March 28, 2024, 11:56:06 AM
Forum Rules: Read This Before Posting


Topic: Equilibrium kinetics  (Read 113871 times)

0 Members and 1 Guest are viewing this topic.

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Equilibrium kinetics
« Reply #30 on: May 05, 2013, 03:53:30 PM »

What about the rate-laws under Section C (Nonelementary rate laws) Reaction (2), Section D and Section E here: http://www.umich.edu/~elements/course/lectures/three/ in the Section 1.2 - Power Model?

Even those. Doesn't matter.


Quote
Those are some very strange rate laws indeed, I can't even begin to understand them ...

When the reaction isn't elementary the law gets complicated.

Quote
Wait, so it is possible for reaction orders/the rate constant to change depending on the concentration?

You mentioned the possibility. I merely said it doesn't matter to our modelling strategy. ODE's will work. :)

But yes, I suppose for non-elementary reactions orders etc. might change. Why, even an entirely different form might apply to different concentration regimes.

Quote
So how do we incorporate that into our model? We need to write the rate constant as a function of the concentrations, then the reaction order for each reactant/product as a function of the concentrations?

Again, all I need is  rate = function(time, concentrations). What function is the domain of determining reaction kinetics and rate laws. A different ball game. Could be empirical. Microkinetics. etc.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #31 on: May 05, 2013, 04:08:19 PM »

What about the rate-laws under Section C (Nonelementary rate laws) Reaction (2), Section D and Section E here: http://www.umich.edu/~elements/course/lectures/three/ in the Section 1.2 - Power Model?

Even those. Doesn't matter.


Quote
Those are some very strange rate laws indeed, I can't even begin to understand them ...

When the reaction isn't elementary the law gets complicated.

Quote
Wait, so it is possible for reaction orders/the rate constant to change depending on the concentration?

You mentioned the possibility. I merely said it doesn't matter to our modelling strategy. ODE's will work. :)

But yes, I suppose for non-elementary reactions orders etc. might change. Why, even an entirely different form might apply to different concentration regimes.

Quote
So how do we incorporate that into our model? We need to write the rate constant as a function of the concentrations, then the reaction order for each reactant/product as a function of the concentrations?

Again, all I need is  rate = function(time, concentrations). What function is the domain of determining reaction kinetics and rate laws. A different ball game. Could be empirical. Microkinetics. etc.

You said "even those" for the non-elementary, but then says it gets complicated. But those rate laws are clearly not expressed as r=kAxBy. OTOH I am not entirely convinced they are referring to a single reaction at all, it could be multiple reactions combined into a single law there? So if we express each reaction separately, will it always take the form r=kAxBy?

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Equilibrium kinetics
« Reply #32 on: May 05, 2013, 04:11:48 PM »
OTOH I am not entirely convinced they are referring to a single reaction at all, it could be multiple reactions combined into a single law there?

What do you understand by non-elementary?

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #33 on: May 05, 2013, 04:37:46 PM »
OTOH I am not entirely convinced they are referring to a single reaction at all, it could be multiple reactions combined into a single law there?

What do you understand by non-elementary?

Multiple reactions combined into a single law :p But let's say we don't have the mechanism for a reaction, we only have the overall thing (but it still goes to completion). Can we express the law in the r=kAxBy form or does it need to be something more complicated?

My instinct when I see something non-elementary like A :rarrow: B :rarrow: C is that surely, just like the ODEs we just wrote, we can write 3 more now, one for each species. But if the rate law for each section of the ODE is not of form kAxBy then that means more work is needed...

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Equilibrium kinetics
« Reply #34 on: May 05, 2013, 11:49:34 PM »
But let's say we don't have the mechanism for a reaction, we only have the overall thing (but it still goes to completion). Can we express the law in the r=kAxBy form or does it need to be something more complicated?

No general answer. Rate law for a non-elementary reaction could be practically anything.

Didn't you post a link with all kinds of complicated rate laws. Were they all in kAxBy form?

Quote
My instinct when I see something non-elementary like A :rarrow: B :rarrow: C is that surely, just like the ODEs we just wrote, we can write 3 more now, one for each species. But if the rate law for each section of the ODE is not of form kAxBy then that means more work is needed...

I don't understand what you want to say.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #35 on: May 06, 2013, 06:30:06 AM »
But let's say we don't have the mechanism for a reaction, we only have the overall thing (but it still goes to completion). Can we express the law in the r=kAxBy form or does it need to be something more complicated?

No general answer. Rate law for a non-elementary reaction could be practically anything.

Didn't you post a link with all kinds of complicated rate laws. Were they all in kAxBy form?

Quote
My instinct when I see something non-elementary like A :rarrow: B :rarrow: C is that surely, just like the ODEs we just wrote, we can write 3 more now, one for each species. But if the rate law for each section of the ODE is not of form kAxBy then that means more work is needed...

I don't understand what you want to say.

OK, but even for a non-elementary law, the rate law will always be a function of the rate constants and concentrations (and maybe some other concentrations), so to put it in our ODE we just have to put that function in the space where I just now put the -k1f·(cA)x section etc. (the function will be put negative if it involves conversion of the substance for which I'm writing the ODE to something else, and will be put positive if it involves production of the substance for which I'm writing the ODE), correct?

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Equilibrium kinetics
« Reply #36 on: May 06, 2013, 07:58:05 AM »

OK, but even for a non-elementary law, the rate law will always be a function of the rate constants and concentrations (and maybe some other concentrations), so to put it in our ODE we just have to put that function in the space where I just now put the -k1f·(cA)x section etc. (the function will be put negative if it involves conversion of the substance for which I'm writing the ODE to something else, and will be put positive if it involves production of the substance for which I'm writing the ODE), correct?

Yes. Correct.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #37 on: May 10, 2013, 04:15:05 PM »
Does it occur, in your experience, that we have to use different rate equations in the middle of calculating concentration/time dependence for a single reaction? (i.e. either the reaction orders or the rate "constants" change at some point)

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3471
  • Mole Snacks: +526/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Equilibrium kinetics
« Reply #38 on: May 11, 2013, 05:08:21 PM »
Why should the kinetics of a reaction depend on the process of calculation?
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #39 on: May 11, 2013, 05:52:24 PM »
Why should the kinetics of a reaction depend on the process of calculation?

Why does this matter? I'm asking whether we observe a single rate law to work for all reactions until the end, or whether in your experience you actually have observed that reactions do not follow the same exact rate law until the end (the end being when all the concentrations are equilibrium concentrations - might be 0 for a reactant in an equilibrium where backward rate constant is 0, i.e. one which goes to completion).

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3471
  • Mole Snacks: +526/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Equilibrium kinetics
« Reply #40 on: May 11, 2013, 08:29:16 PM »
What I meant was: the rate law is an artificial construct based upon a specified mechanism.  If experimental reaction rate data deviate from the prediction of the rate law at any point, it's probably not because of any bizarre behavior, but because the specified mechanism isn't correct.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #41 on: May 12, 2013, 05:42:40 AM »
What I meant was: the rate law is an artificial construct based upon a specified mechanism.  If experimental reaction rate data deviate from the prediction of the rate law at any point, it's probably not because of any bizarre behavior, but because the specified mechanism isn't correct.

OK, so in your experience the deviation would never occur because the rate constants or reaction orders are themselves functions of concentration, but rather because the rate law was inaccurately determined. The question comes down to this:

If we can break up the mechanism into a (correct) set of single steps, each of which will therefore have an elementary rate law, can this rate law always be expressed in the form r=k[A]x[ B]y where k, x and y are constants, with complete accuracy as far as your experience suggests?

Directed to anyone who can answer.
« Last Edit: May 12, 2013, 06:23:28 AM by Borek »

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Equilibrium kinetics
« Reply #42 on: May 12, 2013, 08:03:57 AM »
If we can break up the mechanism into a (correct) set of single steps, each of which will therefore have an elementary rate law, can this rate law always be expressed in the form r=k[A]x[ B]y where k, x and y are constants, with complete accuracy as far as your experience suggests?

With complete accuracy? Never.

Rate laws in this commonly used form (r=k[A]x[ B]y) for elementary steps are yet an approximation. A model.

This isn't like Newton's law that nature is forced to obey it. More like Ohm's Law. Mostly true and a good approximation.

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3471
  • Mole Snacks: +526/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Equilibrium kinetics
« Reply #43 on: May 12, 2013, 08:35:07 AM »
@BD

What do you envision the reaction orders would vary as a function of?
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #44 on: May 12, 2013, 08:56:01 AM »
With complete accuracy? Never.

Rate laws in this commonly used form (r=k[A]x[ B]y) for elementary steps are yet an approximation. A model.

This isn't like Newton's law that nature is forced to obey it. More like Ohm's Law. Mostly true and a good approximation.

Huh ... ok, let me ask a different question then: do we ever observe the rate law to dramatically change at a certain point in the experiment (i.e. discrete change)? Or is it always just a continuous change which we can model in terms of species concentration?

@BD

What do you envision the reaction orders would vary as a function of?

Concentration of any and all species. OR something discrete, if you have ever observed that this occurs (e.g. at some point in the concentration of the reactants or products, the rate law suddenly changes from r=k[A]2[ B]3 to r=k[A]3[ B]2).
« Last Edit: May 12, 2013, 09:23:19 AM by Borek »

Sponsored Links