The Eyring Equation

Introduction

The Eyring equation turns the transition-state picture into a measurable rate constant. It combines how often reactants reach barrier configurations with the flow of those configurations toward products. For a unimolecular step, its familiar form is k = κ(k B T/h)K‡. The symbols are compact, but their meanings matter: K‡ is a standard-state-consistent activation factor, k B T/h is a thermal frequency scale, and κ accounts for departures from ideal one-way crossing. A second-order reaction needs an additional standard-concentration conversion.

Core explanation

Start with an elementary A → products step. Under the quasi-equilibrium assumption, a formal activated-complex population is proportional to the A population: a ‡ = K‡a A. In conventional transition-state theory, the mean forward flux associated with the unstable reaction-coordinate motion contributes the factor k B T/h. Multiplying population by this frequency gives an ideal first-order rate constant, k TST = (k B T/h)K‡. If some counted crossings return to reactants, include a transmission factor κ and write k = κ(k B T/h)K‡. The derivation's detailed statistical mechanics treats the reaction-coordinate momentum separately from the constrained barrier modes; it is not a claim that every activated complex is a persistent molecule vibrating at exactly k B T/h.

At T = 298 K, use k B = 1.380649 × 10⁻²³ J K⁻¹ and h = 6.62607015 × 10⁻³⁴ J s. Their values are fixed in the modern SI; the BIPM SI Brochure specifies them. The frequency factor is (1.380649 × 10⁻²³ J K⁻¹ × 298 K)/(6.62607015 × 10⁻³⁴ J s) ≈ 6.21 × 10¹² s⁻¹. The kelvin and joule units cancel, leaving reciprocal seconds. This factor grows linearly with absolute temperature; most reactions change rate much more strongly with temperature because K‡ also changes.

What does K‡ contain? It is a formal equilibrium factor for population at the dividing surface relative to reactants. If an activation standard Gibbs energy is defined consistently, K‡ = exp(−ΔG‡°/RT). Then k = κ(k B T/h)exp(−ΔG‡°/RT) for a unimolecular step. This expression separates a universal thermal frequency scale from the reaction-specific free-energy penalty of reaching the barrier. A small K‡ means few reactant configurations occupy the bottleneck. The next page expands ΔG‡° into enthalpy and entropy contributions.

For an elementary bimolecular reaction A + B → products, rate is r = k₂c Ac B. The formal activity quotient is K‡ = a ‡/(a Aa B). Under an ideal dilute-solution convention with standard concentration c°, a i = c i/c°. The product-forming flux is κ(k B T/h)c ‡, with c ‡ = c°K‡(c A/c°)(c B/c°) = K‡c Ac B/c°. Therefore k₂ = κ(k B T/h)K‡/c°. If concentrations are in mol L⁻¹ and c° = 1 mol L⁻¹, the units are L mol⁻¹ s⁻¹. The numerical value of c° happens to be one in those units, but the unit factor must not disappear conceptually. A gas reaction may instead be formulated with pressure activities and a standard pressure; conversion to concentration units then uses the ideal-gas relationship and the selected standard state.

The equation presumes the elementary reaction described actually has the proposed dividing surface and quasi-equilibrium supply. If measured kinetics show a non-elementary order, a single A+B transition-state formula may not be the rate law for the whole network. A pre-equilibrium, steady-state intermediate, adsorption or diffusion limitation can alter the observed concentration dependence. The Eyring equation belongs to an elementary rate step; a mechanism combines elementary steps into the observed behaviour.

The coefficient κ needs context. In classical no-recrossing TST, κ is set to one. A dynamic recrossing correction may lower the predicted rate. Quantum tunnelling can instead enhance a reaction relative to a purely classical over-barrier rate, and some authors fold a tunnelling correction into a broader transmission factor. Thus “κ is always below one” is safe only when referring specifically to a classical recrossing fraction. State which convention a calculation uses. The IUPAC reaction-rate theory review discusses these assumptions, and the IUPAC transition-state terminology report makes clear why activation quantities and standard states require care.

Step-by-step reasoning

1. Identify whether the target elementary step is unimolecular or bimolecular. 2. Express the barrier population through a dimensionless, standard-state-defined K‡. 3. Multiply that population by k B T/h to obtain an ideal crossing flux. 4. Include κ only after stating whether it represents classical recrossing, tunnelling or a broader correction. 5. Convert activity-based flux to concentration-based rate units; include 1/c° for an ideal bimolecular step. 6. Check the predicted rate constant's dimensions and compare the assumed mechanism with the observed order.

Visual explanation

Make a two-column flow chart. In the left column, show “reactant population” leading to “barrier population = K‡ × reactant activity.” In the right column, show “crossing frequency = k B T/h” followed by “successful fraction = κ.” Join the columns into the unimolecular result k₁ = κ(k B T/h)K‡. Below it, draw a second line for A+B with two reactant-activity boxes and the concentration conversion k₂ = k₁-like factor divided by c°. Colour the standard-state conversion differently so it is easy to see why second-order units differ.

Real-world analogy

Imagine a narrow turnstile. The number of visitors near it determines how many are available to pass, while its throughput per waiting visitor sets a frequency. Multiplying availability by throughput gives departures per second. The analogy only separates population and flux; molecular barrier crossings are not periodic mechanical turns, and some trajectories may return after crossing.

Real-world example

Computational kinetics commonly estimates a barrier's activation free energy, inserts it into the Eyring expression at a series of temperatures and compares the predicted rate constants with experiment. A one-unit difference in ΔG‡° measured in multiples of RT changes the exponential factor by e. Thus a modest free-energy error can produce a large rate error. Accurate standard-state conversion is especially important when comparing association in solution with a unimolecular isomerisation or with gas-phase data.

Why?

Why does k B T/h have units of frequency? k B T is energy, with units joules. Planck's constant has units joule-seconds. Their ratio therefore has units s⁻¹. It is called a universal thermal frequency factor because the same constants and temperature appear for many reaction types, but the actual rate is not universal: it is multiplied by the often tiny K‡ and by an appropriate transmission factor.

Common misconception

“The Eyring equation gives the same units for every reaction because k B T/h is always s⁻¹.” That ignores how reactant activities convert to concentration-based rate laws. A bimolecular k₂ has reciprocal-concentration per time units and requires a standard-concentration factor. Another error is reading h/(k B T) as a measured lifetime of an activated complex. It is a statistical crossing-timescale factor in the derivation, not a stopwatch observation of a stable species.

Worked example

At 298 K, suppose an elementary unimolecular step has K‡ = 1.0 × 10⁻¹⁵ under a stated convention and a classical recrossing factor κ = 0.80. Using k B T/h = 6.21 × 10¹² s⁻¹, k = 0.80 × 6.21 × 10¹² × 1.0 × 10⁻¹⁵ = 4.97 × 10⁻³ s⁻¹. Its ideal first-order half-life is ln2/k ≈ 139 s, assuming the conditions and mechanism remain constant. The much larger thermal frequency does not mean the reaction is instantaneous: only about one in 10¹⁵ of the relevant statistical configurations contributes to the formal barrier population factor.

Quick check

1. What are the units of k B T/h? Answer: s⁻¹, because k B T is energy and h is energy multiplied by time.

Exam focus

Write the rate law and molecularity before selecting the Eyring form. For a unimolecular elementary step use k₁ = κ(k B T/h)K‡. For an ideal bimolecular step expressed in concentration units, use k₂ = κ(k B T/h)K‡/c° with consistent activities. Calculate k B T/h by checking units rather than memorising an unlabelled number. State the assumptions of quasi-equilibrium and negligible recrossing when using the ideal equation.

Advanced insight

The factor k B T/h arises after integrating the positive momentum component normal to the dividing surface and comparing it with the statistical population of constrained barrier states. The reaction coordinate's unstable mode is omitted from the usual activated-complex partition function to avoid counting it as an ordinary bound vibration. A refined theory may choose a variational dividing surface, add tunnelling or treat solvent dynamics explicitly. These refinements change the physical prediction while retaining the useful population-times-flux structure.

Summary

The Eyring equation connects a formal activation equilibrium factor to a rate through the thermal frequency k B T/h and a transmission correction κ. At 298 K, k B T/h is about 6.21 × 10¹² s⁻¹. For a unimolecular elementary step, k has s⁻¹ units; for a bimolecular step expressed with molar concentrations, the activity-to-concentration conversion supplies 1/c° and gives L mol⁻¹ s⁻¹. The equation applies to a specified elementary mechanism under transition-state assumptions.

Practice questions

1. At fixed K‡ and κ, how does the explicit k B T/h factor change if absolute temperature doubles? Answer: It doubles, although the full rate normally changes by more because K‡ usually depends on temperature.

2. At 298 K, find k₁ if K‡ = 2.0 × 10⁻¹⁴ and κ = 1. Answer: k₁ = 6.21 × 10¹² × 2.0 × 10⁻¹⁴ = 0.124 s⁻¹.

3. With c° = 1.0 mol L⁻¹, what units result from (k B T/h)K‡/c°? Answer: L mol⁻¹ s⁻¹, appropriate for a second-order concentration-based rate constant.

4. Why must a measured non-elementary rate law not be assigned a single overall Eyring molecularity from its balanced equation? Answer: The observed law can reflect multiple elementary steps, intermediates or transport; the overall stoichiometry does not identify the transition-state reactants of the rate-controlling step.

5. Using the rate constant from question 2, what is the first-order half-life of the step? Answer: t½ = ln2/k₁ = 0.693/0.124 s⁻¹ ≈ 5.6 s, assuming the mechanism and conditions stay constant.