Chemistry
Kinetics of Consecutive First-Order Reactions and the Steady-State Approximation
Quick fact
In a consecutive first-order reaction (A → B → C), the intermediate B's concentration peaks at a time given by tmax = ln(k₁/k₂)/(k₁ − k₂), which depends solely on the two rate constants.
Why this is interesting
Ever wondered why a reaction's intermediate concentration first rises, then falls, even though it is continually being formed? That rise and fall is the signature of consecutive reactions—and there is a clever trick that lets chemists simplify the math immensely.
Read the full explanation
Understanding Kinetics of Consecutive First-Order Reactions and the Steady-State Approximation
Think of a reaction that proceeds through a short-lived intermediate: A first turns into B, which then turns into C. Each step is first-order, meaning its rate is proportional to the concentration of the reactant for that step. The rate of change of A is simply −k₁[A]. The rate of change of B has two contributions: it is produced from A (at rate +k₁[A]) and consumed to make C (at rate −k₂[B]). So d[B]/dt = k₁[A] − k₂[B]. Similarly, C is produced from B at rate +k₂[B]. These linked differential equations can be solved exactly to give the concentrations over time. The solution for A is a simple exponential decay: [A] = [A]₀ e^(−k₁t). The concentration of B is the difference of two exponentials: [B](t) = [A]₀ · (k₁/(k₂−k₁)) · (e^(−k₁t) − e^(−k₂t)), with a special form when k₁=k₂. The concentration of C builds up as the sum: [C](t) = [A]₀ · (1 − e^(−k₁t) − (k₁/(k₂−k₁))(e^(−k₁t) − e^(−k₂t))). The key qualitative insight is that B, the intermediate, rises at first (because it is being formed from A faster than it is consumed), reaches a maximum, and then falls as A is depleted. This is the classic bell-shaped curve of an intermediate.
A deeper explanation
The exact solution to consecutive first-order reactions shows that the intermediate B can accumulate to a significant extent if k₁ is much larger than k₂ (B forms fast and decays slow). In many mechanisms, however, the intermediate is highly reactive and short-lived (k₂ is very large compared to k₁). In that case, its concentration remains tiny and almost constant after a brief induction period. This is the basis of the steady-state approximation (SSA): assume that the rate of change of the intermediate's concentration is effectively zero, d[B]/dt = 0. This assumption converts the differential equation for B into an algebraic equation: 0 = k₁[A] − k₂[B] ⇒ [B] = (k₁/k₂)[A]. This simplifies the rate law enormously: the rate of formation of the final product C is d[C]/dt = k₂[B] = k₁[A]. That is, the overall rate is controlled by the first step (the slow, rate-determining step), and the rate constant of the second step cancels out. The SSA is valid only when k₂ k₁, i.e., when the intermediate is consumed much faster than it is formed, so that its concentration never builds up. This approximation is widely used in chemical kinetics, from enzyme catalysis (where the enzyme-substrate complex is treated as a steady-state intermediate) to atmospheric chemistry and combustion mechanisms. It is a powerful tool because it avoids solving the full differential equations while still capturing the rate-determining behavior of the mechanism.