Arrhenius and Eyring Formulae

Introduction

Rate constants often rise with temperature, but the amount and shape of that rise contain mechanistic information. The Arrhenius equation uses an activation energy and a pre-exponential factor. The Eyring equation expresses a related temperature dependence through activation Gibbs energy, enthalpy and entropy. Both are models over a temperature range, and neither proves a single transition state merely because a plot is straight.

Core explanation

The Arrhenius form is k = A exp(−E a/RT), where k is a rate constant, A has the same units as k, E a is activation energy, R is the molar gas constant and T is kelvin. Taking logarithms gives ln k = ln A − E a/(RT). A plot of ln k against 1/T is approximately linear with slope −E a/R if E a and A are constant over the range.

The two-temperature form eliminates A: ln(k₂/k₁) = (E a/R)(1/T₁ − 1/T₂). If T₂ > T₁ and E a > 0, the right side is positive and k₂ > k₁. E a must be in J mol⁻¹ when R is J mol⁻¹ K⁻¹. A large activation energy produces a larger temperature sensitivity, but rate also depends on A.

In conventional transition-state theory for a simple first-order process, k ≈ κ(k B T/h)exp(−ΔG‡/RT), where κ is a transmission factor and ΔG‡ = ΔH‡ − TΔS‡. If κ is treated as constant and near unity, k/T plotted logarithmically against 1/T has slope approximately −ΔH‡/R and intercept related to ΔS‡/R plus ln(k B/h). For higher-order rate constants, standard concentration factors must be handled carefully so units remain correct.

Activation entropy describes the entropy difference between an activated configuration and chosen reactant standard state. A negative ΔS‡ can indicate a more constrained activated arrangement, but assigning a specific molecular geometry from its sign alone is unsafe. Solvation, standard-state choices and coupled equilibria contribute. Activation enthalpy and Arrhenius E a are related but not always numerically identical; their relationship depends on rate-law molecularity and conventions.

A straight Arrhenius line can summarize useful data without specifying the mechanism. Curvature may arise from a change of dominant pathway, catalyst speciation, diffusion limitation, phase transition or temperature-dependent heat capacities. Conversely, two compensating pathways can accidentally look linear over a narrow range. Compare chemistry and conditions, not just fit quality.

Measured k must refer to the same kinetic law at every temperature. A pseudo-first-order k obs may change because an excess reagent's activity or an equilibrium pre-step changes, not simply because the intrinsic elementary barrier changes. Gas pressure, solvent viscosity and pH should be controlled or measured.

At high temperature, a reaction may approach equilibrium or decompose by another route. A two-temperature Arrhenius extrapolation far outside measured range may be unreliable. Report temperature interval, uncertainty in slopes and the rate units. Small temperature spans make E a uncertain because 1/T changes very little.

Step-by-step reasoning

Establish the rate law and measure k at several temperatures. Convert to kelvin and keep a consistent rate unit. Plot ln k versus 1/T for Arrhenius analysis or ln(k/T) versus 1/T for Eyring analysis with appropriate standard-state factors. Inspect residuals and chemical changes before interpreting slope as a single barrier.

Visual explanation

Draw a reaction-coordinate barrier and two plots beneath it. An Arrhenius plot slopes downward with increasing 1/T; an Eyring plot uses ln(k/T). Label the different slope parameters and mark that changing mechanism can bend either line.

Real-world analogy

Crossing a hill becomes easier as travelers have more energy, but the route width and organization also affect how many cross per minute. Arrhenius E a resembles temperature sensitivity of crossing, while Eyring entropy adds a measure of how constrained a successful route is. Molecules do not literally possess a single shared hiking path.

Real-world example

A decomposition is measured from 290 to 320 K. Rate constants increase steadily, and an Arrhenius plot gives an approximately constant slope. This supports reporting an apparent E a over that range. A catalyst added at 330 K may create a new pathway, so the old slope should not be extrapolated through the changed system.

Why?

Temperature-dependent formulae help predict storage stability and reactor rates and can constrain mechanism. Their proper interpretation separates an empirical apparent activation energy from detailed transition-state thermodynamics.

Common misconception

“A higher E a always means a slower reaction” ignores the pre-exponential factor and temperature. Another error inserts Celsius into 1/T, yielding a meaningless slope and sometimes a reversed comparison.

Worked example

Let E a = 50.0 kJ mol⁻¹, T₁ = 300 K and T₂ = 310 K. Then ln(k₂/k₁) = (50,000/8.314)(1/300 − 1/310) ≈ 0.647. Thus k₂/k₁ ≈ e^0.647 ≈ 1.91. A ten-kelvin rise nearly doubles k under a constant-Arrhenius-parameter approximation.

Quick check

1. What is the slope of an Arrhenius ln k versus 1/T plot? Answer: Approximately −E a/R when Arrhenius parameters are constant.

Exam focus

Use kelvin, joules and natural logarithms consistently. Distinguish ln k from ln(k/T) plots. State that activation parameters depend on model, standard state and temperature range.

Advanced insight

The transmission coefficient κ accounts for trajectories that reach the dividing surface but do not become products. Quantum tunneling can also change temperature dependence, especially for light particles. These effects can make a simple Eyring fit an apparent parameterization rather than a literal barrier measurement.

Summary

Arrhenius analysis describes k through A and E a; Eyring analysis relates rate to ΔH‡ and ΔS‡ through a transition-state model. Both require consistent rate laws and units and should be fitted only over conditions with a stable mechanism.

Practice questions

1. What temperature unit belongs in Arrhenius equations? Answer: Kelvin. 2. If E a > 0, what happens to k when T increases under the simple Arrhenius model? Answer: k increases. 3. What does a curved ln k versus 1/T plot suggest? Answer: Temperature-dependent apparent parameters or a change of mechanism or conditions. 4. Does a negative ΔS‡ uniquely prove one transition-state geometry? Answer: No. Solvation, standard states and other effects can influence the measured activation entropy.

Sources

- IUPAC Gold Book: Enthalpy of Activation and Eyring Form. - OpenStax Chemistry 2e: Arrhenius Analysis.