gifts - What to present to the rabbi

My friend is going to meet rabbi, to talk about house management, and she wants to bring some present to him. What you would present to the rabbi, and what you should not to? Please explain why.

electrochemistry - Why is the water reduction of oxygen favored in acidic conditions?

The standard reduction potential of diatomic oxygen in acidic conditions is +1.23 volts. However, the standard reduction potential of diatomic oxygen in basic conditions is only +0.40 volts.

Why is the reduction of diatomic oxygen by acidic media more favored than the reduction of oxygen by water that is basic?

I can think of a few reasons:

1) We are likely going to be forming negative two oxidation state oxygen anions when we oxidize diatomic oxygen. Oxygen adopts an oxidation state of negative two except in peroxides, which aren't very stable, and I don't think the oxygen radical will be very stable in solution.

Can someone provide a less fluffy reason that we're going to be forming negative two oxidation state oxygen atoms? I'd imagine that the electronegativity and electron affinity of oxygen are two reasons it might favor the negative two oxidation state over the negative one oxidation state in solution. I.e. why doesn't this occur to a significant extent?

$\ce{O2 + 2e^- -> 2O^-}$

Or this?

$\ce{O2 + 2e^- -> HOOH}$

2) From the above we can deduce that water is the product; water is pretty stable in either acidic or basic medium - i.e. it won't react to a large extent with either hydronium or hydroxide ion. Plus the oxygen in water has a -2 oxidation state. So if we look at our skeleton half reaction:

$\ce{O2 + 4e^- -> 2H2O}$

We realize that we need a source of protons. And the proton source in acidic medium is likely hydronium ion.

However, in basic medium, the proton source is water. I'd imagine that water is a poorer proton source than hydronium ion.

So would this account for the difference in standard reduction potential of oxygen in media that vary in pH?

I understand that voltage is an extension of K through the Nernst equation; is what I'm seeing here basically that the protonation of a base ($\ce{O^2-}$) is more favored in acidic rather than basic media? Could there also be kinetic factors? In other words, why is this statement true?

Also:

1) What's the mechanism for the reduction of diatomic oxygen?

2) Why is the reduction of protons at high pH less favored than the reduction of protons at low pH? My first guess: you don't find as many protons in basic solutions. But then, protons don't exist by themselves anyway. So what can the proton source be? I'm guessing that in basic pH, there doesn't exist much hydronium ion. So the only two sources of protons are water and hydroxide anion. Heterolytically cleaving the hydroxide anion to form the highly basic oxide anion doesn't seem thermodynamically favorable. Cleaving water to form hydroxide anions ... are we running into a Le Chatlier's Principle effect?

ETA: Or should I be considering proton activities? Is the activity of the proton less in basic solution than in acidic solution?

Answer

If you look at the half equations, you will see when you add them together, you get 1.23 V, 1.22 V, and 1.23 V.

As the mixture becomes more basic, it is more difficult for the oxygen to grab $H^+$ ions and easier for $H_2$ to lose electrons to generate $H^+$ ions.

The mechanism probably involves these combinations: $O_2, H_2O_2, H^+, HO_2^-, H_3O_2^-, H_2O, and\ HO^-$.

kinetics - Rate-limiting in transition state theory

Suppose an initial state $A$ can transition to either state $C$ or state $D$. Suppose further that both of these two processes are rate-limited by a transition to the same intermediate state $B$, as follows:

Note that $\Delta G_{AB}> \Delta G_{BC}>\Delta G_{BD}$.

According to transition state theory, rates are determined by the rate-limiting step. So the rate of each process will be the same,

\begin{align} r_{A\to C}&=r_{A\to B} \\ r_{A\to D}&=r_{A\to B} \end{align}

from which it follows that the number of transitions made during some large $\Delta t$ will also be the same,

\begin{align} N_{A\to C}&=r_{A\to B}\Delta t \\ N_{A\to D}&=r_{A\to B}\Delta t \end{align}

However, if we were to consider the two pathways collectively then we would predict more transitions to be made to state $D$ than $B$ since $\Delta G_{BD}<\Delta G_{BC}$.

How do we resolve this apparent discrepancy?

Answer

In transition state theory (TST, goldbook) one of the necessary assumptions is that reactants and products are in equilibrium. In principle this gives us \begin{align}\ce{ A &<=> [AB]^{$\ddagger$} -> B\\ B &<=> [AB]^{$\ddagger$} -> A\\ A &<=> B, }\end{align} and in addition to this, also \begin{align}\ce{ B &<=> [BC]^{$\ddagger$} -> C & B &<=> [BD]^{$\ddagger$} -> D\\ C &<=> [BC]^{$\ddagger$} -> B & D &<=> [BD]^{$\ddagger$} -> B\\ B &<=> C & B &<=> D, }\end{align} as well as \begin{align}\ce{ C <=> [BC]^{$\ddagger$} -> & B <=> [BD]^{$\ddagger$} -> D\\ D <=> [BD]^{$\ddagger$} -> & B <=> [BC]^{$\ddagger$} -> C\\ C <=> & B <=> D, }\end{align} and $$\ce{ C <=> A <=> D\\ [AB]^{$\ddagger$} <=> [BC]^{$\ddagger$} <=> [BD]^{$\ddagger$} }.$$

You can use TST to estimate rate constants. $$k = \frac{\mathcal{k}_\mathrm{B} T}{\mathrm{h}} \exp\left\{ −\frac{\Delta^\ddagger{}G^\circ}{\mathcal{R} T}\right\}$$

Since all components have to be in equilibrium, you can predict a ratio between them via Boltzmann statistics, hence: $$\frac{N([BC]^{\ddagger})}{N([BD]^{\ddagger})} = \exp\left\{ −\frac{\Delta^\ddagger{}G^\circ([BC]^{\ddagger})-\Delta^\ddagger{}G^\circ([BD]^{\ddagger})}{\mathcal{R} T}\right\}=\frac{N(C)}{N(D)} $$

The preceding equilibrium does not need to be considered to determine the ratio of the products. However, it is important to note, that there are many assumptions necessary for TST to work. For example, at higher temperatures, anharmonic corrections have to be considered. More crucial is the assumption, that the transition can be described as a translatory movement, i.e. it is treated with classical mechanics.

For a system with a preceding equilibrium, it is well possible, that the TST approximation breaks down completely.

fft - Overlap-Add versus Overlap-Save

What differences or other criteria can be used to help decide between using overlap-add and overlap-save for filtering? Both overlap-add and overlap-save are described as algorithms for doing FFT based fast convolution of data streams with FIR filter kernels. What are the latency, computational efficiency or caching locality (etc.) differences, if any? Or are they the same?

Answer

Essentially, OS is slightly more efficient since it does not require the addition of the overlapping transients. However, you may want to use OA if you need to reuse the FFTs with zero-padding rather than repeated samples.

Here is a quick overview from an article I wrote a while ago

Fast convolution refers to the blockwise use of circular convolution to accomplish linear convolution. Fast convolution can be accomplished by OA or OS methods. OS is also known as “overlap- scrap” . In OA filtering, each signal data block contains only as many samples as allows circular convolution to be equivalent to linear convolution. The signal data block is zero-padded prior to the FFT to prevent the filter impulse response from “wrapping around” the end of the sequence. OA filtering adds the input-on transient from one block with the input-off transient from the previous block. In OS filtering, shown in Figure 1, no zero-padding is performed on the input data, thus the circular convolution is not equivalent to linear convolution. The portions that “wrap around” are useless and discarded. To compensate for this, the last part of the previous input block is used as the beginning of the next block. OS requires no addition of transients, making it faster than OA.

bond - Physical intuition behind negative values for wave function?

So a positive and a positive wave function create a bonding orbital where the probability of finding an electron is summed while a positive and a negative create an anti-bonding orbital with a lower electron probability in the region between them leading to a repulsion. My confusion stems from not having any idea as to what a negative wave function is representing - can anyone give me some physical intuition on how this negative wave function could be correlated to something in reality?

(I tried asking this question on physics.SE but it would seem that in physics you just don't talk about or use the wave function until after you square it)

homework - When Z>1 are repulsive forces dominant or are attractive forces dominant?

According to PVreal = ZRT

If z >1 repulsive forces are dominant or are attractive forces? Different teachers are telling contradictory answers and after referring a few books, I have become more confused.

Points for ATTRACTIVE FORCES are dominant:

• At high pressure the molecules do not strike the walls with full impact as they are dragged back by attractive forces.


• When molecules are farther away from each other(as Vreal > Videal) the attractive forces are dominant.


• Repulsive force are dominant when he molecules are almost in contact .

Points for REPULSIVE FORCES are dominant:

• At high pressure the molecules are close each other that attractive forces are negligible but molecular size is significant.Thus repulsive forces are significant


• As volume of real gas is more than volume of ideal gas , the real gas is more difficult to compress.Thus repulsive forces are dominating.

There was a question where P = 12atm T = 350K and Z = 1.2(after calculation). It was given that repulsive force are dominant.But my teacher said it was wrong and it should be attractive forces.

organic chemistry - Ring contraction in a carbocation due to ring strain and back bonding

Today our teacher told us that the following carbocation rearrangement occurs due to back bonding. I could not really follow what he meant. Can someone please explain what is actually happening during the following carbocation rearrangement, and why it happens?

Answer

The rearrangement has little to do with ring strain and a lot to do with resonance stabilization.

The ring strain in a cyclobutane ring is ~26.3 kcal/mol, while the ring strain in a cyclopropane ring is ~27.5 kcal.mol (reference). This isn't much of a difference, certainly not enough to drive a reaction one way or the other.

The reason that cyclobutyl carbocations generally rearrange to cyclopropylcarbinyl carbocations is due to resonance stabilization. Here are some drawings that may help, if you build a model and look at the model that will help even more. The $\ce{C-C}$ bonds in a cyclopropane ring are approximately $\ce{sp^{3.74}}$ hybridized (see here), said differently, they have a lot of p-character in the bond. In fact, the p-character is so high that a cyclopropane compound will absorb bromine much like an olefin. If that cyclopropane high p-character bond (orbital) can align itself with the p-orbital on the carbocation center, it will stabilize the carbocation, just like a p-orbital on an adjacent double bond (the allyl system) stabilizes a carbocation.

here's another representation

Typically we discuss two conformations of the cyclopropane ring with the carbocation p-orbital, a bisected and a perpendicular conformation. In the bisected conformation the cyclopropane $\ce{C-C}$ bond lies in the same plane as the carbocation p-orbital. The carbocation p-orbital and the $\ce{C-C}$ orbital which resembles a p-orbital (because it is high in p-character) overlap! In the perpendicular conformation, the cyclopropane $\ce{C-H}$ bond is in the same plane as the carbocation p-orbital, the $\ce{C-C}$ bond is not - there is no carbocation stabilization by the cyclopropane ring in this conformation. Molecules where the bisected conformation can be attained show stabilization due to the stabilizing overlap between the carbocation p-orbital and the p-like $\ce{C-C}$ cyclopropane ring bond. Sometimes we use resonance structures such as the following to denote this stabilization.