In a pseudo-first-order condition, what is the form of the integrated rate law?

• A. [A]t = [A]0 e^{-kobs t} with kobs = k[B]0
• B. [A]t = [A]0 e^{-kt}
• C. [A]t = [A]0 - kobs t
• D. [A]t = [A]0 /(1 + kobs t)

Answer: (A)

When one reactant is in large excess, its concentration can be treated as constant during the reaction. This makes the rate law effectively first-order in the other reactant: rate = k[A][B] ≈ k[A][B]0 = kobs[A], where kobs = k[B]0. Solving d[A]/dt = -kobs[A] gives [A]t = [A]0 e^{-kobs t}, which is the integrated rate law for pseudo-first-order kinetics. This exponential decay with a constant observed rate kobs shows why the correct form is an exponential function with kobs defined as k times the excess concentration. The other forms correspond to zero-order behavior or true second-order behavior without the excess reactant treated as constant.