Consider this mechanism: Step 1: A + B reversibly and rapidly forms X, with equilibrium reached almost immediately; Step 2: X + B -> C (slow). Which rate law does pre-equilibrium analysis predict for the overall reaction?

Core Concept

The slow step contributes rate = k2[X][B], but X is an intermediate that cannot appear in the observable law. Because Step 1 equilibrates rapidly, the ratio [X]/([A][B]) equals the equilibrium constant K1, so [X] = K1[A][B]. Substituting eliminates X and gives rate = k2K1[A][B]^2, written simply as k[A][B]^2. The squared dependence on B combines its role in forming X with its role in the slow step.