A researcher is studying the mechanism of a complex reaction with two elementary steps: Step 1: A + B → C (fast) Step 2: C + D → E (slow) Which of the following would be the most appropriate rate law for the overall reaction?
- ARate = k[A][B][D]/[C]
- BRate = k[A][B]
- CRate = k[C][D]
- DRate = k[A][B][D]
Show answer & explanationAnswer & explanation
Correct answer: D. Rate = k[A][B][D]
The rate law for an overall reaction is determined by its slow (rate-determining) step. In this case, Step 2 (C + D → E) is the slow step, so the initial rate law would be Rate = k₂[C][D]. However, C is an intermediate, which cannot appear in the overall rate law. Since Step 1 is fast and reversible, we can assume it reaches equilibrium: Keq = [C] / ([A][B]). Therefore, [C] = Keq[A][B]. Substituting this into the rate law for the slow step gives Rate = k₂ * Keq * [A][B][D]. Since k₂ * Keq is a constant, we can combine it into a new rate constant, k_overall, resulting in Rate = k_overall[A][B][D].
Why the other options are wrong
- A. This form is not consistent with the elementary steps and the steady-state or equilibrium approximation.
- B. This would be the rate law if Step 1 were the slow step and C was not an intermediate, or if the overall reaction only involved A and B.
- C. This is the rate law for the slow step, but it includes an intermediate (C), which is not allowed in the overall rate law.
Rate-Determining Step
The rate-determining step (RDS) is the slowest elementary step in a reaction mechanism, and it dictates the overall rate law of the reaction.
- The rate law for the overall reaction is derived from the RDS.
- Intermediates cannot appear in the overall rate law.
- If the RDS involves an intermediate, its concentration must be expressed in terms of reactants using the equilibrium of a preceding fast step.
Memory trick: Slow Step Sets the Speed, No Intermediates.