.

Therefore the Fock operator is also dependent only on the total wave function and any unitary transformations of the canonical orbitals will retain the total energy. However, the canonical orbitals are the only ones for which $\eqref{fock-pseudo}$ holds.


How does chloroform degrade nitrile gloves? Many of us have experienced the failure of nitrile gloves when exposed to chloroform. What's going on at a mechanistic level when this occurs?

I would guess that the chloroform dissolves some of the polymer into its constituent monomers, but I've never heard anything more definite than that. Is it a similar mechanism to what happens when chloroform is left too long in a plastic bottle?

Nitrile gloves are made of nitrile rubber, or poly(butadiene/acrylonitrile). This polymer is highly soluble in chloroform, with some papers I found indicating that one can dissolve up to 18% in mass of nitrile butadiene rubber in chloroform. Moreover, it permeates easily through NBR, meaning we can expect the dissolution to be fast in addition to thermodynamically
favourable.

Finally, I would not expect the mechanism here to be any different from that of any polymer dissolution by a good solvent. The solvent will permeate through the polymer, intercalate between polymer chains, and solvate them (inducing swelling). Once solvated, the network of polymer chains looses its mechanical properties and they can fully separate. (I wish I could find a good existing illustration for that part, but I can't right now… If anyone can, feel free to edit!)


Cathodic protection of aluminium I am planning an experiment for a chemistry class about corrosion. I am going to show how different prevention methods affects the corrosion of iron and aluminium parts using bleach as the oxidizer.

One of the methods I plan on is protecting the parts using cathodic protection, the iron nail will be protected by aluminium but I don't know a good, easily available metal to protect the  aluminium stud. Any ideas?

You might want to look at this link which has a nice general discussion about cathodic protection, with an abbreviated activity series for the metals of interest in cathodic protection.  Typically, aluminum, magnesium and zinc are used as sacrificial anodes for cathodic protection.

Strictly speaking from a thermodynamic standpoint, one would suspect that magnesium would be needed to protect aluminum.  However aluminum boat manufacturers state that zinc can be used for protecting aluminum boats in salt water.

I'm completely unfamiliar with the die casting industry, but they use Al/Zn alloys which, if affordable, may provide an interesting pedagogical component to your corrosion activity (namely, what do scientists do when the properties of the elements on the periodic table don't quite fit what we have in mind).


Aluminum to prevent Iron corrosion I have heard of how zinc is used to prevent corrosion of iron. So why not use aluminium coat instead? After all Aluminum oxide forms a stable coat over the surface.

Aluminum, just like zinc, is already widely used to coat iron surfaces to prevent corrosion (in the elemental form, or in combination with other metals, like zinc). You can deposit aluminum on practically any surface (including plastics). On metals it is done by electrochemical deposition.

There are several companies working with this type of coating nowadays, including AlumiPlate (http://www.alumiplate.com/).


Niobium (V) Fluoride becoming sticky Today I was making a pastille of Niobium Pentafluoride powder. The procedure includes a 6 ton squeezer, precision wipes, silica gel, filter papers and acetone.

The first suspicious part was that on product details, $NbF_5$ was suppossed to be $5\mu m$ powder. Instead it looked like having hair(forgive my bad terminology but I am not a chemist, but a physicist and  I don't know how to describe it).

The procedure is the following.

Acetone and wipe papers$\rightarrow$clean squeezers parts(made of stainless steel)
Powder gets into the squeezer(using a funnel made of the filter paper) and gets pressed
The powder becomes a pastille.
It is stored in a plastic bottle, with Silica gel.

The filter paper is also used to fill the bottle with silica gel(as a funnel) as well as a way to store the pastille while filling the bottle with silica gel.

I have made several pastilles with the exact same recipe(i.e. $B,\;ZnF_2,\;LiF$ etc) and everything went fine. This powder however became very sticky. I have also noticed that while it's stored in the silica gel box, there are some drops inside the bottle.

So I have some questions... First of all why did it became sticky and secondly why the powder wasn't actually a powder?Is it typical for $NbF_5$?

From the properties block on the Wikipedia page for $\ce{NbF4}$:

Niobium (V) fluoride is hygroscopic - it absorbs water from the atmosphere. This hygroscopicity is your main problem. When the powder absorbs water, it probably is becoming sticky.

The added water is probably also providing the extra mobility the ions need to crystallize from the powder. Niobium (V) fluoride forms monoclinic crystals, which on a very small scale could look like hair or needles.


Variation of current with time in gel electrophoresis During gel electrophoresis, if the voltage of the power supply is kept constant, how would the current in the circuit change over time?

I was wondering whether the electrical resistance of the gel would change over time. There will be migration of ions present in the buffer and it will resist the change in pH. But how does the electrical resistance of the gel change? Can someone explain?

I think this might help you a bit:
https://www.researchgate.net/post/SDS_PAGE_should_be_run_at_constant_current_or_constant_voltage


What is the geometrical structure of Pb3O4? I'm confused by the structure of Pb3O4. There are some sites claiming that it is a six membered ring of alternating Lead and Oxygen atoms with another Oxygen atom in the center bonding with 2 out of the 3 Lead atoms.

There are other sources which claim that it is actually two 4 membered rings of alternating lead and oxygen atoms with a lead atom in common between the 2 rings.

Moreover, isn't Lead a metal? Then How can it combine with oxygen in such "bonds" whereas ionic bonds are non-directional?

Maybe below 3d structure of Pb3O4 will help?

3d structure of Pb3O4

Also in the below link of a pdf, just before the discussion section the paragraph clearly states that there are both ionic and covalent bonds present in Pb3O4.

REsearch Paper on Pb3O4 structure

Thus we can conclude that it is Tetragonal, tP28 .(as per the link provided by Maxw in the comment to your question)

Also metals can form covalent bonds. Refer to below question and its answers:
Metals forming covalent bonds


How many liters of hydrogen, H2, are needed to react with 10 liters of nitrogen gas in the reaction forming ammonia? How many liters of hydrogen, H2, are needed to react with 10 liters of nitrogen gas in the reaction forming ammonia?

$$\ce{3 H2(g) + N2(g) -> 2 NH3(g)}$$

My try:: Because we have 10 L of nitrogen gas, we have 10/22.4 moles = 0.446 moles of nitrogen gas, and thus need 0.446 * 3 = 1.338 moles of hydrogen gas. As such, we need 2.676 grams of hydrogen gas, or 2.676/0.09 = 29.8 L of hydrogen gas. However this is not correct. Can someone find my problem?

Note: This reaction occurs at STP.

You are on the right track. Dimensional Analysis is your friend:

$$\frac{10\,\text{L  $\ce{N2}$}}{1} \cdot \frac{1\,\text{mol  $\ce{N2}$}}{22.4 \,\text{L  $\ce{N2}$}}\cdot \frac{3\,\text{mol  $\ce{H2}$}}{1\,\text {mol  $\ce{N2}$}}\cdot \frac{22.4\,\text{L  $\ce{H2}$}}{1\,\text {mol  $\ce{H2}$}}=\,?\,\, \text{L $\ce{H2}$}$$

If you do this, you will realize that the quantities cancel exactly and give you a nice clean number.


Identify the type of structural isomerism Question:

My Approach:

Since both the compounds have the same principal functional group, they can't be functional isomers. Obviously they are different compounds. Now options A and C are ruled out. Further the substituent ketone in first compound and aldehyde in the second compound have the same locant, so the given compounds cannot be positional isomers. So the compounds must be Chain Isomers. Further, we can just say, as there is difference in the chain length, the compounds are chain isomers. But the answer is incorrect.

The solution in my book, for this problem states, since compounds have different functional groups ketone in first compound and aldehyde in the second compound, they are functional isomers.

Now, I am totally confused. Please explain how to attend this question. Is the solution incorrect?

Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. Constitutional or structural isomers are compounds with the same molecular formula but their structural formulas are different. They also called chain isomers because of the possibility of branching in carbon chains (e.g., n-butane and 2-methylpropane). Note that just the twisted versions of the original molecule can be mistaken by novices as structural or chain isomers (e.g., conformers).

Positional isomers: Positional isomers are constitutional isomers that have the same carbon skeleton and the same functional groups. They differ from each other in the location of the functional groups on or in the carbon chain. In other words, in position isomerism, the basic carbon skeleton of isomers remains unchanged, but important groups on the skeleton are moved around on it (e.g., 1-bromopentane, 2-bromopentane, and 3-bromopentane).

Functional isomers: Functional isomers are constitutional isomers bearing different functional groups (e.g., propanoic acid, methyl ethanoate, and 3-hydroxypropanal).

Accordingly, the answer given is correct: These are functional isomers, since each have at least one different functional group changed (keto vs aldehyde).


Predicting incomplete combustion? I understand the concept of incomplete combustion and that it occurs when oxygen is not a large enough supply to allow for complete combustion. My question is: how can we model combustion properly in terms of predicting the ratio of complete to incomplete combustion? Is it possible to use basic kinetic with reaction rate constants to model the rate of each reaction and then obviously model how the changing concentration from both affects each individual rate etc.

Before entering the kinetics realm, it is advisable to check the equilibrium. Besides the straightforward combustion reactions, you have two more, Boudouard reaction, which favors formation of $\ce{CO}$ from $\ce{CO_2}$ and Water-gas shift reaction.

The actual reactions which contribute to the equilibrium depend on the fuel. In case of pure carbon, you at least get rid of the Water-gas shift reaction.

You might need to solve the system iteratively, as the temperature of the equilibrium mixture is result of the combustion reactions, but also affects the position of equilibrium.


Why is tetrathiocyanatocobaltate(II) blue? Why is $\ce{Co(SCN)_{4}^{2-}}$ blue?

I simply can't predict the colors of such inorganic salts. I have learned that $\ce {Co^{2+}}$ salts are pink in color. Also, if the ligand is a strong field ligand ($\ce {SCN-}$ in this case), then the splitting is more and the color should shift to the red side...

Is my concept wrong?

Let us consider this qualitatively with crystal field theory (CFT). Co is a transition metal with $d^7$ configuration. CFT says that depending on the geometry of the metal complex, the d-orbital energies will split into different energy levels. If the ligands (in this case, the four $\ce{SCN-}$ ligands) align with certain orbitals, then the electrons in these orbitals will feel an increased repulsion due to the closeness of the ligands, and increase in energy. The orbitals in-between the aligned axes, will not feel this repulsion, and will not increase in energy. The splitting diagram might look something like this

There is an equation that relates the (splitting)
 energy to wavelength:

$$
\Delta E = \frac{hc}{\lambda}
$$

From this equation, we see that if the splitting energy increases, wavelength will have to decrease, i.e. be shorter than before. This means a shift towards the violet region. As you say, $\ce{SCN-}$ is a strong ligand, meaning it leads to a rather high $\Delta E$. So we could say that $\ce{Co(SCN)_4^{2-}}$ is blue due to the high energy gap between the d-orbitals. Electrons are excited from one of the lower orbitals up to one of the higher orbitals. Upon de-exciting, light is emitted with an energy equal to $\frac{hc}{\lambda}$. This happens to be in the blue/violet region.

Weaker ligands would split the energy levels less, and the complex's color would therefore have a color closer to the red part of the visible spectrum.

Did this make it a little clearer?


Simple Kinetics Clarification I just wanted to check my understanding of this. Given the reaction,

$$ (1) \; \ce{A->B} $$

the corresponding rate law is $ v=k[A] $

Also consider

$$ (2) \; \ce{A <=> B} $$

with forward rate law: $ v_1=k_1[A] $ and reverse: $ v_2=k_2[B] $. This makes the corresponding rate equation $ v=v_1-v_2=k_1[A]-k_2[B] $.

My question is in the difference between the two. I've often seen reactions expressed in both ways. In $ (1) $, is the rate law assuming the reverse reaction not to occur (or at least to be completely negligible? That's why I drew it with $ \ce{->} $ instead of $ \ce{<=>} $. Am I correct in my thinking? In actuality, is it always more correct to use the rate law of $ (2) $? The rate law for $ (1) $ could also correspond to initial reaction velocity, right? This is because there is no $ B $ present at the start of the reaction.

If this is correct, then the integrated rate equations will become substantially more complicated using rate laws of form $ (2) $. I can't use simple equations like $ [A]_t=[A]_0 e^{-kt} $, as this wouldn't take into account the reverse reaction.

Not all reactions can or should be represented with a double arrow.  The general-audience example is that you "can't unbake a cake".  You are right that there are instances that we assume that k2 is negligible (relative to k1) and therefore a reaction is treated as if it is reversible.

Here's a sample plot of how the concentrations of A for the irreversible and reversible reactions compare for k1 = k2 = 1.

No surprises here; the reversible reaction results in an equilibrium concentration of A. We can compare the concentration of A in each of the reactions as a function of kr:

Here I've plotted Log[k2] on the x-axis and the ratio of A in the irreversible reaction to the reversible reaction.  The different lines are at various times (in seconds).  I've kept k1 = 1 and the initial concentration 1 as well.  The two equations predict the same concentration of A in the limit of very small k2, and by very small I suspect this means at least a factor of 10^4 or more (not shown on the graph).

While you are correct that the integrated rate laws get much more complicated for reversible reactions, the one you are considering is still solvable in closed form.


Determining spin of metal complex Is there any way to determine the spin of $\ce{[Fe(OH)6]^{4-}}$ without looking at the spectrochemical series?

Are you asking experimentally or theoretically?
Theoretically speaking for general cases you should know what is the energy of the high-spin and low-spin configuration, and in given state, what is the approximate spin and vibrational entropy so you can estimate an equilibrium constant for given temperature. Now this either needs some calculation, or some sort of database from where you can make this estimations.

Experience (i.e. database or whatever series) says that water is a weak ligand, from that you generally guess that water complexes are generally in high spin state.

Experimentally you can determine spin state by measuring magnetic susceptibility, interpreting EPR, measuring 57Fe Mossbauer spectrum etc


How can iodine(I) cations be produced in this reaction if no additional acid is added? In laboratory, we had to synthesize an aromatic compound and identify it. The chemical reaction involved salicylamid, $\ce{NaI}$ and $\ce{NaClO}$. The solvent is ethanol.

First, we dissolved the salicylamid with the ethanol. Then we added the NaI. When the mixture was homogeneous, we let it cool down to 0 °C. Once the mixture reached that temperature, we quickly added the NaClO.

We figured out that for the salicylamid to react, we needed an $\ce{I+}$.
Here's how we think we can obtain the $\ce{I+}$:

$$\ce{NaI -> Na+ + I-}\tag{1}$$

$$\ce{NaClO -> Na+ + ClO-}\tag{2}$$

$$\ce{ClO- + I- + 2H+ -> I+ + Cl- + H2O}\tag{3}$$

It seems to be the easiest way to obtain the $\ce{I+}$. However, isn't it impossible for the ethanol to release an $\ce{H+}$? Isn't ethanol neutral? How can the ethanol reacts with the $\ce{NaI}$ and the $\ce{NaClO}$?

PS: Sorry if the terms may not be the correct ones. English isn't my first language.

Edit: Since ethanol is only the solvent and isn't involved in the reaction, how can we obtain an $\ce{I+}$?

Is it like this?

$$\ce{2NaI + NaClO -> NaCl + Na2O + 2I+}\tag{1}$$

There are multiple reasons why this works.

While ethanol may be a very weak acid, there should be a non-neglegible concentration of deprotonated ethanol molecules at any time which is enough to give a nonzero fraction of $\ce{I+}$.
As $\ce{I+}$ is consumed in the reaction, protons should again be released. Unfortunately, it is only one per $\ce{I+}$, so the process is not entirely catalytic. However:
Salicylamid is a phenol and phenols are acidic. The phenol will be deprotonated and supply even more excess protons.

You should also pretend you never wrote anything about $\ce{Na2O}$ in ethanolic solution. This will happen:

$$\ce{Na2O + 2 EtOH -> 2 NaOEt + H2O}$$


Derive expression for heat flow from oil bath to metal sphere A spherical ball of iron is put into a hot bath of oil. The iron ball is initially at a low temperature and heat transfer from the oil to the iron will increase the temperature of the iron ball.
Use the concept of heat transfer coefficients to express the heat flow (i.e. the quantity with units J/s) from the oil to the ball. Denote your variables according to the notation in Table 1. Notice that not all variables in Tab. 1 might show up in your expression. Variables not listed in Table 1 must not be used in your final expression.
Table 1: Notations
$$
\begin{array}{ll}
\hline
c_{p, f} & \text{ heat capacity oil} \\
c_{p, s} & \text{heat capacity iron} \\
D & \text{diameter iron ball} \\
h & \text{heat transfer coefﬁcient} \\
k_f & \text{thermal conductivity oil} \\
k_s & \text{thermal conductivity iron} \\
Q & \text{heat ﬂow oil to iron (J/s)} \\
R & \text{radius of the iron ball} \\
T_s & \text{temperature iron ball} \\
  & \text{(can be assumed homogeneous)} \\
T_0 & \text{bulk temperature oil} \\
ρ_s & \text{density iron} \\
ρ_f & \text{density oil} \\
\hline
\end{array}
$$

I'm really unsure of how to solve this problem. I know that the heat flow for a sphere with an inner and outer radius is
$$Q = \frac{4πkR_o}{1 - R_o/R_a}(T_0 - T_a),$$
but here I only have the radius of the whole sphere. So, how do I write the expression for the heat flow then?

Newton's law of cooling is

$$q = A h (T_0 - T_S )$$

With the area of the sphere being

$$A = 4 \pi R^2$$


Can a free radical be created by chemical reaction of non-radical species? Here's my thinking:

Atomic oxygen in its ground state is a free radical because there are two unpaired electrons in p-orbitals.
When species react new orbitals are formed, and perhaps a situation can occur where the amount of degenerate orbitals is changed by a chemical reaction. This would implicate the filling of these orbitals, following the aufbau principle and hunds rule of filling.
Therefore it is thinkable that two reactant species (i.e. not homolytic bond fission) which are not radicals, can react in such a way that the new electron orbitals are filled in such a way that all electrons are no longer paired => radical species formed from non-radical species.
Likewise it is conceivable that a radical species can react with a non-radical species (i.e. not an even number of radical species terminating eachother) to form only non-radical species.

Is my argument and results 3 & 4 correct? Can someone give an example of 3 and/or 4?

Edit: My main takeaway - when considering the feasibility of a radical creation/destruction, look at conservation of spin

As for #3, photoinduced electron transfer (PET) reactions do what you want.

Imagine that 2,4,6-triphenylpyrylium tetrafluoroborate ($\ce{P+BF4-}$) is irradiated at $\lambda = 350~\mathrm{nm}$ in the presence of an alkene and a nucleophile ($\ce{ROH}$).

In the excited singlet state, $\ce{^1P+^*}$ is a strong electron acceptor, that oxidizes the alkene to the corresponding radical cation an is itself turned to a pyranyl radical, $\ce{P^{\cdot}}$. (Here is the first radical.)

Then the alcohol adds to the alkene radical cation, the adduct deprotonates spontaneously. (Here is the second radical.)


On basic mercury salts $\ce{HgCO3.3HgO}$ and $\ce{HgO.Hg(NH2)Cl}$ are called basic mercury carbonate and basic mercury amidochloride, respectively. I found this thread on the former; I've only seen the latter as the product of a qualitative test in one of my textbooks, and Google yields nothing on it.
Is "basic metal salt" a general term for salts of the form $\ce{M_yB_x.M_2O_x}$(M the metal, B some anion with valency y, x the valency of the metal) or is it just for some compounds of mercury? If it's just for mercury, which ones more are there?
Are there any "acidic salt" analogues? If so, what general formula do they have?

There exist plenty of basic salts. For example,

Copper produces a "simple" copper carbonate $\ce{CuCO3}$ as high $\ce{CO2}$ pressure. At lower pressures, the following basic copper carbonates are produced : $\ce{Cu3(CO3)2(OH)2}$ or $\ce{Cu2CO3(OH)2}$ (malachite), depending on the partial $\ce{CO2}$ pressure. Other copper salts produce basic copper salts like : $\ce{Cu2NO3(OH)3}$ , $\ce{Cu4(OH)6Cl2}$, $\ce{Cu4(OH)6SO4}$ .
Zinc produces the two following basic zinc carbonates, according to the chosen pH : $\ce{Zn5(OH)6(CO3)2}$ or $\ce{Zn2CO3(OH)2}$
Iron(III) produces also basic salts like $\ce{FeSO4(OH)}$ which is produced by air oxidation of iron(II) sulfate. Dilution of a ferric acetate solution produces a precipitate of basic acetate $\ce{Fe(CH3COO)(OH)2}$
Aluminum solutions are often hydrolyzed producing basic salts. In the case of acetate, it gives a precipitate of basic acetate $\ce{Al(CH3COO)(OH)2}$
Cobalt chloride solutions treated by cold NaOH solutions give a precipitate of basic chloride $\ce{CoClOH}$ When treated by carbonate solutions they produce a great variety of basic carbonates.
This list is not exhaustive.


Stability of a carbocation, in part of camphor synthesis Can someone explain to me why there is a rearrangement reaction of carbocation 1 to carbocation 2? To me the second carbocation appears to be less stable than the first because it is less substituted.

Thank you in advance

Since mechanistic studies of the camphene/ bornyl acetate system are not available, I will focus on the parent compound norbonyl, and with it explaining the stability of the cation.

For quite some time there has been active debate over the structure of the 2-norbonyl cation. A very brief description of the two sides can be found in the Wikipedia article. A short history of the argument can be found at the homepage of the Royal Society of Chemistry (RSC): "The nonclassical cation: a classic case of conflict" by Mark Peplow.

In 2013 the case was finalized after I. Krossing's group was able to obtain a crystal structure of the compound. Another blog post at the RSC gives some more insight into the matter: "Crystal structure closes classic carbocation case" by Andy Extance. Finally we can speak of the 2-norbonyl cation as a true non-classical cation. The findings are also in good agreement with high-quality calculations. F. Scholz, D. Himmel, F. W. Heinemann, P. v. R. Schleyer, K. Meyer, and I. Krossing, Science, 2013, 341 (6141), 62-64.

This does in principle mean, that at very low temperatures, the 2-norbonyl cation is best described by the following resonance structures (courtesy of wikipedia):

At higher temperature the structure is quite disordered. In solution this means basically, that there is an ultra fast equilibrium (at the rate of molecular vibrations) between several structures. This would also explain the high reactivity of this compound.

Since camphene is a substituted norbornene, similar reactivity can be expected. The reaction is therefore most likely determined by the small steric effects. Since this reaction is highly selective, one can also safely assume, that the thermodynamic favoured product is formed. In this case it is the isobornyl acetate.

I have already mentioned it in the comments, but I am including it here nonetheless. The second row of your image is contains the wrong numeration and therefore depicts the mirror image. Correct is the following depiction:


How is the bond length calculated from the total electronic energy? In the quantum chemistry course that I currently attend, it was said several times that one of the key quantities derived from molecules by means of computation is the total electronic energy $E_\text{el}$, because from this many derived quantities can be calculated, such as bond length or $J$-coupling values for NMR experiments.

However, I've never gotten a good explanation as to how to proceed when I've calculated a value for $E_\text{el}$ and want to find out the bond lengths in the molecule in question.

What are the formulae and/or computation steps necessary for this?

The geometry of a system is what is known (or given) and a corresponding energy is computed.  If you have an energy then you should have the geometry (or in your case, bond length).

The wave function, $\Psi$, in the Schrodiner equation is simply a function of all the positions of the particles within the system (nuclei and electrons).  Typically chemists invoke the Born-Oppenheimer approximation which assumes that the nuclei are stationary (frozen in space) with respect to the very 'fast-moving' electrons.  Therefore, we freeze the nuclei in space and compute the energy of this stationary state.  Move the nuclei in your molecule a little bit and you will get a new energy.

As Wildcat was explaining, optimization procedures essentially moves the nuclei in your molecule around in space, computing an energy for each configuration. The goal of this type of routine is to find an energy that is lower than all the rest.  Take, for example, the $\ce{H2}$ molecule.  It contains one bond length and therefore the only coordinate we need to consider in an 'optimization procedure' is the internuclear separation of the two $\ce{H}$ atoms.  It is the chemist which will choose the initial positions of these atoms.  Say we initially set these atoms to be 200 pm apart.  We compute an energy.  Now, can we do better?  (i.e. can we perturb this geometry in such a way that gives rise to a lower energy?)  Well we can certainly try other bond lengths so we now set our bond length to 150 pm and compute a new energy.  Lo and behold we see that this new energy is lower than the first!  We keep trying different bond lengths until we reach a point where the energy increases no matter what we do with the separation (whether we increase it or decrease it).  Once we are at this point, we have found the lowest energy configuration of $\ce{H2}$.  Of course optimization routines are smarter than doing a brute-force approach to sampling all the possible quantities each geometrical parameter can take on.  Furthermore, second-derivatives are essential for characterizing the nature of the stationary state that was determined (i.e. is the stationary state a true minimum, a transition state or some higher order saddle point?).


Writing Ksp when the ions are on the left of the equilibrium equation? For the equation:
$\ce{Ca^{2+} {(aq)} + ^-OOC-COO- {(aq)} <=> Ca(OOC-COO) {(s)}}$
My book gives $K_{sp} = \pu{3.0e-9 M} = [\ce{Ca^{2+}}] \times [\ce{^-OOC-COO-}]$.
However, as per Chem.Libretexts, for any  $K_c$, the right hand side substances go in the numerator and the left hand side substances go in the denominator.

"It is really important to write down the equilibrium reaction whenever you talk about an equilibrium constant. That is the only way that you can be sure that you have got the expression the right way up - with the right-hand substances on the top and the left-hand ones at the bottom."

So for $K_{sp}$, why is it not the same?
I guess intuitively for a $K_{sp}$ I would probably put the $K_{sp}$ = product of the ions. But is there any other exceptions to when the above rule from Chem.Libretexts is not followed (in regards to K values)?

By convention, the solid is on the left of the side dissolving reaction. If you do it the $K_{sp}$ has the correct form.

The value of the $K_{sp}$ given above suggests that it conforms to this convention.


How to properly dispose of sodium...or store it properly While performing an inventory of my school's chemical storage closet today, my fellow teacher and I found some sodium tucked away in the back corner of our flammables cabinet.  One container appears to have leaked the oil the sodium was being stored in, and we aren't sure about the second container.  We don't know how long it's been there, but neither of us is going to use it, and, honestly, we don't want it.  How do we dispose of this?  Can someone with a little more chemistry know-how point me in the right direction?  Are we stuck with this?

Or...

In the event that the administration won't let us dispose of the sodium, can you tell me how to remedy the storage situation?  We don't feel that there is an adequate amount of oil in the container for proper storage of the remaining sodium, and if we have to keep it I want to make sure it is being stored correctly.

If you handle sodium, the first thing you need is a way to extinguish it if it catches on fire. You should have a bucket of sand ready to use, and additionally a class D fire extinguisher, if available. I'm assuming you know this, but don't let the sodium come into contact with water.

You always want the sodium stored in an inert liquid, usually mineral oil is used for that. Organic solvents like n-hexane are also used for short time periods. You should refill the mineral oil immediately.

The usual way to neutralize sodium is to put small pieces in an organic solvent like n-hexane and slowly add either isopropyl alcohol or ethanol to it. Isopropyl alcohol reacts slower, and is therefore a bit safer. When all the sodium is dissolved, you can add water to make sure there is no elementary sodium left.

How dangerous this is depends a lot on the scale, it's easy to do if you are only getting rid of a bit of sodium you haven't used in your reaction. I have no experience with neutralizing large amounts of sodium, this method should still work but it might take a long time if you are careful as you'll have to do it in small batches.

To be honest, the fact that you asked this question here on the internet indicates to me that you should at least let someone more experienced supervise the procedure.


Conservation of angular momentum in electronic transition My book says the "angular momentum is always conserved in a transition". But if suppose we have a s electron then it's angular momentum is $\hbar\sqrt{l(l+1)}+\hbar\sqrt{s(s+1)}$ $\implies \hbar\frac{\sqrt{3}}{2}$ and suppose it transits to a p electron (which is in accordance with the selection rule as $\Delta l=1$) which has an angular momentum of           $\hbar\sqrt2 + \hbar\frac{\sqrt{3}}{2}$. As the angular momentum of a photon is $\hbar$ then it means (mathematically) that we would have to add $\sqrt2$photons to the atom but that's not possible. Then how is the angular momentum conserved??

Also I can't see why the spin angular momentum of a photon is $\hbar$ as spin angular momentum is $\hbar\sqrt{s(s+1)}$ and as $s=1$ for a photon so it should be $\hbar\sqrt2$.

P.S. the book I am using is Elements of Physical Chemistry by Peter Atkins and Julia de Paula.

I think you have misunderstood some concept. The angular momentum of spin is just $\sqrt{s(s+1)}\hbar$. The spin which we always say is spin projection quantum number in fact not the angular momentum. And I'm not familiar with chemistry, but I know that we can not add up the orbital and spin angular momentum directly because they are not in the same direction. Sometimes we can add up the quantum number. If the spin and orbital angular momentum in opposite direction, we should subtract.

I have some idea about your example. The electron transition is "electron jumps from one energy level to another". So the energy must change. Where does the energy come from? If it is photon, this example has no paradox. But can the energy come from other way? I have no idea.


IUPAC Nomenclature of a dioic acid I have to name the following organic cpd according to IUPAC.
My attempt:
I am unable to select a parent carbon chain as there are 2 carbon chains possible and I don’t know how do we proceed in such a type of compound.
According to the key the answer should be 3,3’ -[oxy bis(ethyleneoxy)]dipropanoic acid
Now, this seems certainly wrong to me as there is no double bond in the carbon chain so ‘ene’ must not be there, or am I missing out on something.
PubChem gives the name of the cpd as 3,3'-((Oxybis(ethane-2,1-diyl))bis(oxy))dipropanoic acid
I would be glad if somebody could explain the reason behind this nomenclature.
Thank You!

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a structure with two or more identical units that are linked by di- or polyvalent groups can be named in two ways:

by ordinary substitutive nomenclature, in which one of the parent structures is chosen as the senior parent structure and the remainder of the structure is expressed by substituent prefixes
by multiplicative nomenclature, in which two or more parent structures are connected by symmetrical or unsymmetrical single or concatenated substituent groups

Both methods yield correct names, which may be used in general nomenclature. However, the IUPAC recommendations define a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations.

P-45.1 MULTIPLICATION OF IDENTICAL SENIOR PARENT STRUCTURES
P-45.1.1 Multiplicative nomenclature is senior to substitutive nomenclature for generating preferred IUPAC names to express multiple occurrences of identical senior parent structures, other than alkanes, in the name of the parent structure (see P-51.3.1). In most cases multiplicative names are shorter than regular substitutive names. A preferred IUPAC name is generated by multiplicative nomenclature when the following criteria for its use are met (see P-51.2.3).
(1) the linking bonds (single or multiple) between the central substituent group of the multiplicative group and all subsequent structural units are identical; and
(2) the multiplicative groups, other than the central multiplicative group, are symmetrically substituted; and
(3) the locants of all substituent groups on the identical parent structures, including suffix groups, are identical.
When these conditions are not met, substitutive nomenclature generates preferred IUPAC names.

The given structure fulfils these requirements; therefore, the preferred IUPAC name is generated by multiplicative nomenclature as follows.

P-15.3.1.3 Multiplicative name formation
Multiplicative names are formed in accordance with the number of occurrences of identical structural units as defined in P-15.3.1.1 and the relationship of the linking multiplicative substituent group to the identical structural units.
When a compound contains identical structural units as defined in P-15.3.1.1 linked by a symmetrical simple, compound, complex, or concatenated multiplicative group (a di- or polyvalent substituent group), it is named by stating successively:
(a) the locants for the positions of substitution of the linking multiplicative substituent atom or group to the identical parent structural unit (the locant 1 is omitted when alone in the name of a mononuclear parent hydride);

Here: 3,3'

(b) the name of the linking multiplicative substituent atom or group;

Here: oxybis(ethane-2,1-diyloxy) for $\ce{-O-CH2-CH2-O-CH2-CH2-O-{}}$

(c) the numerical prefix ‘di’, ‘tri’, etc.; and/or ‘bis-‘, ‘tris-‘, etc., with no elision of the final vowel before the name of the identical parent structural unit;

Here: di

(d) the name of one of the identical structural units including the principal characteristic group and substituents, if any, enclosed in appropriate enclosing marks (see P-16.5).

Here: propanoic acid

The numbering of the identical parent structural unit is retained and, when there is a choice, the locants of the point of substitution by the linking multiplicative substituent groups on the identical parent structure are as low as possible.

Here: 3,3'-(…)dipropanoic acid
Therefore, the preferred IUPAC name is 3,3'-[oxybis(ethane-2,1-diyloxy)]dipropanoic acid.

By way of comparison, the Blue Book contains the similar example 2,2'-[oxybis(ethane-2,1-diyloxy)]diacetic acid.


Why are enthalpies of reaction or Gibbs FE change given in kJ/mol It makes no sense to me that these thermodynamic changes are given in kJ/mol. Reactions can have several compounds on both sides and its not clear what exactly the molar variable refers to…

I've been given a homework where this is the convention followed. My temptation is to just ignore the /mol as senseless, and proceed just considering the energy change. But I'm sure I'm wrong.

Can anyone help me?

Thanks.

It makes perfect sense that the units KJ/mol are used.

If you burn 2 cords of wood you get twice the heat as from 1 cord of wood.

Enthaply (and Gibbs free energy) of the reaction is proportional to the number of moles that react.

If the reaction is written as:

3A + 2B => 6C

Then the change per mole would mean: per 3 moles of A = per 2 moles of B = per 6 moles of C.


Why can some redox half-reactions be combined and some cannot? I mean by "combining" is to make a new half-reaction equation and not an overall equation for a reaction in whole.

For instance, I was trying to arrive at the following half-reaction:

$\ce{MnO2(s) + 4H+(aq) + 2e- -> Mn^{2+}(aq) + 2H2O(l)}\quad E^\circ= 1.23\,\mathrm{V}$

by combining

$\ce{MnO4^{-}(aq) + 8H^+(aq) + 5e- -> Mn^{2+}(aq) + 4H_2O(l)}\quad\quad E^\circ=  1.51 \,\mathrm{V}$

and

$\ce{MnO4^{-}(aq) + 4H+(aq) + 3e- -> MnO2(s) + 2H_2O(l)}\quad\quad E^\circ=
1.7\,\mathrm{V}$

using Hess's Law. My calculation yielded +0.19 V, and not +1.23 V as I would expect.

Why caused this? I know that there are half-reactions that can be constructed in this way (as above, by combining two or more other half-reactions) that result in the correct potential being calculated.

In general, you cannot simply subtract electrode potentials like this to find the potential of another half-reaction. They have to be weighted by $n$ (the stoichiometric coefficient of $\ce{e-}$). In cases where your method works, it is only because $n$ is coincidentally the same for both half-reactions you are combining.

The foolproof way is always to convert electrode potentials to Gibbs free energy changes:
$$\Delta G^\circ = -nFE^\circ$$

So $\Delta G^\circ$ for the half-reaction $\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}$ is

$$(-5)(96485 \text{ C mol}^{-1})(1.51 \text{ V}) = -728.4 \text{ kJ mol}^{-1}$$

and $\Delta G^\circ$ for the half-reaction $\ce{MnO4- + 4H+ + 3e- -> MnO2 + 2H2O}$ is

$$(-3)(96485 \text{ C mol}^{-1})(1.7 \text{ V}) = -492.1 \text{ kJ mol}^{-1}$$

Subtracting the second equation from the first, you will find that $\Delta G^\circ$ for the half-reaction of interest ($\ce{MnO2 + 4H+ + 2e- -> Mn2+ + 2H2O}$) is

$$(-728.4) - (-492.1) = -236.3 \text{ kJ mol}^{-1}$$

which gives:

$$E^\circ = -\frac{-236.3 \text{ kJ mol}^{-1}}{(2)(96485 \text{ C mol}^{-1})} = 1.22 \text{ V.}$$

I presume some rounding errors led to the slight discrepancy between my answer and the answer you want. You may leave out the Faraday constant in the calculations once you are more familiar with them; basically that means you work with the quantity $nE^\circ$ instead of $\Delta G^\circ$. The answer should be the same since they are directly proportional.

For more information you can refer to the redox chapters in any inorganic chemistry textbook (Shriver/Atkins, Housecroft, etc.). These calculations are dealt with under the sections discussing Latimer and Frost diagrams.


Photochemical ligand substitution of metal carbonyls - role of water cooled lamp-jacket? Why, in the photochemical substitution of a ligand in a carbonyl complex, is a water-cooled lamp jacket required to separate the solution and the mercury vapour lamp? It's shown in Figure 22.16 of Atkins' Inorganic Chemistry.

The main point is to stop the heat flow from the lamp, while allowing UV through. You want to do photochemistry, not thermochemistry.


Dissociation of silver chloride in a solution When silver chloride is exposed to light it can dissociate to silver metal and atomic chlorine. It might be a silly question, but could the chlorine go back into the solution as chloride ions ? if so is this likely to occur? does the outcome change if the solution is being boiled?
Thanks!

The answer is yes and no. Silver chloride doesn't "dissociate", it decomposes into silver metal and dissolved chlorine.  The chloride reacts with the water to form hydrochloric acid and hypochlorous acid as shown below:
$$\ce{2AgCl_{(s)} +H_2O_{(l)} ->[h\nu] 2Ag_{(s)} + Cl_2_{(aq)} +H_2O_{(l)}->2Ag_{(s)} + HCl_{(aq)} + HClO_{(aq)}}$$


How can I calculate the molar conductivity of a solution with multiple solutes? I'm running an experiment mixing 2 different solutions, $\pu{5 mL}$ of $\pu{0.1 M}$ $\ce{NaCl}$ with $\pu{5 mL}$ of  $\pu{0.1 M}$ $\ce{KCl}$. I am measuring the conductivity of the total solution in $\pu{mS/cm}$, and also measuring concentration of the electrolytes with electrolyte analyzer, so I will get the individual concentrations of $\ce{Na+, K+, Cl-}$ in the mixed solution.

How do I calculate this mixed solution total molar conductivity given these settings? The literature only shows the calculation for a single solution, and I couldn't find any reference on how to calculate this kind of combination.

If you are interested in the modelling of electrical conductivity of aqueous solutions containing more than a single salt, then there are already some references in literature, for example for the salinity of natural waters (Limnology and Oceanography Methods, doi 10.4319/lom.2008.6.489) or in systems relevant to chemical engineering (Ind. Eng. Chem. Res., doi 10.1021/ie9605903).

(from the latter)


Which is hardest: iron, brass or bone? I was hopping around random wikipedia articles when I came across the article for the Behemoth. In the description for the beast it says:

  His bones are as strong pieces of brass; his bones are like bars of
  iron

So it got me thinking, which of these three substances is hardest: iron, brass or bone?

(I had a quick look at the Mohs scale, which lists iron as 4, but could not find anything for brass or bone.)

These two sources both put bone at a hardness of 5:

http://www.chacha.com/question/how-hard-is-bone-according-to-moh's-hardness-scale

https://answers.yahoo.com/question/index?qid=20110310200841AABwtMj

Whether they are trustworthy is questionable though, so take it as you will.

This source put brass at 3 and iron at 4.5:

http://www.jewelrynotes.com/the-mohs-scale-of-hardness-for-metals-why-it-is-important/

and this image puts brass at 4 and iron at 4-5 (Similar to 4.5):
http://patentimages.storage.googleapis.com/WO2001048807A1/imgf000009_0001.png

While these different sources seem to have conflicting data, I think it would be safe to assume that Brass is the softest of these three materials, Iron comes second, and Bone is the hardest.

Edit: In the description of that monster, the adjective used is 'strong'. You may want to consider how much force each of these materials can withstand instead of how hard they each are :)


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

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

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

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

Eg.:

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

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

or

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

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

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

P.S. Use nitric acid to dissolve Cu.


Predict melting point of water CaCl2 solution I'm doing a science project for my school. I experimentally measured the melting point of water-salt solutions, and I want to compare my results to theoretical melting points, but I encountered a problem.

I used this equation to calculate the melting point depression: $\Delta T = b \cdot i \cdot K_\mathrm{f}$. I also compared my data to a phase diagram. I expected my measured melting point would be on the liquid line. But these two methods give me a different prediction. A phase diagram tells me that $\ce{CaCl2}$ would be much more effective at lowering the freezing point of water while the above equation tells me otherwise.

Here is a comparison: Purple indicates the melting point calculated with the equation. The black line is the liquid line on the phase diagram.

Which method is correct and why is there such a big difference?

(I didn't encounter this problem with $\ce{NaCl}$ solutions, the two methods gave more or less the same predictions)

EDIT: my math was wrong on the graph i posted, here is the corrected graph:

I think there are three issues here. The first is that your purple line of calculating the freezing point doesn't appear to be correct. Let's arbitrarily select $b=1 ~ \mathrm{mol/kg}$ to compare. $i=3$ and $K_\mathrm{f}=1.86 ~ \mathrm{^{o}C/mol}$. The molecular weight of $\ce{CaCl2}$ is $111 ~ \mathrm{g/mol}$. Then, our mass concentration is $b·\mathrm{MW}/(b·\mathrm{MW}+1 ~ \mathrm{kg})=0.1 ~ \mathrm{kg}_\mathrm{salt}/\mathrm{kg}_\mathrm{total}$. The freezing point is $b·i·K_\mathrm{f}=1·3·1.86=5.58 ~ \mathrm{^{o}C}$. This is much closer to the red/black line than the purple one.

The second problem is that that relation is only valid at low concentrations. As the concentration of salt goes up, it has an increasing effect on the activity of the water. As we can see from the phase diagram, once you get to a 6:1 mol ratio of $\ce{Ca^{2+}}$ to $\ce{H2O}$, the interaction is extremely strong. I'm just speculating here, but I would guess six water molecules are forming a weak octahedral coordination complex with the $\ce{Ca^{2+}}$, with the electronegative oxygen ends oriented toward the ion. This effect becomes so strong, that the hydrate becomes the relevant element for consideration above that concentration.

Third, you're ignoring the existence of hydrates. If you find the melting points and the cryoscopic constants of the hydrates, you can actually calculate the right side of the plot the same way.


Altering Electronegativity of metals Can the electronegativity of metals be changed by applying an electric potential?

Electronegativity of an atom is due to the pull of electrons by the nucleus of atom.

Applying electric potential will not change the force the nucleus applies on the electron and hence it will not change the electronegativity.

By applying electric potential, the electron can be pulled out of the atom, not because it decreased the electronegativity but because it could apply enough force on the electron to overcome the electrostatic attraction by the nucleus and hence let it come out of the atom.

ps: If you had this doubt, then probably you are not clear with the actual concept of electronegativity. You might want to read up on that.


Why does ozone have such a large solubility in water? At $25~\mathrm{^\circ C}$, the solubility of ozone is $109~\mathrm{mg~L^{-1}}$. This is $13$ times higher than the solubility of diatomic oxygen. What causes this?

If you look at the molecular structure of ozone below, you can see that it is polar.  Diatomic oxygen is non-polar.  This mild polarity of ozone greatly enhances the solubility of the molecule in a polar solvent like water.


Calculate Rate constant, k, for a consecutive reaction where k1<<k2 I have a consecutive reaction where the first step is the rate determining step so $k_1<<k_2$:

$$\ce{A ->[$k_1$] B ->[$k_2$] C}$$

The rate law can be shown to be:

$$[\ce{C}] = {1 - \exp(-k_1t)} [\ce{A}]_0\tag{eq. 1}$$

My reaction is pseudo first order since the first step is a hydrolysis. If I take the natural log of eq. 1 will I be able to convert it into the following form?:

$$\ln[\ce{C}] = -k_1t + \ln[\ce{A}]_0\tag{eq. 2}$$

... because If I could get it to this form, I will be able to plot $\ln[\ce{C}]$ vs $t$ to get $k$ from the slope.

If I have made a mistake with the natural logarithm (or if it is not possible to convert it to the form of eq. 2), what function do I need to use to find $k$?

My question is specifically: What do I need to plot to determine my rate constant considering I have done a time course NMR experiment?

In a consecutive reaction A-B-C, A decays as $A=A_0\exp(-k_1t)$ with $A_0$ the initial amount (initial B and C are assumed to be zero) and B rate expression $\displaystyle \frac{dB}{dt}=k_1A-k_2B$  which when substituting for $A$ and integrating gives $\displaystyle B=A_0\frac{k_1}{k_2-k_1}\left(\exp(-k_1t) -\exp(-k_2t  \right)$. C is found as $A_0-A-B$

then $\displaystyle C= A_0\left(1+\frac{k_2\exp(-k_1t)-k_1\exp(-k_2t)}{k_1-k_2}  \right) $

In your case $k_1<<k_2$ then $\displaystyle C= A_0\left( 1 - \exp(-k_1t)  \right) $ which is almost your equation.

Taking logs gives $\displaystyle\ln\left(\frac{C}{A_0}\right)=\ln\left(1-\exp(-k_1t)\right)$ and this is what you should use.

[To go further you will have to expand the series for log and exponential but this will lead to a polynomial and will only be an approximation anyway.  The first few terms are $\log(k_1t)  - k_1t/2 + (k_1t)^2/24\cdots$; very messy.]


Can I successfully use a humidifier to distribute caffeine to wake someone up without the usual morning drowsiness? Would I have any success putting coffee in a humidifier? Or perhaps water mixed with caffeine powder?

So I could wake up without a jolting alarm clock, feeling like I already had a cup of coffee?

I understand this will have to be a carefully controlled dose if working with caffeine powder, but is it feasible?

It looks like at least one person/organization has thought of this and patented a delivery mechanism ("Delivery of caffeine through an inhalation route").

That said, you might be able to couple the system outline in the patent documentation with an alarm clock of sorts to trigger the release at a specified time. I'd certainly buy one if the price was right.

Edit: The inhalation part has been investigated. As far as I can tell, the business about administering caffeine to a sleeping person and then having them awaken into a fully alert (or pseudocaffeinated) state has not.


Why does this flame travel (in a circular path, multiple times) We filled a graduated cylinder with Ever Clear and let it overflow. Then we lit the liquid on fire. To our surprise the flame started running around the base of the cylinder. Why does this happen?

http://webmshare.com/m7vb8

A flame is a dynamic phenomenon, using up fuel and oxygen in places and driven by convection and diffusion. Flames are sensitive to factors such as external sound, and can be used to measure frequency and waveform with a Rubens tube.

Under micro-gravity, flames burn quite differently, due to the lack of convection.

So though not a specific answer to your questions, I assume that oxygen is being used up rapidly in one spot, and the flame front moves along as convection brings in fresh oxygen. You could test this hypothesis by piping in just a little air in one spot on the "ring of fire" to see if it stabilizes the flame there.

Thanks for sharing that interesting video!


How to dissolve white dextrin in water? Dextrin is labelled as highly soluble in water. I have white dextrin (analytical from Sigma Aldrich). When I mix it in water, no matter how long I stir and how little dextrin I put, I get a suspension. The dextrin does not dissolve.

I am not a "real" chemist. If I search on Google with keyword such as dextrin water solution, nothing pops up, except that dextrin is highly soluble in water.

I must miss something obvious. I just don't know what. If someone could point me in the right direction...

Try heating the water.

Sugar is also very water soluble, but just dumping sugar into cold water is going to take forever to dissolve. You'll age before you get a clear solution. So, add heat and stir it, and it should dissolve.

PS: Some dextrins are only partially solvable. If sigma aldrich says that your particular dextrin is fully solvable, then it should be. Check the label.


Solving two electron integral numerically Is there any software package in python or C to evaluate two electron integral repulsion integral in quantum chemistry of atoms. I am trying to solve the helium atom by constructing many body basis.Using those basis i am attempting to evaluate matrix elements.Also please suggest some references to understand more about two electron integral.

The canonical text on quantum chemistry is Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory by Szabo and Ostlund. Every electronic structure theorist I have ever met keeps a copy of that book within arms reach of his/her desk.

As for software, I would recommend PySCF. It is an entirely free python library for quantum chemistry with the most critical routines written in C and is an excellent platform to develop on. It's also robust and fast enough to be used for large calculations should you want to scale up. (https://github.com/sunqm/pyscf)

PySCF uses libcint to perform the integrals, and contains a low-level interface to the integral evaluator. An example of how to use these methods to make a df-HF calculation is shown in the docs here: https://sunqm.github.io/pyscf/tutorial.html#molecular-integrals. A complete list of available integral routines from the libcint package is shown here (it's very comprehensive): https://sunqm.github.io/pyscf/gto.html#module-pyscf.gto.moleintor


Toluenesulfonate vs. methylsulfonate Are there any advantages/disadvantages of using toluenesulfonate instead of methylsulfonate as a leaving group in E2 and SN2 reactions? I know they are both excellent leaving groups but how do they compare to each other? Is there a green chemistry aspect to it?

Thank you!

From the point of atom efficiency, methylsulfonate is advantageous compared to the p-toluenesulfonate.
On the reagent itself, the methanesulfonate is more polar than the p-TsOH; so the later dissolves better in less polar organic solvents than the former.  Methanesulfonate even may be used to dissolve some metal salts.
If used to generate a better leaving group, on workup, a methanesulfonate enters the aqueous phase easier than the tosylate.
p-TsOH is a cheap, widely available solid (sometimes sold as hydrate) you may prepare by your own in large quantities; methanesulfonate however is a liquid, hence potentially easier to add continuously, and more expensive (even more so, per mL, the triflates).
Methyl sulfonates are able to alkylate DNA, hence potentially carcinogen.  For tosylates, currently there is not such evidence (but obviously the dust of this strong acid is considered harmful).

As a personal note, among other reasons I found consultation of journals like Organic Process Research & Development just under this aspect (including green chemistry) as useful in two ways: i) publications describing how running a particular reaction within a chain of others works well / better than an other -- especially on larger scales of application--, and ii) the sections in line of "Some Items of Interest to Process R&D Chemists and Engineers".

Addendum:  There are dedicated compilations to figure out scope and limitations of chemical reagents.  To mention only two, Fieser & Fieser and EROS edited by Leo Paquette (there is an e-EROS as electronic ressource, too) are multivolume references.


Deriving the partition function for a harmonic oscillator This question is in reference to an answer I posted here yesterday.

In it I derived the partition function for a harmonic oscillator as follows

$$q = \sum_{j}e^{-\frac{\epsilon_j}{kT}}$$

For the harmonic, oscillator $\epsilon_j = (\frac{1}{2}+j)\hbar \omega$ for $j \in \{ 0,1,2.. \}$

Note that $\epsilon_0 \neq 0$ there exists a zero point energy.

Let's write out a few terms $$q = e^{-\frac{\hbar \omega/2}{kT}} + e^{-\frac{\hbar \omega3/2}{kT}} + e^{-\frac{\hbar \omega5/2}{kT}} +..... $$

factoring out $e^{-\frac{\hbar \omega/2}{kT}}$

$$q = e^{-\frac{\hbar \omega/2}{kT}} \left( 1+ e^{-\frac{\hbar \omega}{kT}} + e^{-\frac{2\hbar \omega}{kT}} +.....\right) $$

The sum in the bracket takes the form of a geometric series whose sum converges as shown below $$1+x+x^2+... = \frac{1}{(1-x)} $$
 herein, $ x \equiv e^{-\frac{\hbar\omega}{kT}} $

Putting all of this together

$$q = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{(1-e^{-\frac{\hbar\omega}{kT}})}$$

I happened to check Atkins Physical chemistry (10th edition) for the same derivation earlier today.

On page 620, the vibrational partition function using the harmonic oscillator approximation is given as
$$q = \frac{1}{1-e^{-\beta h c \nu'}}$$, $\beta$ is $\frac{1}{kT}$ and $\nu'$ is wave number

This result was derived in brief illustration 15B.1 on page 613 using a uniform ladder. However, in that illustration the uniform ladder starts at 0, but the harmonic oscillator has a zero point energy (which I accounted for in my derivation).

I discussed this with my instructor and he pointed out that as $T \rightarrow 0$ $q \rightarrow 1 $ (since only one state is thermally accessible.

The result derived in Atkins' does that indeed, but mine goes $q \rightarrow 0$.

Now, in the formalism developed in Atkins the do set the ground state energies to zero (on page 605), and basically add non-zero ground state energies to a calculated $\langle \epsilon \rangle$, which is fine.

Anyway, I would love it if someone could weigh in on this and help me resolve this conceptual mess in my head.

I'm confused why you're interpreting the partition function as a count of states. It can't be a count; it's continuous.

The zero point energy doesn't actually matter because you can just shift the energy scale so that it starts at zero. The main point of zero point energy is that the ground state of the harmonic oscillator is such that there is energy, and the system is not stationary. I'm going to use it below anyway because you are.

You can get the answer you want, but you'll want to look at the probability $P_i$ of being in state $i$, where the partition function is used to normalize:

$$P_{i} = \frac{\epsilon_{i}}{q} = \frac{\mathrm{e}^{-\beta\hbar\omega (i + 1)/2}}{\sum_{i}\mathrm{e}^{-\beta\hbar\omega (i+1)/2}}$$

Substitution with your convergent sum:

$$P_{i} = \mathrm{e}^{-\beta\hbar\omega (i+1)/2} \frac{1 - \mathrm{e}^{-\beta\hbar\omega}}{\mathrm{e}^{-\beta\hbar\omega/2}} = \mathrm{e}^{-\beta\hbar\omega i}(1- e^{-\beta\hbar\omega})$$

For $T\rightarrow 0$, $P_{i} = \delta_{i0}$, which is exactly what you are looking for.


Why does the position of equilibrium depend only on the relative Gibbs energies of the products and the reactants? The position of equilibrium, I read, is dependent only on the relative Gibbs energies of the products and the reactants. Why is this so? Surely one must take into account the entropy of mixing of the products and the reactants, which wouldn't be dependent on the Gibbs energies, to work out the lowest point (energy wise) and find the position of equilibrium. Why, therefore, is the position of equilibrium between products and reactants only dependent on the relative gibbs energies of the products and reactants.
Thanks!

Why does the position of equilibrium depend only on the relative Gibbs energies of the products and the reactants?

The short answer is that the approximate entropy of mixing is a simple function of the reaction quotient Q, so it is implicitly considered when calculating the equilibrium constant from $\Delta_\mathrm{r} G$.

  Surely one must take into account the entropy of mixing of the products and the reactants, which wouldn't be dependent on the Gibbs energies, to work out the lowest point (energy wise) and find the position of equilibrium.

Yes, the entropy of mixing is important in describing an equilibrium and figuring out the equilibrium constant. Without the entropy of mixing, all reactions would go to completion, see Why does equilibrium exist? and Why can't a reaction go to completion?.

The Gibbs energy of reaction has an enthalpy and an entropy component:

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

The enthalpy does not depend on concentrations (or reaction quotient Q), but the entropy does:

$$\Delta_\mathrm{r} S = \Delta_\mathrm{r} S^\circ - R \ln Q$$

The term $R \ln Q$ is the entropy of mixing (relative to everything being at standard state, i.e. $Q = 1$). Having both reactants and products present results in a higher entropy than having just reactants or just products.

Where is the mixing?

There is no actual mixing, just making reactants or products, but the expression is the same. You could do a thought experiment where the reaction proceeds via an enzyme that takes up reactants from one container and puts them into another (so reactants and products would never be in the same container). In a second step, you would mix the two containers so that reactants and products are in the same container.

Equilibrium and the entropy of mixing

As a result of the concentration-dependent entropy term, the Gibbs energy of reaction changes (increases) as the reaction goes towards equilibrium. When it reaches equilibrium, it is zero. The expression for $\Delta_\mathrm{r} G$ is:

$$\tag{1}\Delta_\mathrm{r} G = \Delta_\mathrm{r} H - T \Delta_\mathrm{r} S$$ $$ = \Delta_\mathrm{r} H^\circ - T (\Delta_\mathrm{r} S^\circ - R \ln Q)$$ $$ = \Delta_\mathrm{r} G^\circ + R T \ln Q$$

At equilibrium, $Q = K$ (that's the special name we give to $Q$ for a set of concentrations that is at equilibrium) and $\Delta_\mathrm{r} G = 0$. If we make these substitutions in equation (1), we get:

$$\tag{2}0 = \Delta_\mathrm{r} G^\circ + R T \ln K$$

Solving for $\Delta_\mathrm{r} G^\circ$, we get the well-known relationship between standard Gibbs energy of reaction and the equilibrium constant:

$$\Delta_\mathrm{r} G^\circ = - R T \ln K$$

The left-hand side does not make any reference to concentration dependence (it refers to a single state, the standard state), whereas the right-hand side contains the concentration-dependent term (entropy of mixing times temperature, again compared to the standard state where Q is one).


What does it mean for a reaction to favor the reactants/products? (This example is purely hypothetical.)

You have the reaction $$\ce{H2(g) + O2(g) <=> H2O2(g)}$$

at $T = 500\ \mathrm{K}$. The reaction reaches equilibrium at the following concentrations:

$$\ce{[H2]} = \ce{[O2]} = 5 \times 10^{-3}\ \mathrm{mol\ dm^{-3}}$$
$$\ce{[H2O2]} = 4\times10^{-5}\ \mathrm{mol\ dm^{-3}}$$

This gives $$K_c = \left(\frac{\ce{[H2O2]}}{\ce{[H2][O2]}}\right) = 1.6$$

However, the total concentration of the reactants is $250$ times higher than the concentration of the product.

Still, as $K_c > 1$, per definition, the products are favored.

This seems counterintuitive to me, and my Chemistry teacher couldn't really explain it to me, so I was hoping someone here could explain why, even when the reactants are so much more plentiful, the products are considered to be favored.

You may not have had kinetics (reaction rates) yet, but I find it really useful in understanding equilibria.

Kinetics of a reaction

Molecules in solution (gas or liquid) move randomly. Because they need to bump into each other to react, the probability of reaction increases as concentrations of reactants increase. The rate of the reaction is then proportional to the concentrations of reactants:

$$\ce{A + B \rightarrow C}$$
$$\mathrm{rate} = k_\mathrm{forward}\ce{[A][B]}.$$

Likewise, the reverse reaction has a rate dependent on concentration of product:

$$\mathrm{rate} = k_\mathrm{reverse}\ce{[C]}.$$

From Kinetics to Equilibrium

At equilibrium, the forward rate is equal to the reverse rate, and concentrations of both reactants and products remain unchanged. $$k_\mathrm{forward}\ce{[A][B]}=k_\mathrm{reverse}\ce{[C]}$$ We can rearrange to get: $$\frac{k_\mathrm{forward}} {k_\mathrm{reverse}} = \ce{\frac{[C]}{[A][B]}}.$$ We can substitute the right side of the equation to be the equilibrium constant, and we get $$\frac{k_\mathrm{forward}} {k_\mathrm{reverse}}=K_c$$ Thus, when we say that $K_c>1$ we are also saying that $k_\mathrm{forward} > k_\mathrm{reverse}$ and inversely if $K_c<1$ then $k_\mathrm{reverse} > k_\mathrm{forward}.$ To favor either the reactants or the products in equilibrium is to say the formation of either the reactants or products is favored, as indicated by the rate constants.


Relationship between vapour pressure and saturated vapour pressure I need clarity on saturated vapour pressure in a closed system at equilibrium. How does saturated vapour pressure relate to vapour pressure?

If you're at equilibrium, you're saturated.   If you're saturated, you're at equilibrium.  The end.

If things aren't at equilibrium, the situation is different.  Consider a dish of water in the middle of the Sahara desert.  The amount of water in the atomsphere in the Sahara desert is generally nonzero, say in the range of 1 $\mathrm g~{kg}^{-1}$.  Even though the water vapor pressure (even in the Sahara!) is nonzero, it is way less than "saturated".  Therefore it isn't at equilibrium, and the pan of water quickly evaporates.


What makes ABS plastic turn yellow? If you owned old kitchen appliances or old computer you might notice that they the once cream white plastic become increasingly yellow, old and disgusting. I have read that UV lighting can damage plastics as they disrupt the structure of plastics but what really is at play here? What chemical reactions are happening? Are such damages reversible?

What you're specifically describing is typically known as UV degradation.  The color change is due to the oxidation of the polymers that make up the plastic.  These chemical reactions are demonstrated in satisfying detail here.  Under the section titled "Photodegradation" is where you can find the intricacies of the polymer-oxidation process.  And yes, such damages are reversible.  As you will read, the oxidation process results in the formation of free radicals, which are "atoms or groups of atoms with an odd (unpaired) number of electrons," to quote Wikipedia's definition.  The formation of these radicals is the reason for the color change (since certain light gets reflected because of these new molecules, in many cases yellow light, etc...)

To get rid of the radicals, you can add hydrogen peroxide, which allows hydrogen to bond with the radical (hydrogen has an additional electron), thus de-yellowing the plastic. This is demonstrated here.

As a further note, UV degradation is not the only process that makes plastics turn yellow, as you will probably read in that paper.  There are also processes like biodegradation and mechanical degradation that are similar to photodegradation but still have their own cool processes.

Your initial question was "What makes ABS plastic turn yellow."  I should point out that the first few words of the second link read "Manufacturers add bromine to ABS plastic to act as flame retardant. Over time, those chemicals react to the plastic's basic polymers and turn yellow."  Ergo, there is probably more to the yellowing of ABS plastic in particular than just photodegradation.

Lastly, I should point out that the yellowing of plastic needn't happen in the presence of oxygen.  But again, that is addressed beautifully in the paper.


Copper plating with electrolyte not working (Cu(OH)2 vs CuSO4) I've been experimenting using electroplating, and I recently made a solution by placing two copper pieces connected to a voltage source, in water. I suspect that this created copper(II) hydroxide, as it matches the color of that specific solution (green-blue) and doesn't stay dissolved in the water (it sinks to the bottom after a day).
A lot of people use copper(II) sulfate for an electrolyte during copper electrolysis, so I assumed $\ce{Cu(OH)2}$ would work the same, but if doesn't seem to be. A coin was attached to the negative wire, copper was attached to the positive wire, but attempting to copper plate a coin in this solution has failed.
Why is that? Both copper(II) hydroxide and sulfate have $\ce{Cu^2+}$. Shouldn't that be attracted to the negative wire, and append the copper to the desired item?
Why does $\ce{Cu(OH)2}$ not behave like $\ce{CuSO4}$ during electrolysis?
I don't know too much about chemistry so if you explain it in simple terms, I would appreciate it!

If you are going to do anything differently than is normally done, you should understand what you are going to do and why it should work.(*)
The essential (but often not sufficient) conditions for successful electroplating by any metal are:

The used metal salts/compounds are soluble, so ions can migrate within the solution in applied electrostatic gradient. That is not the case of copper in copper hydroxide nor alkalic copper carbonates, which are insoluble.
The metal salt has sufficient concentration, so electrolytic current reacts primarily with this salt, at least on the cathode.
The current density on the plated electrode is low enough ( typically $\pu{10 mA/cm^2}$).

The solution composition for successful creation of shiny metal surface can be tricky, based on intensive empirical reasearch, involving solution modifiers like sugar or soluble polymers. The goal is mostly decreasing ion mobility and/or the rate of electrode reaction to support even metal deposition.

(*) It does not fully apply in the basic research, which uses often "what if.." approach, but this is not the case.


Interpretation of an H NMR I am fairly new to NMR and require assistance to analyze and interpret the NMR I have provided (sorry for it being vertical). This is an unknown compound that I have derived during one of my lab classes and it would help me a lot to know and understand what compound this is.

This is what I have interpreted so far:

The compound should have between 12 - 13 hydrogens, probably 13 if there is an alcohol involved and no saturated bonding.

the multiple peaks at slightly below 4 ppm can be a ring but also can be a $\ce{CH2}$ near an oxygen and also bonded to a $\ce{CH3}$ group.

The multiple peaks at around 1.5 ppm are also $\ce{CH2}$ group bonded to another $\ce{CH2}$ group and further away from any oxygen

The triplet at slightly below 1 ppm is a $\ce{CH3}$ group attached to a $\ce{CH2}$ group

However, my interpretations don't seem to add up and I am still figuring out the double peak.

You have now given yourself some information to go on with, and indeed have identified most of the important fragments. Now, the tricky bit is putting them all together! Let's work through what you've got.

Let's first assume that the sample is pure, and the integrals are correct to within reasonable error. This then gives you 5 obvious environments with ratio 3:3:4:1:1. Environments with 4 equivalent protons are not that common, so we might suspect there is some overlap here, and that we have more than one group. So, I count 12 protons.

The peak at 0.9 integrates to 3 - it is most likely a methyl group, and is closely resembling a triplet, so we can make the guess that it is next to a CH2. (It is in fact not a true triplet, and is immediately recognisable as a terminal methyl for a longer alkyl chain - see this link for more information). But for now let's call it a triplet.

At 1.2, we also have a -CH3 (integrates as 3), which is a doublet, so must be next to a -CH.

The multiplet mess around 1.4 must contain at least one -CH2 (to couple to the methyl at 0.9), which will leave us with another -CH2, or perhaps two -CH groups.

That broad peak at 1.72 is characteristic of an alcohol peak. It is also indicative of a sample which, if not run almost neat, is very concentrated and in a dry solvent. Its integral is not quite correct, but probably because of some excess water in the sample. But it is broad, and in the correct region for an alcohol. You could confirm this by doing a D2O shake, and seeing this peak disappear.

And, at 3.8 you have a -CH which is coupled to more than one thing. You have reasonably proposed that it is adjacent to a -CH3 and -CH2- Its chemical shift suggests that it is near some electronegative group or deshielding group. Knowing you have an alcohol might help narrow the possibilities here. The chemical shift is not indicative of a ring on its own. An alkyl ring would still be to higher field.

So, let's summarise what you know:

CH3CH2-
CH3-CH(OH)-CH2-

I think there is perhaps only one way to connect these fragments....
The key to these types of questions is that once you propose a structure, you can then go back and rationalise what you see to test the structure. Sometimes, you might have to come up with several structures before you get the correct one. But it usually best to start with the simplest, most probable and start eliminating from that.

If you still have problems with understanding any of this, ask some more, and we can go from there.


How do I figure out the relative polarity of organic compounds? It's a question on my quiz which I guessed right and couldn't do the reasoning:

Based on the intermolecular forces, predict the boiling points of the following substances and list them in the order of increasing boiling point. State your reasons accordingly.

Dimethyl ether $(\ce{CH3OCH3})$
Ethanol $(\ce{CH3CH2OH})$
Propane $(\ce{CH3CH2CH3})$

Right away, I see the hydrogen bond in ethanol, so it must have the highest boiling point.
I couldn't decide on the other two, so I drew lewis structures (kind of) for both

I ignored the hydrogens because they don't affect the polarity (correct me if my lewis strctures are wrong).
This is what I guessed, and I was right. When my teacher takes this question up, he said that propane is non polar, which means it only interacts via london forces, which means it has the lowest boiling points out of these three.
But how is propane non polar though, when I asked, he said "I don't have an answer for this, life is hard."
I mean, if I draw the structure in the bad way, it does look symmetrical and balanced out:

But if I draw it in this way, it doesn't help me when I'm trying to figure out the polarity, because everything will look non polar. I can even say water is in a linear shape because the H's and the O line up perfectly in a line, which is not the correct picture of reality.
My biggest question is, how do I figure out if something is polar or not.

General steps to figure out the polarity (or at least relative polarity) and subsequently the relative boiling point of compounds:

Acquire the correct geometry
Estimate dipole moments
Sum up estimated dipole moments
Check for other intermolecular interactions

Acquire Correct Geometry

This step requires some knowledge about the geometry of hydrocarbon compounds. But it seems that you already know about the tetrahedral bonding centers of saturated hydrocarbons. This is then important later on for the dipole forces. Never try to linearize the carbon backbone just to try to make sense of this, it will do more damage than good.

Estimate Dipole Moments

The dipole moments that you have to estimate will be those along bonds. There is a larger dipole moment in the ether than the hydrocarbon without heteroatoms, because the oxygen atom is strongly electronegative. As such, there will be a dipole along the $\ce{C-O}$ bond.

Sum Up Estimated Dipole Moments

These dipoles can be understood as vectors along bonds, pointing from positive to negative partial charges, and can be summed up for the total dipole moment of a molecule. The larger the absolute value of this dipole vector is, the larger the dipole-dipole interactions between molecules: Subsequently, the higher the forces between the molecules of this substance.

Check for Other Intermolecular Interactions

One other main intermolecular action is, as you already pointed out in your question, hydrogen bonding. This one is quite strong, but isn't found for every compound out there. Basically, it is when a highly polarized $\ce{H-X}$ bond can be formed, where $\ce{X}$ can be oxygen, nitrogen, chlorine or some other electronegative atom (with respect to carbon). The hydrogen that is sandwiched in the hydrogen bond has to be (formally) bonded to such an electronegative atom as well, otherwise it wouldn't be as free to create those bonds (as you also have realized, because the hydrocarbon hydrogens are really quite boring from an intermolecular interaction perspective).

Result

Based on the steps above, we can generate a table:

The more "Yes" a substance has, the higher the intermolecular forces. Based on these data, assumed by us, we can generate a list (from highest boiling point to lowest):

Ethanol
Dimethyl ether
Propane


Is there a catalyst that will reduce an alcohol to an alkane? There are many reduction methods of going from an alcohol to an alkane: http://www.transtutors.com/chemistry-homework-help/hydrocarbons/reduction-of-alcohols.aspx

And there are catalytic methods of adding hydrogen  gas to aldehydes and keystones to convert them into alcohols: http://jmcct.com/products-services/product_p404.html

However, I have not found a catalytic method of converting an alcohol into an alkane using a catalyst and possibly hydrogen. The fewer non-catalytic entities the better.

Here is a method for converting secondary and tertiary alcohols into the corresponding alkane using $\ce{InCl3}$ as a catalyst.  Chlorodiphenylsilane is the only reagent used in addition to the catalyst. The method works in the presence of common functional groups leaving them untouched.

Since primary alcohols are unreactive, this reaction can be used to selectively reduce secondary and tertiary alcohols in their presence.  An abstract showing a number of examples of the reaction can be found here.


Why does water diffuse from a lower solute concentration to a higher one? I'm studying biology and this came up in the context of passive transport. I understand why, in general a gas moves down a concentration gradient (the random movement of gas particles and probability). But I don't understand why the concentration of some solute (ions, salt, etc) would affect the movement of water. Why does water move from an area with a lower concentration of solute to an area with a higher concentration of solute?

There are two possibilities. The first is that it is simple dilution, but for more likely is that osmosis is occurring. In this case the concentrated solution and the water, or a  dilute solution, are separated by a semi-permeable membrane, i.e. one that allows passage of small water molecule but not large solute ones.  To lower the overall energy (as represented as the chemical potential) water moves through the membrane from the dilute solution to the more concentrated one. In a cell this can lead to the cell swelling and possibly bursting if left uncontrolled. There is some more explanation in the answer to this question Entropy as the driving force for osmosis


Why doesn't nitrogen react with oxygen at room temperature? I already know that nitrogen's triple bond requires a lot of energy to be broken, so the activation energy is very high, but the book also lists the endothermic energy change as a reason.

Does the positive enthalpy change have something to do with the reaction not being spontaneous i.e. due to positive free energy?
Is there a link between $E_a$ and free energy when deciding whether a reaction will occur?

Yes, you are right, the reason primarily looking at it from a theoretical point of view is the high bond dissociation energy of $\ce{N#N}$

Let's look at this aspect from every point of view

Theoretical- High bond dissociation energy of $\ce{N#N}$
Kinetics- We have the Arrhenius equation given below, which states the relationship between the rate constant and activation energy of the reaction.

$$k = A \mathrm e^{-E_\mathrm a/(R T)}$$
So essentially, this relation shows that if activation energy is higher, then the rate is slower. However, if we increase the temperature sufficiently, then the reaction does start to happen significantly. (Hence, oxygen and nitrogen do react when lightning strikes)

Thermodynamics- We have the very famous relation of thermodynamics

$$\Delta G= \Delta H-T \Delta S$$ So now lets again look at each term here

$H$- Enthalpy of the reaction, it is very high, as we discussed before, the bond dissociation energy of $\ce{N#N}$ is very high.
$S$- Entropy change won't be any significant as the reactants and products both are gases, so this is not an entropy-driven reaction.

Thus
, in every way possible we can conclude that nitrogen and oxygen do not normally react at room temperature.


How do I prepare the following compound from diphenylmethane? I am having trouble with the preparation of this product from diphenylmethane as a starting material. One of my thoughts was to attach the butene into the structure, but I don't see how it is viable. So I would like some suggestions on how to proceed, and also with the name of the compound (butenediphenylmethane?).

Here's a mechanism that I thought of (It's more theoretical than practical):

Bromination in the first step will occur at the benzyl position due to a stable free radical. Sodium Propynide then adds more carbon atoms via the SN2 mechanism, which are then reduced with Lindlar's catalyst to give the cis-alkene.

The name of this compound should be (Z)-1,1-diphenylbut-2-ene, but as pointed out in the comments, the preferred IUPAC name (PIN) seems to be (Z)-but-2-ene-1,1-diyldibenzene (verified by ChemDraw)


Can a s orbital overlap with any p orbital to form a sigma bond? In internet resources, only $p_z$ orbitals are shown to overlap with $s$ orbitals to form sigma bond. But my teacher says that according to Valence Bond Theory, $s$ orbital overlaps with all three $p_x$, $p_y$ and $p_z$ orbitals to form sigma bonds. I am really confused about this. I heard that the concept changes when it comes to molecular orbital theory. Only similar orbitals (like $1s$-$1s$,$2p_x$-$2p_x$,$3p_y$-$3p_y$) form bonds but ($2s$-$2p_z$) do not.
Please clarify if there is any difference in concepts of overlapping between 'Molecular orbital theory' and 'Valence Bond Theory'.

$\ce{Px}$, $\ce{Py}$, and $\ce{Pz}$ orbitals can all overlap with an $\ce{S}$ orbital to form a bond.  It simply depends upon along which axis (X, Y or Z) the bond will be directed. These concepts of overlap do not change between Molecular Orbital theory and Valence Bond theory.