# Rate law for A + B → C → P

I need some clarification for the following assignment:

Derive the rate law for

$$\ce{A + B -> C -> P}$$

when $$\ce{A + B -> C}$$ is the slowest step and very slow.

My understanding is $$\ce{A + B -> C}$$ is the rate-determining step, so the rate law would just be

$$\mathrm{rate} = k[\ce{A}][\ce{B}].$$

I'm not sure if the question is asking about steady state approximation (SSA) or just a simple rate law. I know SSA happens if you assume the second step $$(\ce{C -> P})$$ is faster than $$\ce{A + B -> C},$$ meaning $$\ce{A + B -> C}$$ is the slow step. I'm not so sure if my understanding of the question is correct.

The rate is determined by the slowest step, as you correctly assumed. The transition state of C->P is assumed to occur significantly faster than the first step in Organic Chemistry at least.

Unless you have an energy diagram for the reaction or any Le Chat, I think you have it right as rate = k[A][B].