rate determining step

[1800RAW]

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is it always necessarily the slowest step? and by slowest i mean the largest activation energy.

take for instance a rxn involving the formation of an intermediate. the formation of the intermediate has an Ea of lets say 20J. The intermediate then reacts to form the product w/ an Ea of 10J, however, the activated complex is of higher energy going from intermediate-prod than react-inter.

so basically the first step has a higher activation energy than the 2nd step, but the second step has a higher energy TS than the first. all the sources ive checked out say the RDS is the step w/ the highest energy TS, but in this case, forming that TS is technically not the slowest step, which would make the RDS not the slowest step.

is this correct?

 

[sbdento]

The slow step is the one with the highest activation E, the rate determining is the highest E. Usually they are the same step, but they can be different. If I am reading your hypothetical situation correctly the first step would be the slow step (highest activation E) and the second would be the RDS(highest E).

 

[Kahr]

The slow step is the one with the highest activation E, the rate determining is the highest E. Usually they are the same step, but they can be different. If I am reading your hypothetical situation correctly the first step would be the slow step (highest activation E) and the second would be the RDS(highest E).

I'm surprised you didn't tell him the RDS is the hill with the highest peak 😉

 

[sbdento]

I'm surprised you didn't tell him the RDS is the hill with the highest peak 😉

That's the wording I wanted, I like your style 😉

 

[ZFT]

is it always necessarily the slowest step? and by slowest i mean the largest activation energy.

take for instance(a hypothetical scenario) a rxn involving the formation of an intermediate. the formation of the intermediate has an Ea of lets say 20J. The intermediate then reacts to form the product w/ an Ea of 10J, however, the activated complex is of higher energy going from intermediate-prod than react-inter.

so basically the first step has a higher activation energy than the 2nd step, but the second step has a higher energy TS than the first. all the sources ive checked out say the RDS is the step w/ the highest energy TS, but in this case, forming that TS is technically not the slowest step, which would make the RDS not the slowest step.

is this correct?

interesting question...

i believe this is impossible. see if your sources say Transition State energy = Activation energy.

This hypothetical scenario is only hypothetical, so the answer is yes, the RDS always has the highest Ea (or TS).

 

[tendram]

There is no way for for the first TS energy (the reaction with the higher Ea) to be lower than the second TS energy since both TS's are at the top of each Ea peak on a graph. The reaction with the highest Ea will always have the highest TS since you are putting more energy into the reaction to get to the TS.