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### Topic: Equilibrium kinetics  (Read 124055 times)

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#### curiouscat

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##### Re: Equilibrium kinetics
« Reply #165 on: June 01, 2013, 01:41:19 PM »

It would also require a modification from the RA*V ODE if we use activities instead of concentrations (multiplying activity by volume does not reach number of moles).

Why would you multiply activity by volume.

#### Big-Daddy

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##### Re: Equilibrium kinetics
« Reply #166 on: June 01, 2013, 02:12:27 PM »
Why would you multiply activity by volume.

You can't, that's my point. Though our ODE might still be fine at constant volume if we replace all concentration terms directly with activity terms, it's not so simple to replace the concentration terms with activity terms in the non-constant volume form of the ODE:

$$\frac{d \left ( c_A \cdot V \right ) }{dt}= R_A \cdot V \\$$

RA was the function in terms of rate that we spent so long figuring out.

#### curiouscat

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##### Re: Equilibrium kinetics
« Reply #167 on: June 01, 2013, 05:27:52 PM »
No clue.

#### Big-Daddy

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##### Re: Equilibrium kinetics
« Reply #168 on: June 02, 2013, 11:25:54 AM »
OK, let's leave the activities thing for the time being unless someone does know the answer.

Often the system is stiff and solving it is one gigantic headache.

You said this a while back. Is that when you're trying to find analytical solutions? The equations as we wrote them now are exact, I think, so there must be some approximations you make most of the time to reach analytical solutions.

Or even when looking for numerical solutions, the ODEs still throw up some problems?

#### curiouscat

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##### Re: Equilibrium kinetics
« Reply #169 on: June 02, 2013, 12:05:08 PM »
OK, let's leave the activities thing for the time being unless someone does know the answer.

Often the system is stiff and solving it is one gigantic headache.

You said this a while back. Is that when you're trying to find analytical solutions?

What do you understand by a "stiff" system?

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The equations as we wrote them now are exact, I think, so there must be some approximations you make most of the time to reach analytical solutions.

Quasi Steady State Assumption often. Equilibriated steps sometimes. Depends.

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Or even when looking for numerical solutions, the ODEs still throw up some problems?

Quite often.

#### Big-Daddy

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##### Re: Equilibrium kinetics
« Reply #170 on: June 02, 2013, 01:16:55 PM »
What do you understand by a "stiff" system?

Something that resists solution by commonly known or straightforward (/obvious) methods.

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Quite often.

Ah. Can you give an example of one which provides problems for numerical solution?

Equilibriated steps sometimes.

Where can I look this particular one up? (what assumption is taken, when and why we can take it, how it makes the solution easier) The steady state approximation is already known to me.

#### curiouscat

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##### Re: Equilibrium kinetics
« Reply #171 on: June 02, 2013, 01:34:41 PM »
What do you understand by a "stiff" system?

Something that resists solution by commonly known or straightforward (/obvious) methods.

Why don't you read it up. That answer is practically useless.

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Quote
Quite often.

Ah. Can you give an example of one which provides problems for numerical solution?

If I gave you one what will you do with it.

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Equilibriated steps sometimes.

Where can I look this particular one up?

Google's a good start.

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The steady state approximation is already known to me.

Good.

#### Big-Daddy

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##### Re: Equilibrium kinetics
« Reply #172 on: June 02, 2013, 07:14:36 PM »
Why don't you read it up. That answer is practically useless.

What I read up contained some quite detailed maths which I mostly haven't come across. But from several internet searches it seems like actually quite a complicated field. Overall, my impression on a basic level is that we call the ODE system stiff if, of the different mathematically possible real solutions, a very wide range and/or difference exists (e.g. one solution is 109, another 10-9) meaning that in predicting estimates for iteration we will find it very cumbersome to yield both answers which could be correct.

If I gave you one what will you do with it.

Mainly just wanted to see if the ODE itself looked any different from the ones we've already studied. Or if there is a visibly obvious reason from the ODE why it should be stiff. I guess I could just have asked. Sorry, if that would have been simpler for you.

Google's a good start.

I Googled "chemical kinetics equilibriated steps" and found nothing which seems to be of relevance. There was some stuff I found very interesting on a Wikipedia page that came up as a result (http://en.wikipedia.org/wiki/Detailed_balance), which is vaguely to do with kinetics (it postulates that each reaction should actually be written as an equilibrium, i.e. reversible, and then throws up some details, interesting for me at least, about what is or is not possible if we treat reactions are irreversible), but I don't see what ostensibly is the approximation you mean by 'equilibriated steps'. On the Wikipedia page, the "approximation" is to make a series (cycle) of irreversible reactions into equilibria so that they are actually possible, but this is of course the opposite of an approximation, and in any case you've already shown me how to write the ODEs for equilibria as opposed to irreversible, the page doesn't go into approximations.

My question was a little open-ended before so I'll govern it now. Are there any more approximations I should be aware of, e.g. in the first few years of undergraduate study of kinetics? (Other than steady state) I bring this up because some of the stuff on the Wikipedia page I linked to is beyond me. If that is what you mean by equilibriated steps maybe it's better to ignore it for now.

#### curiouscat

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##### Re: Equilibrium kinetics
« Reply #173 on: June 03, 2013, 01:14:27 AM »
Why don't you read it up. That answer is practically useless.

What I read up contained some quite detailed maths which I mostly haven't come across. But from several internet searches it seems like actually quite a complicated field. Overall, my impression on a basic level is that we call the ODE system stiff if, of the different mathematically possible real solutions, a very wide range and/or difference exists (e.g. one solution is 109, another 10-9) meaning that in predicting estimates for iteration we will find it very cumbersome to yield both answers which could be correct.

Well, somewhat OK I guess but in any case that should answer your own question "You said this a while back. Is that when you're trying to find analytical solutions? "

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If I gave you one what will you do with it.

Mainly just wanted to see if the ODE itself looked any different from the ones we've already studied. Or if there is a visibly obvious reason from the ODE why it should be stiff. I guess I could just have asked. Sorry, if that would have been simpler for you.

No it won't look any different. A wide variation in magnitudes of constants, yes.  More pertinently, I've been trying (quite unsuccessfully I admit) to coax you away from the idea that you can do much quantitative science by just "looking".  The only good ways to appreciate kinetics, ODE's, stiffness etc. is by actually getting your hands dirty. Of all the questions you've asked I don't recall many where you've actually been attempting any numerical solution?

Quote
My question was a little open-ended before so I'll govern it now. Are there any more approximations I should be aware of, e.g. in the first few years of undergraduate study of kinetics? (Other than steady state) I bring this up because some of the stuff on the Wikipedia page I linked to is beyond me. If that is what you mean by equilibriated steps maybe it's better to ignore it for now.

Sometimes fast steps are assumed to be at equilibrium. Can't think of any others.

If solving numerically, you don't need approximations mostly.

#### Big-Daddy

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##### Re: Equilibrium kinetics
« Reply #174 on: June 03, 2013, 05:20:33 AM »
No it won't look any different. A wide variation in magnitudes of constants, yes.  More pertinently, I've been trying (quite unsuccessfully I admit) to coax you away from the idea that you can do much quantitative science by just "looking".  The only good ways to appreciate kinetics, ODE's, stiffness etc. is by actually getting your hands dirty. Of all the questions you've asked I don't recall many where you've actually been attempting any numerical solution?

Solving ODEs simultaneously is going to be quite difficult for me. Until I can do it on a computer I'm not going to do it by hand. Although I can handle very basic cases analytically, these are practically trivial compared with what we've been talking about, and I don't see much point in going through a rote solution using a computer which just requires me to type out what we've been talking about here and then plugging in some initial values? I'll get there eventually but not sure I see the hurry in comparison to understanding the formulation. I wouldn't even know whether the answer that comes out is actually correct.

Sometimes fast steps are assumed to be at equilibrium. Can't think of any others.

If solving numerically, you don't need approximations mostly.

OK, thanks.

#### curiouscat

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##### Re: Equilibrium kinetics
« Reply #175 on: June 03, 2013, 05:28:51 AM »
No it won't look any different. A wide variation in magnitudes of constants, yes.  More pertinently, I've been trying (quite unsuccessfully I admit) to coax you away from the idea that you can do much quantitative science by just "looking".  The only good ways to appreciate kinetics, ODE's, stiffness etc. is by actually getting your hands dirty. Of all the questions you've asked I don't recall many where you've actually been attempting any numerical solution?

Solving ODEs simultaneously is going to be quite difficult for me. Until I can do it on a computer I'm not going to do it by hand. Although I can handle very basic cases analytically, these are practically trivial compared with what we've been talking about, and I don't see much point in going through a rote solution using a computer which just requires me to type out what we've been talking about here and then plugging in some initial values? I'll get there eventually but not sure I see the hurry in comparison to understanding the formulation.

To me that's absolutely the wrong, totally inefficient approach. But you are entitled to having your opinion.

And if you think a computer solution is "a rote solution", I urge you to correct that misconception. There's nothing rote about solving a good kinetics problem, with or without a computer.