CHEMISTRY COMMUNITY

I know this is late but when given multiple steps is the overall rate always reliant on the slow step even if the slow step is also the first step and an equilibrium?

You are correct that the rate law is always dependent on the slow step.

Step 1 does not always have to be fast with step 2 being slow. But when this happens, you are able to use the pre-equilibrium approach. The pre-equilibrium approach works under these circumstances because when step 2 is the slow step, there is a buildup of the product of step 1. This buildup then causes some of that "product" to covert back to reactant, creating a "pseudo-equilibrium".