The Eyring Equation

Introduction

The Eyring equation turns a thermodynamic activation free energy into a rate estimate. Its exponential factor describes the rarity of an activated arrangement, and its temperature-dependent prefactor gives a characteristic crossing scale. The equation is especially useful when comparing related elementary reactions or interpreting temperature-dependent rate data. It must be applied with the correct reaction molecularity, standard state and awareness of recrossing or tunnelling.

Core explanation

For a unimolecular elementary step under conventional transition-state theory, a common form is k = κ(kBT/h) exp(−ΔG‡/RT) . Here kB is Boltzmann's constant, h is Planck's constant, R is the molar gas constant, T is absolute temperature, ΔG‡ is a molar activation Gibbs energy under a stated convention, and κ is a transmission coefficient accounting for specified dynamical effects. If κ is set to one, the expression gives the conventional no-recrossing TST estimate. The exponential argument must be dimensionless, so ΔG‡ and RT must use consistent energy units.

Since ΔG‡ = ΔH‡ − TΔS‡, the same expression can be written k = κ(kBT/h) exp(ΔS‡/R) exp(−ΔH‡/RT), assuming the activation parameters are defined consistently. This form separates an enthalpic exponential from an entropy factor, but the factors are not independent physical “causes” that can be adjusted arbitrarily. Both arise from the same state-counting model. A negative activation entropy reduces the rate relative to an otherwise similar enthalpy barrier, often reflecting a more constrained activated arrangement.

At about room temperature, kBT/h is of order 10¹²–10¹³ s⁻¹. That large prefactor does not imply fast reaction if the activated fraction exp(−ΔG‡/RT) is tiny. For example, a barrier of tens of kilojoules per mole can reduce the rate by many orders of magnitude. A barrier error of a few kilojoules per mole can therefore have a large effect on predicted k. Always report the temperature alongside a rate estimate.

Units require attention. The displayed formula directly yields s⁻¹ for a unimolecular process when ΔG‡ is defined for that convention. A bimolecular rate constant typically has concentration⁻¹ time⁻¹ units, so a standard concentration or pressure factor enters. Quoting a bimolecular ΔG‡ without its standard state can lead to a wrong numerical comparison. IUPAC notes that experimentally inferred activation Gibbs energies depend on the units and standard-state choices used for rate constants.

An Eyring plot uses ln(k/T) versus 1/T. If κ and activation enthalpy and entropy are approximately constant over a narrow temperature range, its slope is −ΔH‡/R and its intercept includes ln(kB/h), ΔS‡/R and ln κ. Strong curvature may reflect changing mechanism, heat capacities, conformer populations, tunnelling or temperature-dependent κ. A straight line is compatible with the simple model but does not by itself prove one microscopic pathway.

The equation also needs a valid bottleneck model. Recrossing makes κ below the ideal classical value for a given dividing surface, while tunnelling can enhance passage through a barrier. It is often misleading to fold several unrelated corrections into one unexplained κ. For a multi-step catalytic or enzymatic mechanism, an Eyring barrier inferred from the overall rate may be an effective parameter rather than the energy difference to one computed saddle.

Step-by-step reasoning

Identify the elementary step and its molecularity. Choose T and a consistent standard-state activation Gibbs energy. Compute kBT/h, then ΔG‡/(RT), take the exponential and multiply. State whether κ is assumed one or estimated by a dynamical model. Check the final units. For temperature-series data, plot ln(k/T) against 1/T only after confirming a comparable rate law and mechanism across temperatures.

Visual explanation

Place three blocks in a rate diagram: crossing scale kBT/h, activated population exp(−ΔG‡/RT), and dynamical correction κ. Multiplying the blocks gives k. Plot ln(k/T) vertically against 1/T horizontally and label the ideal straight-line slope −ΔH‡/R. Add a curved alternative to show that changing conformer populations or tunnelling can spoil the simple linear interpretation.

Real-world analogy

A very large number of attempts per second does not guarantee frequent success if the fraction of attempts in the required arrangement is tiny. The Eyring prefactor resembles the attempt scale and the exponential the rare fraction. Real molecules are not workers making independent deliberate attempts; the analogy omits multidimensional dynamics, recrossing and environmental fluctuations.

Real-world example

Consider two related conformational isomerisations with the same approximate crossing prefactor at 298 K. If one has a standard activation Gibbs energy 5 kJ mol⁻¹ higher, its rate is lower by approximately exp[−5000/(8.314×298)] ≈ 0.13, about an eightfold difference under equal κ. This illustrates the exponential sensitivity that makes small barrier errors consequential in computational rate predictions.

Why?

Why does ΔG‡ appear rather than only electronic energy? Both energetic and entropic rarity affect the activated population. Why does kBT/h have time⁻¹ units? kBT is energy and h is energy times time. Why specify molecularity? Rate-constant units and standard-state factors change between unimolecular and bimolecular steps. Why keep κ explicit? Crossing a dividing surface need not mean successful reaction.

Common misconception

The prefactor kBT/h is not a universal measured vibration frequency for the reaction. The Eyring equation is a statistical rate model with assumptions. Another error is to use Celsius in RT or mix kJ mol⁻¹ with R in J mol⁻¹ K⁻¹ without conversion. A third is reading an Eyring-plot slope as an electronic barrier; it corresponds to an activation enthalpy only under the stated approximations.

Worked example

Question: At 300 K a unimolecular step has ΔG‡ = 75.0 kJ mol⁻¹ and κ = 1. Estimate k using kBT/h ≈ 6.25×10¹² s⁻¹ and R = 8.314 J mol⁻¹ K⁻¹.

Reasoning: Convert the barrier to 75,000 J mol⁻¹. The exponent is −75,000/(8.314×300) ≈ −30.07. Its exponential is approximately 8.7×10⁻¹⁴. Multiplying by 6.25×10¹² s⁻¹ gives approximately 0.54 s⁻¹. This is a rounded TST estimate; changing κ or the barrier changes the prediction.

Answer: k ≈ 0.54 s⁻¹ under the given unimolecular, no-recrossing assumptions.

Quick check

1. Why must ΔG‡ and RT have the same energy units in the Eyring equation? Answer: Their ratio is an exponent and therefore must be dimensionless.

Exam focus

Write the unimolecular Eyring equation with all constants and units. Use Kelvin and a consistent molar energy unit. Explain the meaning of ΔG‡ and κ, distinguish unimolecular from bimolecular standard-state factors, and interpret a simple Eyring plot only under stable-mechanism assumptions.

Advanced insight

An effective activation free energy inferred by rearranging the Eyring equation depends on the chosen prefactor and transmission convention. If κ varies with temperature, a fitted activation entropy may absorb dynamics rather than represent only configurational ordering. Tunnelling often produces pronounced temperature and isotope dependence, while conformer reweighting can curve an Eyring plot even if each individual conformer obeys a simple expression. Comparing rates across conditions therefore tests the kinetic model, not just the numerical barrier.

Summary

The Eyring equation estimates an elementary rate from a temperature-dependent crossing prefactor and an exponential activation Gibbs penalty, optionally corrected by κ. Its exponential sensitivity makes barrier accuracy important. Temperature, molecularity, standard state and dynamical assumptions must be specified. Activation enthalpy and entropy can be inferred from temperature trends only when the mechanism and correction factors remain suitably stable.

Practice questions

1. What happens to k if ΔG‡ rises while T and κ remain fixed? Answer: k falls exponentially through exp(−ΔG‡/RT).

2. What are the units of kBT/h? Answer: Reciprocal time, usually s⁻¹.

3. Why cannot the displayed unimolecular Eyring formula be used unchanged for every bimolecular rate constant? Answer: Bimolecular constants have different units and require a concentration or pressure standard-state factor.

4. What could cause curvature in an Eyring plot? Answer: Mechanism changes, temperature-dependent κ, tunnelling, heat-capacity effects or shifting conformer populations.

Sources: IUPAC Gold Book, transition-state theory; IUPAC Gold Book, Gibbs energy of activation; Physical Chemistry teaching text, Eyring equation.