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Offline Big-Daddy

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Re: Equilibrium kinetics
« Reply #15 on: April 30, 2013, 04:39:25 PM »
Out of pure curiosity, what is your current level? HS? College? etc.

Again, pure curiosity. Have you exposure to how to solve ODE's simultaneously numerically?

Officially high school, but I read college books in my spare time  ::)

No I have no knowledge of solving ODEs simultaneously.

Offline Corribus

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Re: Equilibrium kinetics
« Reply #16 on: April 30, 2013, 04:43:50 PM »
Hmmm, but in the video I don't see where the extra terms are for the backward reaction ... to clarify:

A+B  ::equil:: C with k1
::equil:: 2E with k2

But nowhere does there appear to be a k1,rev or k2,rev. And how can we arrive at final ODEs to solve, for equilibrium, which do not include the equilibrium constants?
Typically approximations are used, such as the steady-state approximation.  The validity of such approximations depends a lot on the relative values of the various rate constants.  Without approximations, the math rapidly becomes unwieldy, even for apparently simple systems.
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 curiouscat

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Re: Equilibrium kinetics
« Reply #17 on: May 01, 2013, 11:38:19 AM »

Hmmm, but in the video I don't see where the extra terms are for the backward reaction ... to clarify:

A+B  ::equil:: C with k1
::equil:: 2E with k2

But nowhere does there appear to be a k1,rev or k2,rev. And how can we arrive at final ODEs to solve, for equilibrium, which do not include the equilibrium constants?

Ok, I'll try again:

::equil:: B k1f & k1r

::equil:: C k2f & k2r

[tex]
\frac{dc_A}{dt}=-k_{1f}  \cdot c_A + k_{1r} \cdot c_B \\

\frac{dc_B}{dt}=k_{1f} \cdot c_A - k_{1r} \cdot c_B -k_{2f} \cdot c_B + k_{2r} \cdot c_C \\

\frac{dc_C}{dt}=k_{2f} \cdot c_B - k_{2r} \cdot c_C \\
[/tex]

At t=0 c_A, c_B, c_C all are known.

Solve as an initial value problem.

These are your "final ODE's" and they contain no explicit equilibrium constants. No matter how many equations the same approach applies. You can write one ODE for each species.

Offline Big-Daddy

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Re: Equilibrium kinetics
« Reply #18 on: May 02, 2013, 07:40:02 PM »

Hmmm, but in the video I don't see where the extra terms are for the backward reaction ... to clarify:

A+B  ::equil:: C with k1
::equil:: 2E with k2

But nowhere does there appear to be a k1,rev or k2,rev. And how can we arrive at final ODEs to solve, for equilibrium, which do not include the equilibrium constants?

Ok, I'll try again:

::equil:: B k1f & k1r

::equil:: C k2f & k2r

[tex]
\frac{dc_A}{dt}=-k_{1f}  \cdot c_A + k_{1r} \cdot c_B \\

\frac{dc_B}{dt}=k_{1f} \cdot c_A - k_{1r} \cdot c_B -k_{2f} \cdot c_B + k_{2r} \cdot c_C \\

\frac{dc_C}{dt}=k_{2f} \cdot c_B - k_{2r} \cdot c_C \\
[/tex]

At t=0 c_A, c_B, c_C all are known.

Solve as an initial value problem.

These are your "final ODE's" and they contain no explicit equilibrium constants. No matter how many equations the same approach applies. You can write one ODE for each species.

OK I think I've got it.

How would we write the term, in the ODE for A, for a reaction in which the participation is nA  ::equil:: B (i.e. stoichiometric coefficient)?
[tex]
-n k_{1f}  \cdot c_A
[/tex]

Or what? And what if the rate of this equilibrium with respect to A is not first order?

Offline dariusrickard

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Re: Equilibrium kinetics
« Reply #19 on: May 03, 2013, 08:32:11 AM »
nice thread. thanks for sharing that information. it was very informative.

Offline curiouscat

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Re: Equilibrium kinetics
« Reply #20 on: May 04, 2013, 02:16:30 PM »
OK I think I've got it.

Good.

Quote
How would we write the term, in the ODE for A, for a reaction in which the participation is nA  ::equil:: B (i.e. stoichiometric coefficient)?
[tex]
-n k_{1f}  \cdot c_A
[/tex]

Yes.


Quote
And what if the rate of this equilibrium with respect to A is not first order?

If it isn't, it isn't.

Offline Big-Daddy

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Re: Equilibrium kinetics
« Reply #21 on: May 04, 2013, 06:35:14 PM »
If it isn't, it isn't.

OK let's take a look at a more complicated example because I'm not sure how this will work.

In a general case 3A+5B  ::equil:: C, forward rate is 2nd-order in A and 3rd-order in B (not normal I know but let's just go with it). How do we write the parts of the ODE for cA that would be contributed by this equilibrium? My concern is that this time rate depends on both cA and cB ...

Offline curiouscat

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Re: Equilibrium kinetics
« Reply #22 on: May 04, 2013, 10:43:51 PM »

In a general case 3A+5B  ::equil:: C, forward rate is 2nd-order in A and 3rd-order in B (not normal I know but let's just go with it).

Write me a rate expression. You know how to do it for single equaions, right?

Quote
My concern is that this time rate depends on both cA and cB ...

Doesn't matter.

Offline Big-Daddy

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Re: Equilibrium kinetics
« Reply #23 on: May 05, 2013, 06:03:29 AM »

Write me a rate expression. You know how to do it for single equaions, right?


Well I've never seen it before so I'm probably wrong but I'll give it a shot. :( If it were just 3A  ::equil:: C rate equation is:

[tex]
\frac{dc_A}{dt}=-k_{1f}  \cdot (c_A)^x + \frac{1}{3} k_{1r} \cdot (c_C)^n \\
[/tex]

So 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.

Offline curiouscat

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Re: Equilibrium kinetics
« Reply #24 on: May 05, 2013, 08:49:40 AM »

Write me a rate expression. You know how to do it for single equaions, right?


Well I've never seen it before so I'm probably wrong but I'll give it a shot. :( If it were just 3A  ::equil:: C rate equation is:

[tex]
\frac{dc_A}{dt}=-k_{1f}  \cdot (c_A)^x + \frac{1}{3} k_{1r} \cdot (c_C)^n \\
[/tex]

So 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.

Ok, fine. So we can still solve this set of ODE's as an initial value problem, once you write similar ones for changes of cB, cC and cD.

No problems.  Essentially you've learnt now how to formulate unsteady state and equilibrium concentrations for a set of any  number of reactions (so long as you know individual rate expressions).

 

Offline Big-Daddy

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Re: Equilibrium kinetics
« Reply #25 on: May 05, 2013, 08:54:44 AM »
Ok, fine. So we can still solve this set of ODE's as an initial value problem, once you write similar ones for changes of cB, cC and cD.

No problems.  Essentially you've learnt now how to formulate unsteady state and equilibrium concentrations for a set of any  number of reactions (so long as you know individual rate expressions).

Thanks :) If were to solve for a very high value of t, or even take the limit t=∞, we would land up with the equilibrium concentrations? Meaning that, given the rate constants and initial concentrations, we can work out the equilibrium constant from how far the reaction goes?

Offline curiouscat

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Re: Equilibrium kinetics
« Reply #26 on: May 05, 2013, 09:00:10 AM »
[
Thanks :) If were to solve for a very high value of t, or even take the limit t=∞, we would land up with the equilibrium concentrations? Meaning that, given the rate constants and initial concentrations, we can work out the equilibrium constant from how far the reaction goes?

Yes. Right.

Assuming an ideal world. Often measured rate constants are bad and kf/kr may not evaluate to predicted thermodynamic Keq.

Often the system is stiff and solving it is one gigantic headache. If you only want equilibrium concentrations sometimes n simultaneous non-linear equations is all you need.  Tad less work than solving ODE's. (but not always)

Real life calculations are a mess.


Offline Big-Daddy

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Re: Equilibrium kinetics
« Reply #27 on: May 05, 2013, 11:14:10 AM »
[
Thanks :) If were to solve for a very high value of t, or even take the limit t=∞, we would land up with the equilibrium concentrations? Meaning that, given the rate constants and initial concentrations, we can work out the equilibrium constant from how far the reaction goes?

Yes. Right.

Assuming an ideal world. Often measured rate constants are bad and kf/kr may not evaluate to predicted thermodynamic Keq.

Often the system is stiff and solving it is one gigantic headache. If you only want equilibrium concentrations sometimes n simultaneous non-linear equations is all you need.  Tad less work than solving ODE's. (but not always)

Real life calculations are a mess.

Can any rate expression always be written in the format we discussed? Or are there some cases where even the differential rate equation can't be described like that?

Edit: And could the problem arise that the rate constant itself changes with time? Or the reaction orders? In which case Keq might not be that which we predict by taking the lim(t :rarrow: ∞). Whereas the mass balance/charge balance/equilibria system will always work for a system of equilibria ...
« Last Edit: May 05, 2013, 11:42:03 AM by Big-Daddy »

Offline curiouscat

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Re: Equilibrium kinetics
« Reply #28 on: May 05, 2013, 03:29:59 PM »

Can any rate expression always be written in the format we discussed?

Yes. AFAIK.

Quote
Or are there some cases where even the differential rate equation can't be described like that?

No. AFAIK.

Quote
Edit: And could the problem arise that the rate constant itself changes with time?

Doesn't matter.

Quote
Or the reaction orders?

Again, doesn't matter. All we need is rate = fn(cA,cB,..cN,t). Normally no t.




Quote
In which case Keq might not be that which we predict by taking the lim(t :rarrow: ∞). Whereas the mass balance/charge balance/equilibria system will always work for a system of equilibria ...

I don't know what you mean by that.

Offline Big-Daddy

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Re: Equilibrium kinetics
« Reply #29 on: May 05, 2013, 03:44:40 PM »

Can any rate expression always be written in the format we discussed?

Yes. AFAIK.

Quote
Or are there some cases where even the differential rate equation can't be described like that?

No. AFAIK.

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

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

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