hey so this is for rate law if the first step is the slow step. it is also an equilibrium. I just want to check if this is right.
[Acetophenone]+[Ethoxide]<-k1-->
k-1
[Intermediate]+[Ethanol]
Just want to make sure this will be the rate law: k1/k-1 [Acetophone][Ethoxide]
or do I leave out k-1?
To determine the rate law for a reaction, we need to look at the slow step of the reaction, as it is generally the rate-determining step. The slow step is the step with the highest activation energy and determines the overall rate of the reaction.
In the given reaction, the slow step is the forward reaction of the formation of the intermediate:
[Acetophenone] + [Ethoxide] <-k1->
k-1
[Intermediate] + [Ethanol]
Since this step is at equilibrium, we can write the equilibrium expression for this reaction:
K_eq = [Intermediate] / ([Acetophenone] * [Ethoxide])
Now, for the rate law, it is based on the stoichiometry and the exponents of the reactant concentrations, as determined experimentally. Since the concentration of the intermediate is not included in the rate law, we can assume it is a reactive intermediate and assume it has a constant concentration.
Therefore, the rate law for this reaction would be:
rate = k * [Acetophenone] * [Ethoxide]
Where k represents the rate constant. The rate constant k is related to the rate constant k1 of the slow step as follows:
k = k1 / k-1
So the rate law can be represented as:
rate = (k1/k-1) * [Acetophenone] * [Ethoxide]
To summarize, the rate law for the given reaction involving the slow step at equilibrium is:
rate = (k1/k-1) * [Acetophenone] * [Ethoxide]