Given the reaction 2 w + x → y + z, and the following mechanism:Step 1: w + x → y (slow)Step 2: w + y → z (fast)What is the rate law for this reaction?

• A. rate = k[x][y]
• B. rate = k[w]^2[x]
• C. rate = k[w][y]
• D. rate = k[w][x]

Answer: (D)

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. For Step 1: w + x → y, the rate law is rate = k[w][x], because the rate depends on the concentration of the reactants involved in this slow step.
3. Check if any intermediate species appear in the rate law. Here, y is an intermediate formed in Step 1 and consumed in Step 2, so it should not appear in the overall rate law.
4. Confirm that the rate law matches the stoichiometry of the slow step, not the overall balanced equation, because the slow step determines the kinetics.
5. Conclude that the rate law for the overall reaction is rate = k[w][x], consistent with the slow step and excluding intermediates.