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DJF90
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If you have [H+] or [OH-] in your rate equation it means that there is some form of proton transfer in the rate determining step. Take for instance
the reaction of acetone with iodine under acidic conditions. By varying the concentration of acid, acetone and iodine (one variable at a time
please!), we can show after a little bit of interpretation of the data that the rate equation is:
k = [Me2C=O][H+]
Notice how there's no [I2] in the rate equation... this is because iodine does not participate in the rate determining step, i.e. no matter how much
iodine you add to the reaction, it will not change the rate at which the reaction proceeds. If you consider the mechanism for the reaction, the reason
becomes obvious:
The first step of the reaction is enolisation of the ketone, in which acid participates. This is the slowest step of the whole reaction and as such is
the rate determining step. The second step is a very rapid nucleophilic attack by the enol onto iodine.
I don't know how useful this will be to you, but I really think you're getting too far ahead of yourself. I'm all up for people being keen chemists (I
was one such person myself) but be careful that you learn everything correctly as making a mistake whilst teaching yourself will make like more
difficult at a later date, as opposed to making life easy as is probably half the motivation for teaching yourself in the first place.
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Rich_Insane
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Well I understand what the rate law is and how some things participate in a reaction. I just need confirmation on how to find the exponent (aka
[A]^2[B] = 3rd order reaction). I have an idea, but I'm not sure I'm right. I also still am confused at how to determine whether something is a second
order or third order, or whatever just my looking at the graph. As far as I know, 1st order has something to do with ln, and 2nd order is exponential.
I'm pretty shaky on redox reactions. Especially ones that don't involve metals in elemental form becoming cations. So to verify, please tell me if I'm
correct witht he following reaction:
NH3 + H2O --> (NH4+) + OH-
So the H2O has a charge of 2- and the NH3 has a charge of 0. What happens is that the NH3 loses an electron and is protonated by water to form NH4+
(which would involve one electron being transfered to H2O to drop off the proton), and that electron goes to OH. Basically NH3 is getting oxidized and
water is being reduced... right?
In most reactions, the H+ is present to balance out the oxygens right? (for example, with Cr2O7, those 7 oxygens are balanced out by 14 H+ to form
7H2O). What is the purpose of a hydroxide balanced redox reaction then?
@ DF90: I'm actually taking the class. However since none of the high schools nearby willt ake me as they are grossly overcrowded, I am forced to take
it online. I'm more worried about memorizing all these formulas and remembering them on the test day. Do you seriously have to memorize them all?
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Lambda-Eyde
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That's not a redox reaction, that's an acid-base reaction. The N in ammonia has a -3 charge on both sides of the arrow. H is +1 and O is -2.
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Rich_Insane
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The book describes it as one, I'm aware it isn't one. I think they want to know the transfer in charge.
Alright, so my class is finished (took the final exam). I got a B overall, but my exam was.....OK. It seems as though I need to memorize some
important equations. Is there anyone who can tell me what formulas I should definitely memorize for the AP Exam (which is next week) ?
[Edited on 2-5-2010 by Rich_Insane]
[Edited on 2-5-2010 by Rich_Insane]
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Rich_Insane
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Quick question here, it's a day before the exam, but I have to just open my mind up to some information that I forgot. I know that one can find the
spontaneity of a reaction with Gibb's free energy (dG = dH -TdS), but when you have an equilibrium reaction like the Haber Bosch process, how would
one find which side of the reaction is favored? Doesn't it have something to do with the equilibrium constant? I'm pretty sure I know this, but I
can't recall how one would find that out. Does anyone have some general tips for the exam? How difficult is it?
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JohnWW
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The pressure and temperature greatly affect the yield, as per equilibrium constants, of NH3 in the Haber process for producing NH3 from reaction of N2
with H2. Higher pressures favor NH3, as the result of Avogadro's law.
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Skyjumper
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Test was today for americans. The chemistry department was empty today.
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Rich_Insane
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Well, I took the test. I'm not sure whether it was easy or hard, but all the equilibrium questions were easy. NH3 is favored because nature favors
higher pressures right? So the side that creates the most mols of gas will be favored. Al higher volumes and lower pressures, the reactants will be
favored.... right?
The free response was fairly easy. I sort of forgot all my calorimeter stuff so one of them was difficult, but all the stoichiometry and reactions
were simple. Sucks that they didn't have a question about synthesis. I'm adamant I got at least 3, but I'm really hoping for a 5.
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Magpie
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Where did you get this idea? What is your basis for saying this?
The single most important condition for a successful synthesis is good mixing - Nicodem
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entropy51
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Magpie, since Nature abhors a vacuum she must favor high pressures, no?
| Quote: | | I have a professor who's really nice to me who might let me take O Chem Year 1 right now, in 7th grade, but I really, really don't want to considering
how bad my memory at chemistry is. |
But Richie, a year ago you said that you were already taking organic
chem in college, didn't you????
Ah, here's some of it: Quote: Originally posted by Rich_Insane  | I'm supposed to take the SAT every year after this. I took it in 7th grade, and from now on I have to take it every year. It pisses me off.
I might just study until college and do experimentation there. I mean at least I'll have the knowledge right? I dropped out of O Chem in the 2nd term
(I'd be taking it in High school anyways). It was actually pretty good, because now I have ways to make my life easier when it comes by.
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You said you were past seventh grade a year ago, but above you said you're in seventh
grade. You're full of it, aren't you? Are you insane, or just a pathological liar??
[Edited on 13-5-2010 by entropy51]
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Rich_Insane
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| Quote: |
You said you were past seventh grade a year ago, but above you said you're in seventh grade. You're full of it, aren't you? Are you insane, or just a
pathological liar?? |
Summer between 7th and 8th. The SAT was beginning of the year in 7th grade. My bad if that was confusing. It's not a big deal, a lot of people I knew
took it too. Also I didn't take O Chem, I was auditing the class to see what it was like. I took that in the summer. The O Chem was a
summer session which I mentioned before. Therefore I was in between 7th and 8th, so I wasn't sure what to say. That's the background. That's not the
issue here either.
I learned that the Haber Bosch process was volume dependent, or rather pressure dependent, and if I remembered correctly, the higher pressure if
favored. Lower volume = higher pressure and more moles of gas = higher pressure.
[Edited on 13-5-2010 by Rich_Insane]
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Magpie
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| Quote: Originally posted by Rich_Insane |
I learned that the Haber Bosch process was volume dependent, or rather pressure dependent, and if I remembered correctly, the higher pressure if
favored. Lower volume = higher pressure and more moles of gas = higher pressure.
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Isn't this the Haber process?:
N2 + 3H2 <---> 2NH3
If so, it is turning 4 moles of gas into 2 moles of gas. According to Le Chatelier's principle higher pressure would favor the formation of less
volume, ie, the NH3.
Nature likes entropy (if it "likes" anything at all), which would favor higher volumes and lower pressures. But it is man's machinery and energy
input that is providing the pressure for the Haber process, not Nature.
The single most important condition for a successful synthesis is good mixing - Nicodem
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Rich_Insane
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Thanks Magpie. I probably got that backwards. I always thought higher pressure was higher entropy. I'd assume that man-made equipment does something
to react N2 and H2 to form NH3, because if you just released some H2 gas into the air, I highly doubt that ammonia would form all of a sudden.
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DJF90
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Rich_Insane: Yes, you have to learn by heart alot of equations. If you're clever enough you can just learn the fundamentals and work the rest out from
first principles etc. Thermodynamics, quantum mechanics, statistical thermodynamics, kinetics, spectroscopy, electrochemistry etc all have fairly
large numbers of equations to learn. Its a pretty tough life being a chemist.
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Rich_Insane
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The only thing I have real trouble on at this point is thermodynamics, because I get confused and mix up properties of entropy with enthalpy and vice
versa. I also need to figure out how to find which way an equilibrium tips. If I am right, k is derived from the law of mass action (products
concentration/reactants concentration), which is derived from experimental data. If I am not mistaken, when k is larger than 1, the products are
favored. When it is less than one, the reactants are favored. At one, both exist in the same amounts. Is that all? What equations are absolutely
essential to memorize (I've already got Gibb's free energy, the Nernst equation, and that one equation used for buffers, I can also do most
equilibrium constants)?
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Magpie
lab constructor
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| Quote: Originally posted by Magpie |
Nature likes entropy (if it "likes" anything at all), which would favor higher volumes and lower pressures. |
Actually I don't think this statement is accurate. For reversible (frictionless/equilibrium) processes entropy does not change with either compression
or expansion.
The single most important condition for a successful synthesis is good mixing - Nicodem
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