Given the reaction mechanism:Step 1: NO_2 + F_2 → NO_2F + F (slow)Step 2: F + NO_2 → NO_2F (fast)What is the rate law for the overall reaction based on this mechanism?

• A. rate = k[NO_2]^2[F_2]
• B. rate = k[NO_2F][F_2]
• C. rate = k[NO_2][F_2]
• D. rate = k[F_2]^2

Answer: (C)

Verified step by step guidance

1. Identify the rate-determining step (RDS) in the mechanism. Since Step 1 is labeled as slow, it controls the overall reaction rate.
2. Write the rate law based on the RDS. The rate depends on the reactants involved in Step 1, so the initial rate law is: \(\text{rate} = k[NO_2][F_2]\).
3. Check if any intermediate species appear in the rate law from the RDS. Here, the intermediate is F, which is produced in Step 1 and consumed in Step 2.
4. Since the intermediate F appears only in the mechanism and not in the overall reaction, express its concentration in terms of reactants using the fast equilibrium assumption from Step 2.
5. In this case, Step 2 is fast and consumes F quickly, so the concentration of F remains low and does not affect the rate law derived from the slow step. Therefore, the overall rate law remains \(\text{rate} = k[NO_2][F_2]\).