Consecutive Reaction Mechanisms

Introduction

Most reactions proceed through several elementary steps, with intermediates formed and consumed along the way. The simplest multistep scheme is a chain of two first-order steps, A → I → P. Solving it exactly shows how an intermediate's concentration rises and falls, why product formation can lag behind reactant loss, and how the idea of a rate-determining step emerges naturally from the mathematics.

Core explanation

The scheme. A → I with rate constant k₁, then I → P with rate constant k₂. The rate equations are

d[A]/dt = −k₁[A] d[I]/dt = k₁[A] − k₂[I] d[P]/dt = k₂[I]

and mass balance gives [A] + [I] + [P] = [A]₀ at all times, with only A present initially.

Solutions. A decays by simple first-order kinetics: [A] = [A]₀e^(−k₁t). Solving the equation for I (when k₁ ≠ k₂) gives

[I] = [A]₀ k₁/(k₂ − k₁) × (e^(−k₁t) − e^(−k₂t))

and [P] follows from mass balance.

Shape of the intermediate curve. [I] starts at zero, rises while formation exceeds loss, reaches a maximum when k₁[A] = k₂[I], and then decays. The maximum occurs at

t max = ln(k₁/k₂) ÷ (k₁ − k₂)

Product curve. [P] rises slowly at first — its initial gradient is zero because no I is present at t = 0 — then accelerates and finally levels off at [A]₀. This S-shaped start is an induction period, a tell-tale sign that the product is formed via an intermediate.

Limiting cases. - k₂ ≫ k₁ (second step fast). I is consumed almost as soon as it forms, so [I] stays tiny. P appears at essentially the rate A disappears: d[P]/dt ≈ k₁[A]. The first step is rate-determining. This is the situation in which the steady-state approximation, treated separately, works well. - k₁ ≫ k₂ (first step fast). A is converted rapidly and almost completely into I, which accumulates to near [A]₀ and then slowly turns into P. The second step is rate-determining, and I may be isolable.

The rate-determining step. In either limit, the overall rate of product formation is governed by the slower step. The idea only holds cleanly when the rate constants differ by a large factor; when they are similar, both steps shape the kinetics and the full solution is needed.

Identifiability. The expression for [I] is symmetric in a subtle way: if only [P] or [A]-derived data are available, swapping k₁ and k₂ can reproduce the same product curve. Direct observation of the intermediate, for example spectroscopically, is often needed to decide which constant belongs to which step.

Formulae

[A] = [A]₀e^(−k₁t); [I] = [A]₀k₁(e^(−k₁t) − e^(−k₂t))/(k₂ − k₁); t max = ln(k₁/k₂)/(k₁ − k₂).

Step-by-step reasoning

1. Write one rate equation for each species from the elementary steps. 2. Solve the first-order equation for A. 3. Substitute [A] into the equation for I and solve. 4. Use mass balance to obtain [P]. 5. Examine the limits k₂ ≫ k₁ and k₁ ≫ k₂ to identify the rate-determining step.

Visual explanation

Draw three curves on one graph: [A] decaying exponentially from [A]₀, [I] forming a hump that starts and ends at zero, and [P] rising with an S-shaped start towards [A]₀. The hump's peak lies exactly where the [P] curve is steepest.

Real-world analogy

A two-stage car wash: cars (A) enter the wash bay (I) and then the drying bay before leaving (P). If drying is quick, cars never queue in between. If drying is slow, a queue of washed-but-wet cars builds up, then gradually clears.

Real-world example

Radioactive decay chains follow exactly this mathematics. Polonium-218 decays to lead-214, which decays to bismuth-214. The amount of lead-214 in a freshly separated polonium sample rises, peaks and falls, and the time of the maximum can be predicted from the two half-lives.

Why?

Why does [P] start with zero gradient? Product forms only from I, and at t = 0 there is no I. The product formation rate k₂[I] must therefore start at zero and grow as the intermediate accumulates.

Common misconception

"The intermediate's maximum concentration occurs when A is half used up." The time of the maximum depends on the ratio of both rate constants, not on a fixed fraction of A consumed.

Worked example

Question: For A → I → P with k₁ = 0.20 s⁻¹ and k₂ = 0.10 s⁻¹, when does [I] reach its maximum?

Reasoning: t max = ln(k₁/k₂)/(k₁ − k₂) = ln 2 ÷ 0.10 = 0.693 ÷ 0.10.

Answer: t max ≈ 6.9 s.

Quick check

1. In A → I → P, which step is rate-determining if k₁ = 0.01 s⁻¹ and k₂ = 50 s⁻¹? Answer: The first step, because it is far slower, so the intermediate is consumed as fast as it forms.

Exam focus

Be able to write the three rate equations and sketch the three concentration curves with correct features: exponential decay of A, a hump for I and an induction period for P. Explain the rate-determining step in terms of relative rate constants, not simply "the step with the largest activation energy".

Advanced insight

Longer sequences of consecutive steps produce progressively longer induction periods and sharper product curves. In polymer degradation and in multi-step drug metabolism, fitting such chains to data can show whether an unseen intermediate must exist. Numerical integration handles cases where steps are second order and no analytical solution is available.

Summary

In A → I → P, A decays exponentially, the intermediate rises to a maximum when its formation and loss rates are equal, and the product shows an induction period. When one step is much slower, it becomes rate-determining: if the second step is fast, I stays scarce; if the first is fast, I accumulates. Observing the intermediate helps assign the rate constants unambiguously.

Practice questions

1. Write the rate equation for the intermediate I in A → I → P. Answer: d[I]/dt = k₁[A] − k₂[I]. 2. At the moment [I] is at its maximum, what relationship holds? Answer: The formation and loss rates are equal, so k₁[A] = k₂[I] and d[I]/dt = 0. 3. Describe [I] against time when k₁ ≫ k₂. Answer: I rises quickly almost to [A]₀ and then decays slowly as it forms product. 4. Why is an induction period evidence for an intermediate? Answer: Product formation begins slowly because it depends on an intermediate that must first build up from zero.