July 15, 2024, 01:15:53 PM
Forum Rules: Read This Before Posting


Topic: Equilibrium kinetics  (Read 124032 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 #60 on: May 12, 2013, 05:04:34 PM »
In general, whenever you have asked I have always responded with the hard work. When you asked me to give attempts for the ODEs earlier this thread, I did, and used LaTeX as you suggest. Perhaps your last line - which I embolden - is the main idea though.

Could be. I don't know. All I can  tell you is you've frustrated at least two of us here. 

Maybe I am wrong about my advice. I might have been too harsh. Sorry.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #61 on: May 12, 2013, 05:44:06 PM »
In general, whenever you have asked I have always responded with the hard work. When you asked me to give attempts for the ODEs earlier this thread, I did, and used LaTeX as you suggest. Perhaps your last line - which I embolden - is the main idea though.

Could be. I don't know. All I can  tell you is you've frustrated at least two of us here. 

Maybe I am wrong about my advice. I might have been too harsh. Sorry.

Yeah, don't worry, I have got the main picture. I think the advice is right. I will come back later.

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3513
  • Mole Snacks: +538/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Equilibrium kinetics
« Reply #62 on: May 12, 2013, 09:03:41 PM »
Meh, explaining difficult concepts - and asking pointed, clear questions - is not always easy over an internet forum.  I have no issues with Big Daddy's lines of conversation.  However BD I do think you would be better off focusing on perfecting your understanding of the basics rather than trying to pinpoint all the places and scenarios where widely used models fail.  Don't get me wrong - there is value to this at times.  However if you push too hard all you'll end up doing is confusing yourself.  It does become counterproductive.
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 #63 on: May 13, 2013, 04:21:15 AM »
To be honest, I feel I have learnt quite a lot in my time here, from the three of you (and others). I think I have gained a lot from the discussions which seem to you like running round in circles, and though I may take a long while to understand the concept, when I do I usually won't ask again.

Nevertheless clearly you are frustrated, and I can see what you mean, that asking good questions is an art. As a (reasonably immature :P)  teenager that is my biggest problem. I will try to improve it.

If you can explain what problem exactly there is in my questions that frustrates you (as Corribus has done), it would help me a lot. I know you feel that discussion is running around in circles but it has been a long time since I felt that on this forum.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #64 on: May 14, 2013, 04:56:08 PM »
in the case of aA + bB  ::equil:: cC + dD, I might suggest:

[tex]
\frac{dc_A}{dt}=-\frac{b}{a} k_{1f}  \cdot (c_A)^x \cdot (c_B)^y + \frac{c}{a} \frac{d}{a} k_{1r} \cdot (c_C)^n \cdot (c_D)^m \\
[/tex]

x is the order with respect to A, y with respect to B, n with respect to C, m with respect to D; a, b, c, d are the stoichiometric coefficients on A to D respectively.

I'm thinking now this doesn't work. The stoichiometric coefficients should be in the brackets:

[tex]
\frac{dc_A}{dt}=-k_{1f}  \cdot (c_A)^x \cdot (\frac{b}{a} \cdot c_B)^y + k_{1r} \cdot (\frac{c}{a} \cdot c_C)^n \cdot (\frac{d}{a} \cdot c_D)^m \\
[/tex]

No? I just get the feeling that if an order goes to 0 the coefficient shouldn't be having an effect on the rate either.

I am not sure yet what I did wrong to frustrate you, but still your help would be much appreciated.

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Equilibrium kinetics
« Reply #65 on: May 14, 2013, 11:40:29 PM »

I'm thinking now this doesn't work. The stoichiometric coefficients should be in the brackets:

[tex]
\frac{dc_A}{dt}=-k_{1f}  \cdot (c_A)^x \cdot (\frac{b}{a} \cdot c_B)^y + k_{1r} \cdot (\frac{c}{a} \cdot c_C)^n \cdot (\frac{d}{a} \cdot c_D)^m \\
[/tex]

No? I just get the feeling that if an order goes to 0 the coefficient shouldn't be having an effect on the rate either.

I am not sure yet what I did wrong to frustrate you, but still your help would be much appreciated.

I'll try to help but I really don't understand what you are saying here.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #66 on: May 15, 2013, 05:37:10 AM »
I'll try to help but I really don't understand what you are saying here.

If you look at the two ODEs I've written:

Version 1:
[tex]
\frac{dc_A}{dt}=-\frac{b}{a} k_{1f}  \cdot (c_A)^x \cdot (c_B)^y + \frac{c}{a} \frac{d}{a} k_{1r} \cdot (c_C)^n \cdot (c_D)^m \\
[/tex]

Version 2:
[tex]
\frac{dc_A}{dt}=-k_{1f}  \cdot (c_A)^x \cdot (\frac{b}{a} \cdot c_B)^y + k_{1r} \cdot (\frac{c}{a} \cdot c_C)^n \cdot (\frac{d}{a} \cdot c_D)^m \\
[/tex]

Which is correct, for the single equilibrium aA+bB  ::equil:: cC+dD (where the order with respect to A is x, to B is y, to C is n, to D is m)? Although the first one originally seemed OK I now am leaning towards the second because I would guess that the stoichiometric coefficients should also end up cancelling down to 1 (as they will if they are inside the brackets to the power of order, but not if they are outside) if the species is zero-order. Not sure though.

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Equilibrium kinetics
« Reply #67 on: May 15, 2013, 07:40:26 AM »
Which is correct, for the single equilibrium aA+bB  ::equil:: cC+dD (where the order with respect to A is x, to B is y, to C is n, to D is m)? Although the first one originally seemed OK I now am leaning towards the second because I would guess that the stoichiometric coefficients should also end up cancelling down to 1 (as they will if they are inside the brackets to the power of order, but not if they are outside) if the species is zero-order. Not sure though.

Don't think it matters. They only differ in their constants. Use either.


Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #68 on: May 15, 2013, 08:07:50 AM »
Don't think it matters. They only differ in their constants.

That would mean I would be writing different equations for the rate of change of [A] with time. Imagine for instance that the rate is not dependent on [B] so that y=0 (from my equations). In the first version of the equation, the stoichiometric ratio (b/a) continues to affect the rate. In the second version of the equation, the stoichiometric ratio with respect to B also cancels out. So d[A]/dt is just dependent on [A], no more on [B].
« Last Edit: May 15, 2013, 09:32:37 AM by Big-Daddy »

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Equilibrium kinetics
« Reply #69 on: May 15, 2013, 09:38:00 AM »

That would mean I would be writing different equations for the rate of change of [A] with time. Imagine for instance that the rate is not dependent on [B] so that y=0 (from my equations). In the first version of the equation, the stoichiometric ratio (b/a) continues to affect the rate. In the second version of the equation, the stoichiometric ratio with respect to B also cancels out. So d[A]/dt is just dependent on [A], no more on [B].

Write down both equations for y=0. k1f in first form is not equal to k1f in the second form. They are related by a constant.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #70 on: May 15, 2013, 11:07:34 AM »
k1f in first form is not equal to k1f in the second form. They are related by a constant.

Ah, got it now. That constant that relates them is the product of the stoichiometric coefficients that are not cancelled.

But if I look up values of the rate constant, which of the equations do they normally refer to? My second equation, I'm guessing/hoping ... (My available values of rate constants don't tell me whether they are corrected for stoichiometry or not.)

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Equilibrium kinetics
« Reply #71 on: May 15, 2013, 02:23:00 PM »

Ah, got it now. That constant that relates them is the product of the stoichiometric coefficients that are not cancelled.


Right

Quote
But if I look up values of the rate constant, which of the equations do they normally refer to?

And where will you be looking it up? Show me a literature example that uses your contrived variable order rate equation.

e.g. If I meet a unicorn will it be right handed or left handed?  :)
« Last Edit: May 15, 2013, 03:08:44 PM by curiouscat »

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Equilibrium kinetics
« Reply #72 on: May 15, 2013, 05:13:15 PM »
And where will you be looking it up? Show me a literature example that uses your contrived variable order rate equation.

Don't know what rate equation the constant below uses. All I want is to be able to formulate a rate equation in terms of the constants available, which I imagine relate to the reaction itself with no stoichiometry.

Sucrose (aq)  :rarrow: glucose (aq) + fructose (aq) [acidic conditions, 298.15 K]
k=6.0·10-5 s-1

Better example, to look at the stoichiometry, might be:

2 N2O5 (g) :rarrow: 4 NO2 (g) + O2 (g) [298.15 K]
k=3.38·10-5 s-1

Both values come from Atkins' "Physical Chemistry", 8th edition, page 1020 (Table 22.1: Kinetic data for first-order reactions).

Are you saying that the definition of the rate constant (i.e. whether corrected for stoichiometry or not) differs from source to source? Doesn't that lead to massive confusion, when the rate constant is not accompanied by something clarifying whether it is corrected or not?

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Equilibrium kinetics
« Reply #73 on: May 15, 2013, 11:34:35 PM »
These are simple 1st order reactions. Where's your confusion?

Write down (equations, not words) which different ways you think these equations might be misinterpreted?

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Equilibrium kinetics
« Reply #74 on: May 15, 2013, 11:38:48 PM »

Are you saying that the definition of the rate constant (i.e. whether corrected for stoichiometry or not) differs from source to source? Doesn't that lead to massive confusion, when the rate constant is not accompanied by something clarifying whether it is corrected or not?

There's no "massive" confusion, but yes things can get tricky at times. Which is why it is considered good practice to write down rate expressions explicitly  when potential for confusion exists.

Sponsored Links