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Offline sodium.dioxid

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This will blow your mind
« on: May 11, 2012, 04:14:48 PM »
Say we have tow containers, one on the left and one on the right (each one is 4L).

Left container has two molecules of A (rate=kA2)

Right container has one molecule of B and one molecule of C (rate=kBC)

Let's do the math (assume they both have the same rate constant, k=1)

left container: rate=[.5molecule/L]2= 0.25
right container: rate=[.25molecule/L][.25molecule/L]= 0.0625

What?!! Why are they not the same?!!

As you can see, there is nothing different about these two reactions other than the fact that the left reaction has two of the same molecules reacting. This difference causes us to use the different rate laws that we used. But why is the math not agreeing with the reality and nature of these reactions, as evident in the incompatible answers?

Offline Jorriss

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Re: This will blow your mind
« Reply #1 on: May 11, 2012, 04:41:21 PM »
First off, chemical kinetics is macroscopic. You don't apply the equations to a container with two molecules.

Second, it's not that surprising. Suppose we have a container with enough molecules for kinetics to be meaningful. For your first reaction, supposing it is an elementary reaction, we can safely say r=k[A]^2 and for the second, again assuming elementary reactions, r=k[ B][C].

You would not suspect the rates to be the same given the same concentrations. In the first case, A is reacting with itself to form some other molecule - the key is that every molecule in the container is A. It doesn't need to worry about bumping into the wrong molecule.

In the second case, B and C react but B and C have to waste a lot of time bumping into the wrong partner before they meet eachother and react.

The problem here is viewing it as two single molecules hitting each other. If its just two molecules, then of course the rates could be the same but those rates =/= k [A]^2 and k[ B][C], respectively.

Also, not all rates of the form r=[A][ B] are elementary and there are built in approximations and such. You know the distinction between a composite and elementary reaction, yes?
« Last Edit: May 11, 2012, 04:58:47 PM by Jorriss »

Offline sodium.dioxid

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Re: This will blow your mind
« Reply #2 on: May 11, 2012, 05:12:10 PM »
First off, chemical kinetics is macroscopic. You don't apply the equations to a container with two molecules.

The problem here is viewing it as two single molecules hitting each other. If its just two molecules, then of course the rates could be the same but those rates =/= k [A]^2 and k[ B][C], respectively.

Good stuff!

Also, not all rates of the form r=[A][ B] are elementary and there are built in approximations and such. You know the distinction between a composite and elementary reaction, yes?

Yes, this I know.

Offline juanrga

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Re: This will blow your mind
« Reply #3 on: May 12, 2012, 06:50:40 AM »
Your rate = k A2 is incorrect because you are counting the same molecule twice. Rewrite it as

rate = k A2 = kV nA nA

where kV includes the volume factors.

If your rate was right, when there is only a single molecule nA=1 in the container your equation predicts reaction with rate = kV, which is impossible.

A detailed analysis of the reactive collisions (A+A :rarrow: products) gives

rate = kV nA (nA - 1)

This equation gives the correct answer, rate=0, when there is only one molecule in the container.

If you use the correct rate equation for molecules of A you would get the same rates

rate(left) = rate(right).

Finally, notice that nA is of order of Avogadro number in macroscopic chemical kinetics. In macroscopic chemical kinetics (nA - 1) can be safely approximated by nA, and then one obtains the usual (macroscopic) rate in textbooks

rate = k A2
« Last Edit: May 12, 2012, 07:12:09 AM by juanrga »
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Offline sodium.dioxid

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Re: This will blow your mind
« Reply #4 on: May 12, 2012, 02:16:09 PM »
Your rate = k A2 is incorrect because you are counting the same molecule twice. Rewrite it as

rate = k A2 = kV nA nA

where kV includes the volume factors.

If your rate was right, when there is only a single molecule nA=1 in the container your equation predicts reaction with rate = kV, which is impossible.

A detailed analysis of the reactive collisions (A+A :rarrow: products) gives

rate = kV nA (nA - 1)

This equation gives the correct answer, rate=0, when there is only one molecule in the container.

If you use the correct rate equation for molecules of A you would get the same rates

rate(left) = rate(right).

Finally, notice that nA is of order of Avogadro number in macroscopic chemical kinetics. In macroscopic chemical kinetics (nA - 1) can be safely approximated by nA, and then one obtains the usual (macroscopic) rate in textbooks

rate = k A2

Thank you so much! I am looking for a rigorous chemistry textbook that gets into these kinds of details. Do you know of any books like this?

Offline Jorriss

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Re: This will blow your mind
« Reply #5 on: May 12, 2012, 02:18:13 PM »

Thank you so much! I am looking for a rigorous chemistry textbook that gets into these kinds of details. Do you know of any books like this?
There's lots of books but some would be useless based on your background.

What physical chem/math courses do you have under your belt at the moment?

Offline sodium.dioxid

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Re: This will blow your mind
« Reply #6 on: May 12, 2012, 03:33:39 PM »

Thank you so much! I am looking for a rigorous chemistry textbook that gets into these kinds of details. Do you know of any books like this?
There's lots of books but some would be useless based on your background.

What physical chem/math courses do you have under your belt at the moment?

I only have these:

Principles of Chemistry 1 & 2
Calculus 1 & 2 (I loved the "proofy" nature of these classes)
introductory statistics (I didn't really get too much out of this class)

I am actually studying for the PCAT. Right now, I am going over kinetics because I didn't understand it when I was taking the course. I am aware that the PCAT does not demand this level of detail; however, I am finding it hard to take the rate laws for what they are (that's just my nature). I don't find this stuff too obvious. Maybe there is some statistics behind their derivations.

Offline Jorriss

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Re: This will blow your mind
« Reply #7 on: May 12, 2012, 04:40:09 PM »

Thank you so much! I am looking for a rigorous chemistry textbook that gets into these kinds of details. Do you know of any books like this?
There's lots of books but some would be useless based on your background.

What physical chem/math courses do you have under your belt at the moment?

I only have these:

Principles of Chemistry 1 & 2
Calculus 1 & 2 (I loved the "proofy" nature of these classes)
introductory statistics (I didn't really get too much out of this class)

I am actually studying for the PCAT. Right now, I am going over kinetics because I didn't understand it when I was taking the course. I am aware that the PCAT does not demand this level of detail; however, I am finding it hard to take the rate laws for what they are (that's just my nature). I don't find this stuff too obvious. Maybe there is some statistics behind their derivations.
With calc I and II, you could get a Physical chemistry textbook like Levine and read the chapter on chemical kinetics. There are more specialized text that cover the theory but they would not be of use yet.

Offline sodium.dioxid

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Re: This will blow your mind
« Reply #8 on: May 12, 2012, 10:12:59 PM »

Thank you so much! I am looking for a rigorous chemistry textbook that gets into these kinds of details. Do you know of any books like this?
There's lots of books but some would be useless based on your background.

What physical chem/math courses do you have under your belt at the moment?

I only have these:

Principles of Chemistry 1 & 2
Calculus 1 & 2 (I loved the "proofy" nature of these classes)
introductory statistics (I didn't really get too much out of this class)

I am actually studying for the PCAT. Right now, I am going over kinetics because I didn't understand it when I was taking the course. I am aware that the PCAT does not demand this level of detail; however, I am finding it hard to take the rate laws for what they are (that's just my nature). I don't find this stuff too obvious. Maybe there is some statistics behind their derivations.
With calc I and II, you could get a Physical chemistry textbook like Levine and read the chapter on chemical kinetics. There are more specialized text that cover the theory but they would not be of use yet.

I was able to skim over the section. It looks like a good book. However, I wasn't able to get an answer for the following:

Why does doubling concentration doubles the collision frequency?
If molecules of C and A are bouncing in a container. Why does the collision frequency between them double when we double the molecules of C? Can this fact be described by mathematical language (theory)?

Sure, experiment leads to the conclusion that rate ∝ [A][C]. But, do we take this for granted? I actually intuitively thought that increasing either one and holding the other one constant would lead to exponential increase in rate.

Is it reasonable to conclude that this phenomenon is only an experimental fact (and cannot be proven mathematically)?
« Last Edit: May 12, 2012, 10:29:59 PM by sodium.dioxid »

Offline Jorriss

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Re: This will blow your mind
« Reply #9 on: May 13, 2012, 03:44:33 AM »
Why does doubling concentration doubles the collision frequency?
If molecules of C and A are bouncing in a container. Why does the collision frequency between them double when we double the molecules of C? Can this fact be described by mathematical language (theory)?

Sure, experiment leads to the conclusion that rate ∝ [A][C]. But, do we take this for granted? I actually intuitively thought that increasing either one and holding the other one constant would lead to exponential increase in rate.

Is it reasonable to conclude that this phenomenon is only an experimental fact (and cannot be proven mathematically)?
Your wording seems to think that experiment is second to math. You don't 'prove' anything mathematical in the physical sciences - the highest aim of theory is to explain experiment. Being an experimental fact is a very meaningful fact.

That being said, intuitively, I do not know why you think increasing concentration would increase the rate exponentially. If one has two types of molecules, A and B, and species, say, A is doubled, then why would you think it can collide with B exponentially as many times? The new molecules will have the same geometric constraints, distribution of energies, etc. The only difference is that there are more, so they *Ignore me, I am impatient* into eachother in proportion to the amount that was added.

Offline juanrga

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Re: This will blow your mind
« Reply #10 on: May 13, 2012, 06:33:49 AM »
The cemetery is full of skeletons of people with mathematical 'proofs' discredited by nature in labs.

As Jorriss correctly points, chemistry is one of the experimental sciences. Said that, I want to emphasize that it is possible to derive the general rate law from more fundamental theory. It is then found that the rate is not proportional to the product of concentrations [A][C] but to the product of activities aA aC plus additional terms.

But this requires a background far beyond Levine.
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Offline sodium.dioxid

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Re: This will blow your mind
« Reply #11 on: May 14, 2012, 07:32:13 AM »
I just had an epiphany. :o
I knew that if I didn't give up on this I would be able to understand it on a mathematical level (why concentrations affect the rate the way they do).
With that, I now can "imagine" the rate law with all the different factors at play simultaneously (temperature, steric factor, activation energy, concentration, ...); I have a full sense for it. ;D

And that's what I sought out to do all along.

A combination of things helped: I went back and read over these posts again + I found something called "mean free path" in my Physics textbook.

Huge thanks to you, Jorriss and Juanrga. You guys were sent down from heaven to help us mortals out.

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