Rate law for A + B → C → P

Question

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.

Answer 1

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].