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Kinetics Exact Solutions

 
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Mar12-13, 08:53 PM   #18
 
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Kinetics Exact Solutions


Quote by Big-Daddy View Post
What is x there?

The problem is I'm not really sure how to start in terms of combining the rate constants with the equilibrium constant in general; I've heard K=k1/k2, but I can't see that this would be true if the reaction were not "elementary" (i.e. if stoichiometric coefficients do not necessarily equal orders of reaction). So on the whole, I don't know how to start! (with reaching my differential equation)

The equation I wrote is of the same form as mfb's, merely including a further constant (where he put -C, I put C0-C to denote the possibility of C0 not being 0).
The formulae in which x appeared are just taken from the Wikipedia article that you quoted. Since I was saying that this was just the same 'distance from equilibrium' variable viz. ([A]-[Aeq]) from my post #14 where I was using your terminology I thought you would recognise.

K=k1/k2 is true of the elementary mechanism you have. That mechanism has only two rate constants. If the mechanism is not elementary it will have more rate constants, practically by defnition. The equlibrium constant is then the product of all the forward constants, divided by the product of all the backwards constants, as you can easily deduce by the same sort of reasoning as for the elementary reactions in the Wiki article.

Now don't ask me how you do the calculations for even more complicated mechanisms, because the answer is to do it for simpler ones first and if you can do it for them you can do it for more complicated ones, the only new thing is longer formulae, but no novelty of principle.
 
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