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Course: AP®︎/College Chemistry>Unit 5

Lesson 4: Reaction mechanisms

The pre-equilibrium approximation

The pre-equilibrium approximation is used to find the rate law for a reaction with a fast and reversible initial step. In this method, we first write the rate law based on the slow (rate-determining) step. Then, to eliminate any intermediates from the rate law, we use the fast initial step to solve for the intermediate concentration(s) in terms of reactant and/or product concentrations. Created by Jay.

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• At he says, "...we can do that by assuming that the first elementary step in our mechanism comes to a fast equilibrium." How do we know that's a valid assumption to make?
• If the reaction is fast, then it'll reach its equilibrium concentrations quickly.

Hope that helps.
• Hello, why do k_2 * k_1 / k_-1 = k ?
• (k^2 * k^1)/k-1 would lead t another constant. In the video, he gave the other constant k. The k still stands for (k^2 * k^1)/k-1
• Why is k-1 written that way? It's just a subscript/marker and it's not literally subtracting 1 from k, right?
• Yeah it's a subscript and it's just there to label the rate constant. In this instance if the forward rate reaction constant is k1, then the reverse will be its opposite; k-1.

Hope that helps.
• I don't understand why it is necessary to use this approximation. Isn't the overall rate just equal to the rate determining step, which in this example would be step 2? And if step 2 is an elementary reaction, why can't we just say rate = k [NOBr2][NO]?
• The slow elementary step here, the rate determining step, includes an intermediate, NOBr2. Rate laws are generally not written with intermediates. Jay explains this at . Conceivably this is done because we are able to change the rate the reaction by varying the concentrations of the reactants, but are unable to directly control the intermediate. So it's more useful to us to have the rate law in terms of the reactants than the intermediates.

Hope that helps.
• can a reaction not have a reverse reaction? Which might be equivalent to say k_{-1}=0. If that is possible, how can we get the overall rate law?
• Technically all chemical reactions are reversible (if they remain in a mixture) and therefore have rate laws. In reality there are certain reactions which favor one side of the reaction so overwhelming that they are practically irreversible. A classic example of this is strong acid/base reactions. But for the purposes of constructing an overall rate law, all elementary reactions have an reverse rate law we can utilize.

Hope that helps.
• so at with the (k2*k1)/k-1 stuff, did you simplify it to "k" or "K"
(1 vote)
• I'm pretty sure it's a lowercase k. But the main point is that all three constants combine to form another single constant.
• Isn't the overall reaction an elementary step already? Wouldn't it be a termolecular reaction? A + A + B -> products or 2A + B -> products?
(1 vote)
• So it looks like you're confusing the order of the reaction with molecularity. The order of the reaction just tells us the sum of the chemical species' exponents in a rate law, but molecularity applies to elementary reactions and tells us how many atoms/molecules are colliding in that one step.

So we would say that this reaction is third order overall because of the exponents in the final rate law. But the overall reaction is composed of two elementary reactions. Both steps are bimolecular because they involve the collision of two reactant molecules.

Hope that helps.