The Activated Complex and Its Equilibrium Constant

Introduction

The rate equation in transition-state theory contains a symbol often written K‡. It resembles an equilibrium constant, but the activated complex is not a flask of stable material that can be separated and weighed. The symbol summarises how frequently thermal reactants visit configurations near the dividing surface. Its usefulness comes from a quasi-equilibrium approximation: even while products accumulate, the reactant basin can supply barrier configurations in a nearly thermal way. Care with concentrations, activities and standard states prevents a common mistake about the units of the resulting rate constant.

Core explanation

For a unimolecular elementary process, write the formal scheme A ⇌ A‡ → products. The double arrow represents an equilibrium-like population relation between A and configurations at the barrier, not a claim that A‡ is a persistent chemical species. If activities are used, a dimensionless formal factor can be written K‡ = a(A‡)/a(A). For an ideal dilute description at a consistent standard concentration c°, a(A) ≈ c A/c° and a(A‡) is defined by the corresponding barrier-region population. Then K‡ is the ratio of these dimensionless quantities. The words “corresponding barrier-region population” matter: its value depends on exactly how a thin region or dividing surface is defined. The final rate formula combines that population factor with a crossing frequency so that arbitrary bookkeeping thickness cancels in a proper derivation.

For a bimolecular step A + B → products, the formal scheme is A + B ⇌ [AB]‡ → products and the activity expression is K‡ = a([AB]‡)/(a A a B). In an ideal solution with the same c° for each species, this becomes K‡ = (c ‡/c°)/[(c A/c°)(c B/c°)] = c ‡c°/(c Ac B). Thus a dimensionless K‡ can coexist with a second-order rate constant once the standard-concentration factor is restored. If one writes a bare concentration quotient c ‡/(c Ac B), that quotient has reciprocal-concentration units; it is not the same symbol without a convention. For gases, a standard-pressure convention and partial-pressure activities give a parallel account. The IUPAC treatment of transition-state terminology stresses the importance of standard-state specification in activation quantities.

Why call this relation quasi-equilibrium rather than ordinary equilibrium? The activated complex is crossed and depleted into products. It cannot generally sit in an independently measurable reservoir. Yet if reactant collisions and energy exchange replace barrier configurations rapidly compared with net loss of reactants, their instantaneous population may be approximated from the thermal reactant distribution. The IUPAC Gold Book definition states that conventional transition-state theory assumes activated complexes form in equilibrium with reactants. “Quasi” reminds us that this is a local approximation within a reacting system, not a statement that the overall reaction has reached equilibrium.

The equilibrium-factor language connects kinetics to thermodynamics. A large positive activation free energy corresponds to a small K‡ and a sparsely occupied bottleneck. The following pages use K‡ = exp(−ΔG‡°/RT) with a specified standard state and combine it with the crossing flux. That relation is a formal statistical-mechanical one: it does not mean a calorimeter directly measures the free energy of a bottle of activated complex. It also depends on which conformations and modes contribute to the reactant and barrier ensembles. The same potential-energy maximum can give different rates if one pathway has a narrow, highly organised barrier region and another has many accessible barrier configurations.

Think of an elementary substitution in solution. Two reactants must come together, so the barrier population depends on both activities. Doubling c A at fixed c B approximately doubles the concentration of encounter configurations under ideal conditions, while doubling both approximately quadruples it. That is the concentration origin of a second-order elementary rate law. It is not enough to know the overall balanced equation: only the actual elementary step determines how many reactant factors enter the barrier population. For a multistep mechanism, a measured rate law may involve intermediates, saturation or inhibition instead.

The formal notation should not conceal dynamics. K‡ describes an equilibrium-like population at the chosen dividing surface; a transmission coefficient describes whether crossings commit to products. A barrier may be thermodynamically accessible but crossed repeatedly without net reaction. Conversely, quantum tunnelling can yield product formation that a classical over-the-barrier picture misses. Keeping population and commitment separate helps diagnose why an ideal TST rate differs from an experiment.

Step-by-step reasoning

1. Identify the proposed elementary step and count its reactant particles; do not infer molecularity from the overall equation. 2. Write a formal activated-complex relation, such as A ⇌ A‡ or A + B ⇌ [AB]‡. 3. Choose and state standard concentration or pressure before writing activities. 4. Form the dimensionless quotient from the barrier activity divided by reactant activities. 5. If converting to concentration units, carry every factor of c° explicitly. 6. Combine the population factor with crossing frequency and transmission probability when predicting a rate.

Visual explanation

Draw two reactant valleys feeding a narrow mountain pass marked [AB]‡, followed by a product valley. Under the drawing write a A × a B on the reactant side and a ‡ at the pass. Place a bracket around their ratio and label it K‡. Draw a second arrow from the pass toward products labeled “crossing and commitment.” This separates the equilibrium-like population calculation from the dynamical question of whether a crossing succeeds.

Real-world analogy

At a train station, the number of people standing at a platform at any instant depends on how many arrive and how long they spend there. A snapshot count can describe the platform population without implying that the platform is a permanent home. The activated-complex factor similarly concerns a fleeting population at a bottleneck. The analogy is limited because molecular configurations are distributed statistically, and a dividing surface may have no literal spatial platform.

Real-world example

When computational chemists compare two candidate reaction paths, they can estimate partition functions for reactants and for constrained configurations near each candidate transition state. A path with a somewhat higher energy barrier may still compete if it has more accessible orientations or vibrational states at the bottleneck. The comparison requires consistent standard states and a compatible definition of the dividing surface. Experimental product ratios over temperature can then test whether the modeled relative activation free energies are plausible.

Why?

Why does a bimolecular activation quotient involve two reactant activities? The probability of forming a specific encounter configuration depends on finding both A and B. Under the ideal elementary-step assumption, the reactant-state statistical weight contains one factor for each independent reactant. A single-reactant isomerisation has one factor instead. This distinction later appears in the units: a bimolecular rate constant must convert c Ac B to concentration per time, whereas a unimolecular constant converts c A alone.

Common misconception

“K‡ is an experimentally measured ordinary equilibrium constant for a stable transition-state molecule.” The activated complex is a formal ensemble of barrier configurations, and its equilibrium relation is an approximation. Another common error is writing K‡ = c ‡/(c Ac B) and calling it dimensionless while concentrations are in mol L⁻¹. Use activities or include the standard concentration. Finally, a large K‡ alone does not guarantee rapid product formation if recrossing or other dynamical effects suppress commitment.

Worked example

At a stated standard concentration c° = 1.0 mol L⁻¹, suppose an idealised bimolecular model assigns c A = 0.20 mol L⁻¹, c B = 0.50 mol L⁻¹ and a formal barrier-region concentration c ‡ = 2.0 × 10⁻⁶ mol L⁻¹ under its chosen bookkeeping convention. The dimensionless activities are a A = 0.20, a B = 0.50 and a ‡ = 2.0 × 10⁻⁶. Therefore K‡ = a ‡/(a Aa B) = (2.0 × 10⁻⁶)/(0.20 × 0.50) = 2.0 × 10⁻⁵. The numerical value belongs to this standard-state and barrier-region convention; the exercise shows the activity algebra, not a directly isolated concentration of stable [AB]‡.

Quick check

1. If a A doubles while a B and K‡ stay fixed in the idealised elementary A + B scheme, what happens to a ‡? Answer: It doubles because a ‡ = K‡a Aa B.

Exam focus

Distinguish a formal activation quotient from an ordinary measurable equilibrium between stable species. State standard states whenever you use ΔG‡° or K‡. For A + B, write K‡ = a ‡/(a Aa B) and demonstrate the c° factor if using concentrations. Explain why the activated-complex population supplies only one part of a rate prediction and why an overall stoichiometric equation cannot determine molecularity by itself.

Advanced insight

In a microscopic derivation, a dividing surface has one fewer configurational degree of freedom than the full reactant space; momentum along the reaction coordinate provides the crossing flux. Naively assigning an ordinary three-dimensional concentration to a zero-thickness surface is therefore delicate. Different texts define a thin layer, a constrained partition function or a standard-state-normalised K‡, then combine it with k B T/h in a self-consistent way. Physical predictions should not depend on an arbitrary layer thickness. These bookkeeping choices explain why apparently different printed forms of the Eyring equation can agree after molecularity and standard-state factors are handled consistently.

Summary

The activated complex is a formal ensemble of barrier configurations supplied by thermal reactants. Quasi-equilibrium lets its population be related to reactant activities by K‡ even while the whole reaction remains far from equilibrium. For A + B, a dimensionless activity quotient includes two reactant factors and an explicit standard-state conversion if concentrations are substituted. Population at the barrier and successful passage into products are separate parts of the rate calculation.

Practice questions

1. Write the formal activity quotient for A + B ⇌ [AB]‡. Answer: K‡ = a([AB]‡)/(a Aa B), using activities referred to specified standard states.

2. With c° = 1 mol L⁻¹, c A = 0.10 mol L⁻¹, c B = 0.20 mol L⁻¹ and c ‡ = 4.0 × 10⁻⁷ mol L⁻¹, find K‡. Answer: K‡ = c ‡c°/(c Ac B) = 4.0 × 10⁻⁷/(0.10 × 0.20) = 2.0 × 10⁻⁵.

3. Why is [AB]‡ not automatically a reaction intermediate? Answer: An intermediate occupies a local stable basin; [AB]‡ denotes fleeting configurations near a dividing surface or barrier.

4. What additional idea beyond K‡ is needed to estimate how many barrier visits produce products? Answer: The crossing flux and, where recrossing occurs, a transmission or commitment factor are needed.

5. Is c ‡/(c Ac B) dimensionless when every c is measured in mol L⁻¹? Answer: No. It has units L mol⁻¹; multiply by the chosen c° to make the standard-state activity quotient dimensionless.