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Use the same concept behind the pre-equilibrium approximation to derive expressions for the intermediates in your rate laws. For example, for the reaction with elementary steps (fast step; ) and (slow step; ), treat the first step as if the intermediate and the reactant NO were in equilibrium as represented by the double-sided arrow in the equation. The forward and reverse rates are equal at equilibrium so , where is the rate constant for the reverse of the first elementary step. Plug that into the rate law for the slow step to get the overall rate law , where k is the overall rate constant and is equal to .
Intermediates are products that are formed in one reaction that are then used in another reaction to create the new products wanted. So therefore they are made and consumed so that means they should not be present in the final rate law.
Yes. Intermediates are molecules that are formed in one step of an overall reaction and consumed in one of the following steps. It will not show up in the overall reaction as it will cancel out once you add up the steps of the equation.
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