Steady state and pre-equilibrium
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Steady state and pre-equilibrium
What is the difference between the steady-state and pre-equilibrium? Don't you assume that the intermediate will be formed and completely used up in both?
Re: Steady state and pre-equilibrium
We use the pre-equilibrium method when the first step is faster than the second. Why? A sort of bottleneck is created when the first reaction moves at a second rate then the second, because the products of the first reaction are being consumed at a slower speed that they're being created. This allowed the products to be converted into their reactants again, and the reaction reaches an equilibrium.
We use the steady state method when the first reaction is slower than the second. This is because there is no bottleneck created, since the second equation, which is fast, consumes the products as they are created from the first reaction. Therefore, the first slow reaction controls the rate of the overall reaction.
We use the steady state method when the first reaction is slower than the second. This is because there is no bottleneck created, since the second equation, which is fast, consumes the products as they are created from the first reaction. Therefore, the first slow reaction controls the rate of the overall reaction.
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Re: Steady state and pre-equilibrium
I want to note that Lavelle said in his lecture that we essentially will not use the steady state method, but rather solely the pre-equilibrium method due to its easier procedural steps as well as its brevity in comparison to the steady state method.
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