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

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.

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.