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### Topic: Equilibrium kinetics  (Read 110977 times)

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#### Big-Daddy

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• Mole Snacks: +28/-94 ##### Re: Equilibrium kinetics
« Reply #90 on: May 16, 2013, 11:23:29 AM »
Yep.

And if it were an equilibrium then:

$$\frac{dc_{Species}}{dt}= v_{Species} \cdot k_{1f} \cdot ( c_A)^x \cdot (c_B)^y \cdot (c_C)^z - v_{Species} \cdot k_{1r} \cdot (c_D)^n \cdot (c_E)^m \cdot (c_F)^p \\$$

vSpecies continuing to be the stoichiometric coefficient on the species you're writing this ODE for, and n,m,p being orders of the reverse reaction wrt D, E, F respectively. Same equation whatever x,y,z,n,m,p are, same equation whatever the stoichiometric coefficients of other species are. Correct?

#### Big-Daddy

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« Reply #91 on: May 16, 2013, 11:36:17 AM »
And if the species is involved in multiple rate equations, you would just add on more sections like those above to the ODE for that species, with vSpecies for each section being the stoichiometric coefficient of the species in that reaction.

So if in addition to the equilibrium above I had 3A  H happening at the same time, which is now first-order in A, I'd have as my total ODE for A:

$$\frac{dc_{Species}}{dt}= v_{Species} \cdot k_{1f} \cdot ( c_A)^x \cdot (c_B)^y \cdot (c_C)^z - v_{Species} \cdot k_{1r} \cdot (c_D)^n \cdot (c_E)^m \cdot (c_F)^p - 3 \cdot k_{2} \cdot ( c_A) \\$$

(vSpecies here is the stoichiometric coefficient on A in the equilibrium reaction discussed above.)
« Last Edit: May 16, 2013, 01:25:32 PM by Big-Daddy »

#### curiouscat

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« Reply #92 on: May 16, 2013, 12:28:12 PM »
Yes. Mostly right.

#### Big-Daddy

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« Reply #93 on: May 16, 2013, 12:49:23 PM »
Yes. Mostly right.

Good. I will take "mostly right" as "all right so far". To check: the bit we add to the ODE for cA, for the reaction 3A  H, still follows an elementary rate law (given that A's conversion to G is elementary, so we can really break this up into 3A G, G H), and if it's first order we can write this contribution

$$-3 \cdot k_{2} \cdot (c_A)$$
« Last Edit: May 16, 2013, 02:23:33 PM by Big-Daddy »

#### curiouscat

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« Reply #94 on: May 16, 2013, 01:05:15 PM »
Yes. Mostly right.

Good. I will take "mostly right" as "all right so far". So long as you add a k2f...

#### Big-Daddy

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« Reply #95 on: May 16, 2013, 01:28:15 PM »
So long as you add a k2f...

Ahh! Sorry, silly mistake. Edited in So even if 3A  H overall must be non-elementary, the process 3A G is elementary and thus we can always write its contribution (to the ODE for A) in the form:

$$-k_{2} \cdot 3 \cdot (c_A)^t$$

Where t is the order of this new reaction 3A G wrt to A.

#### curiouscat

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« Reply #96 on: May 16, 2013, 01:51:21 PM »
Yes.

#### Big-Daddy

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« Reply #97 on: May 16, 2013, 02:22:53 PM »
Yes.

Thanks very much for the help.

96 posts! You must be exhausted Sorry if I have frustrated you too much over the course of the last few dozen posts.

#### curiouscat

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« Reply #98 on: May 16, 2013, 02:29:22 PM »
96 posts! You must be exhausted Sorry if I have frustrated you too much over the course of the last few dozen posts.

I wish there was a prize for the longest attempt at the Socratic Method. #### Big-Daddy

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« Reply #99 on: May 16, 2013, 03:52:09 PM »
I wish there was a prize for the longest attempt at the Socratic Method. Sorry! I learn best by example.

#### Big-Daddy

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« Reply #100 on: May 18, 2013, 07:03:37 AM »
Another query:

If the volume of the system is changing, how do we incorporate that? Something to do with molar volumes, and their dependence on pressure/temperature, for each species?

Edit: I'm not sure if this is what Borek would call a secondary or irrelevant field, but I think I have learnt the main matter so unless this is vastly too complicated for me I would like to give it a try.

#### curiouscat

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« Reply #101 on: May 18, 2013, 08:12:35 AM »
Another query:

If the volume of the system is changing, how do we incorporate that? Something to do with molar volumes, and their dependence on pressure/temperature, for each species?

For variable volume instead of using
$$\frac{dc_A}{dt}= R_A \\$$

Use the more general form:

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

V is reactor volume,  R is rate of production for that particular species.

#### curiouscat

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« Reply #102 on: May 18, 2013, 08:15:56 AM »

but I think I have learnt the main matter

Brave. Very brave. #### Big-Daddy

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« Reply #103 on: May 18, 2013, 08:34:37 AM »
$$\frac{d \left ( c_A \cdot V \right ) }{dt}= R_A \cdot V \\$$

V is reactor volume,  R is rate of production for that particular species.

Looks like we're now working in number of moles. That seems reasonable.

Since the volume is changing, does the volume have to be expressed as a function of the different concentrations or number of moles of the different species? Or is it simply its own term, not expressed in terms of the species? After all, how would you predict how much the volume changes for a certain change in number of moles, without having expressed the volume as a function of the number of moles ...

#### curiouscat

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« Reply #104 on: May 18, 2013, 10:06:30 AM »

Since the volume is changing, does the volume have to be expressed as a function of the different concentrations or number of moles of the different species? Or is it simply its own term, not expressed in terms of the species? After all, how would you predict how much the volume changes for a certain change in number of moles, without having expressed the volume as a function of the number of moles ...

You tell me. Why is your volume changing? Each case is different. Express it as a function of whatever is making it change.