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Topic: Equilibrium position and Collision theory  (Read 6739 times)

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Offline hobnob

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Equilibrium position and Collision theory
« on: March 26, 2010, 09:01:58 AM »
Hi there

I'm trying to adapt a collision theory simulation of rates of reaction so it can handle an equilibrium reaction. It's a simple setup with two sets of reactant particles mingling in a gaseous state to form a gaseous product (think 2NO2 <-> N2O4). My problem is that I can't work out the appropriate way to model the backward reaction using collision theory. What causes the product to split back up into reactants?

I've tried three variants:1) Whenever a product particle hits any other particle, if the total energy is greater than the activation energy (+ enthalpy) then a reaction occurs; 2) Whenever a product particle hits any other particle, if the energy of the particle it hits is greater then a reaction occurs; 3) Reactions occur spontaneously without any collision, whenever the number of product particles is greater than the equilibrium position.

None of these is satisfactory. 1) and 2) both produce an equilibrium, but it doesn't satisfy Le Chatelier: changing temperature and/or concentration has no noticeable effect on the equilibrium position. 3) produces the correct equilibrium directly, but doesn't look realistic, and in particular doesn't produce a realistic path to the equilibrium position.

Sorry for the long description. Does anyone have any links to some resource that will derive Le Chatelier from collision theory or some other method?

Thanks
Hob

Offline Borek

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Re: Equilibrium position and Collision theory
« Reply #1 on: March 26, 2010, 09:16:50 AM »
Are you sure your simulation works correctly for the forward reaction? I mean: it satisfies all conditions that you have listed.

Are you sure problem is not with the backward reaction per se, but with the fact that backward reaction is of the first order? Would your simulation work for the first order forward reaction?
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Offline hobnob

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Re: Equilibrium position and Collision theory
« Reply #2 on: March 26, 2010, 09:49:12 AM »
The forward reaction works fine - with no equilibrium it just runs to completion at an expected rate, which varies according to temperature and concentration pretty much as you'd expect. We've been using (and selling!) it for years so we're fairly happy with it.

Offline hobnob

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Re: Equilibrium position and Collision theory
« Reply #3 on: March 26, 2010, 09:53:24 AM »
Are you sure problem is not with the backward reaction per se, but with the fact that backward reaction is of the first order? Would your simulation work for the first order forward reaction?

Sorry, to answer this question, yes, I'm almost certain that this is exactly the problem. I just don't know how collision theory deals with first-order reactions. Having said that, I'm fairly sure that if I set it up using either of my two initial methods it would behave as a first-order reaction, or close to it. And just to make it clear, I do get an equilibrium, it's just not obeying Le Chatelier.

Thanks for the reply

Offline Borek

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Re: Equilibrium position and Collision theory
« Reply #4 on: March 26, 2010, 03:27:28 PM »
I would think in terms of N2O4 colliding with other molecules - and some of these collisions can trigger N2O4 decomposition. But it can be completely off, I don't know details of your simulation.

Can't ypou simulate a A + B <-> C + D? this way there will be no problem with first order reaction and you will use working approach for both forward and backward simulation.
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Offline hobnob

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Re: Equilibrium position and Collision theory
« Reply #5 on: March 27, 2010, 10:44:43 AM »
Can't ypou simulate a A + B <-> C + D? this way there will be no problem with first order reaction and you will use working approach for both forward and backward simulation.

The thought did occur to me :) I'd love to do that and we might go ahead, but unfortunately I'm tied to keeping it in line with the previous version. Also, there's a learning point in the effect of pressure on 2-1 reactions, which according to Le Chatelier should move towards more product when the pressure increases. Given that the Haber process is about the only thing anyone learns about with regard to chemical equilibrium at this level, it's important to keep the concept in.

I've made some progress by moving to a model where the backward reaction occurs whenever there is a collision between two product particles and their energy after the collision is above the activation energy. This is giving me good temperature dependence for the exothermic reaction but not for endothermic for some reason, and no pressure dependence at all. I've got a plan to try introducing a delay factor so that the backward reaction only occurs a shirt time after the collision, and only if the energy hasn't been reduced again - this should make the reaction less likely to occur at greater concentrations as there's more chance of a second collision 'deactivating' the particle. But I'm aware this is a hack!

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