mississippichem

Yeah, plotting [math]ln(k)[/math] versus [math]\frac{1}{T}[/math] gives [math]E_a[/math] as the gradient of the resultant line. Then one could solve for [math]A[/math] i guess. Oh, so find [math]k[/math] at two different temperatures, then rearrange to [math]lnk=-(\frac{E_a}{RT})+lnA[/math]. Subtracting the equations from two diiferent temperatures would give: [math]ln \frac{k_2 }{k_1 } = \frac{-E_a }{R} \left[\frac{1}{T_2 } - \frac{1}{T_1 }\right][/math] which conveniently eliminates the [math]A[/math] term. Cool Thanks for your help guys, you set me on the right course. Anyone know a different mathematical treatment to arrive at activation energy?

[math]k=Ae^{\frac{-E_a}{RT}} [/math] -where A is the collision frequency factor -e is the natural log base -R is the gas constant [math](8.314510 \times 10^{-3} kJ \ K^{-1} \ mol^{-1})[/math] -k is the rate constant for the reaction -Ea is the activation energy Is there a way to calculate the collision frequency factor, A, without just rearranging [assuming activation energy is unknown]? The collision frequency factor is supposed to be specific for every reaction, so there must be a seperate way to calculate it for it to be of any use.

Gravity, not quite vaccum-like though. Only in the superficial, non-technical sense.

Alright: [ce]so \ E^o=2.86 V \ for \ 3KClO_4 + 8Al -> 4Al_2O_3 + 3KCl[/ce] [math]\Delta G^o = -nFE^o[/math] [math]where \ F=96485.338 \ Cmol^{-1}[/math] [math]so \ \Delta G^o = -221 \ kJmol^{-1}[/math] I don't have the standard entropy values for some of the species so I can't calculate total enthalpy but i'd be willing to bet that entropy makes only a small contribution and this reaction is quite exothermic. Must be kept cold and in static free conditions, but dude, I really wouldn't want to store this stuff. The only way to ensure saftey is not to store it.

UC is right that should provide your answer. Also keep in mind though: Will there be a significant difference between the electron donating/withdrawing activities between a carboxylic acid group, methyl ester group, or a carboxylate (conjugate base) group? Also, which of these groups allows the most number of [significantly contributing] resonance forms in the non-aromatic intermediate?

pretty sensitive. This mixture has the potential to explode spontaneously The finer the aluminum powder the easier it will detonate. A you fairly familiar with redox chemistry? If so, I'll explain why quantitatively; if not I won't bore you. Either way, be careful with this mixture. I tend to work with a good degree of caution and would not want to store this mixture. I would prepare this mixture and detonate it on the spot. NASA used to use a similar mixture for their solid fueled rockets, but they stored it in a "polymer matrix" to prevent an ignition event from getting out of hand.

Awesome, I read an article about that, and I've always been curious about the physics of those solutions. Couldn't remember if it was crypt-222 or 18-crown-6 (macrocycle nomenclature sucks). Any shots on my question?

interesting observation. Agreed Perhaps the aromatic ring helps deal with some of the electron density, just a thought; but there is still that bond length problem. I would really love to see some molecular modeling calculations for that intermediate. Anyone with a copy of Spartan willing to waste a couple of hours?

Here's an obscure one, -This compound displays extreme valency (coordination number is two digits) -The ligands in this compound are perhaps the stangest ever -The metal center often takes oxidation states of +2 or +4 -The compound is binary in the sense that there is one metal center and one ligand type. -The ligands, in their atomic form have very high ionization energies. Could this be a solution of a group I metal in ammonia known as an electride salt? ex. [Na(NH3)x]+ e-

Ironically enough, I think it could be said that the death penalty upholds the high value our society ascribes to human life. When someone is convicted of wrongfully taking the lives of others, I feel that most peoples' gut reaction is, "so take his life". No I don't think the death penalty is an effective deterrent, but I think it keeps society in the mindset that human life is a valuable thing; not to be taken lightly.

Its beacause at the end of each period the p-orbitals are full with 6e- and the aufbau principle states that the next n-level s-orbital must be filled. There are three available ml quantum #'s for the set of p-orbitals each p-orbital can hold 2 e-, so then the pauli exclusion principle takes over and electrons must be filled under a different angular momentum quantum number, i.e. the next available s-orbital.

Big Banks have been spooked as of recent due to the recent financial legislation that puts restrictions on them trading derivatives. They new bill created a clearing house that slows down the derivatives securities transactions. That along side with a base interest rate of 1%, keeps big lenders' money locked up. Their is little motivation to lend with such low interest rates. Just take a lok at the S&P 500 on any given day, it's a blood bath out their; not all bad, but extremely volatile. A lot of these companies want that money on their balance sheets when quarterly reports come out, so when potential investors analyze the corperation's financials values like ROE (return on equity), DER (debt to equity ratio) and straight out cash reserves the company will appear fundamentally sound. This is part of a long list of accounting tricks that every CFO who is worth his salt knows. All that to say, alot of these companies are not prepared to acquire the risk (or investor perception of risk) of new employment. Lokk up the concept of liquidity trap. Thst is the situation we are in our 1% base interest rate. At face value it has nothing to do with jobs, but it's effects are far reaching. Hope for a healthy bit of inflation; that is what we need. (never thought I would say that)

There's nothing wrong with the mechanism you posted. I was just saying that I would endorse the mechanism you posted over the direct decarboxylation. There is alot of resonance present in the various intermediates, this tends to hinder the decarboxylations by reducing the electron density on the carboxyl group. Those pesky decarboxylations can stump even the wisest of chemists Another complicating factor is that the decarboxylations are highly favored by entropy assuming the temperature is high enough for the resulting CO2 to exist as a gas (which it always is). Alot of variables to consider. Its one of those borderline cases where entropy could beat out enthalpy. I feel that a calorimetric analysis is needed. Got any grant money?

Okay, how does that make words evil?

What is it that you understand that we don't? Genuinely curious

I think I could see this same transformation occur through direct aldol condensation with benzaldehyde followed by an elimination. But I fell the decarboxylation type mechanism is thermodynamically far fetched. I'd like to see the BDE value to break that C-C bond. Decarboxylations are finicky reactions. Seemingly similar substrates sometimes yield very different results. -interesting post, not enough posts of this type in chemistry forums.

Not so sure about a good solvent, but I wouldn't want to dissolve any chlorate or perchlorate salts in an alcohol. Perchlorates are the slowest oxidizers or the chloro-oxoanion series but their stand reduction potentials are still around ~(+1.2) V. Maybe acetonitrile or acetic acid would be good solvents.

I think your link is broken. I can't access the web page.

Beautiful, you should go to art school. Then maybe you could find an art forum to post on.

Men's bathroom law should've been included in John Locke's "Natural and Inaliable Rights of Man". The right to empty one's bowels without fear of an audience or the most hated "inter-stall conversationist".

Adding alkali metals to anhydrous ammonia produces what basically amounts to a solution of alkali metal cation (I) and a solvated electron. These solutions are often and unofficially reffered to as "electride" solutions. My question is: what happens to the solvated electron upon photo-excitation? For example, what kind of UV-vis or EPR spectrum might arise from characterization of this solution. The physics of these solutions are poorly understood at best as far as I know, (correct me if I'm misinformed). On a side note, these solutions undergo interesting color shifts as concentration changes. The completely saturated solution takes on a golden metallic luster for the sodium analogue, I smell lack of a conductive band gap!?

This is actually quite a tricky question. I believe "D" is the answer. "A" looks good at first site, but EAS occuring at "A" would result in the product being meta-substituted with reference to the methyl group. D works because it is para to the alkoxy activating group. I don't really know why,but activating groups tend to slightly perfer the para postion to the ortho. It must have something to do with steric hinderence in the kinetics. I've just noticed in literature that the para isomer tends to be perfered over the ortho when they compete.

I once heard the laws of thermodynamics stated as: 0) Its not what you think 1) You can't win 2) You can't even break even 3) Don't even try Funny, yes. True, even more so.

Cool, let me add that the 3 p-orbitals, 5 d-orbitals, or 7 f-orbitals are said to be degenerate, that is the correct term to imply they have equivalent energy. Thats an important term to know as it is used extensivley in basic chemistry/ atomic physics.

Different numbers of protons and neutrons in Be and B would cause their 1s orbitals to be of different energies. This highlights the concept of effective nuclear charge, basically how much the nucleus "pulls" on the electrons. As far as different p orbitals: All the energies of the electrons in the same p-orbitals (when n is constant) will be equivalent. For example, electrons in 3p orbitals (x,y,and z) will have equivalent energy.