Slow step vs Rate determining step

[premac]

Hey guys,
Looking at the reaction coordinate diagram, I know that RDS is the one with the highest E, and slow step is the one with the highest Ea.
When we're determining the rate law, what if RDS and slow step are different?
Which step would I choose to be the rate law?
RDS and Slow step are usually the same, but if the second peak happens to be higher than the first peak, RDS and slow step would be different.

I'm a bit confused on this, so I will be waiting for your replies!

Thank you~

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[Msmouth]

The RDS and slow step don't necessarily have to be the first step. They are both the same thing. If the 2nd step has the highest energy then it is the RDS and also the Slow step.
That's what I understood from last semester....someone correct me if I'm wrong.

 

[pkoushik4]

It's not the slowest step because at steady state, the net rate of each step is proportional to the net rate of every other step. It is better described as the most demanding step. If such a step exists
‣ The free energy change due to this step is essentially equal to the free energy
change for the overall, macroscopically observable reaction

 

[FeralisExtremum]

The RDS and slow step don't necessarily have to be the first step. They are both the same thing. If the 2nd step has the highest energy then it is the RDS and also the Slow step.

This is usually true, but not always. It is possible for the RDS and the slow step to be separate steps. The rate determining step is whichever "hill" is the highest overall in the reaction coordinate diagram, aka the one with the highest energy value. The slow step is whichever step has the largest Ea (activation energy) value. So they can be different steps, but this doesn't occur frequently.

To answer your question premac: in such a situation the rate determining step, not the slow step, is the one that determines the rate law.