Eyring Plots and Activation Parameters

Introduction

A rate constant measured at only one temperature gives limited information about a reaction barrier. Repeating the measurement across temperatures tests whether a transition-state model describes the reaction and can estimate activation enthalpy and entropy. The Eyring plot uses ln(k/T) on the vertical axis and 1/T on the horizontal axis. Dividing k by T before taking the logarithm removes the explicit temperature factor k B T/h, leaving a slope tied to ΔH‡° when the model assumptions hold over the chosen range.

Core explanation

For an elementary unimolecular step, write k = κ(k B T/h)exp(ΔS‡°/R)exp(−ΔH‡°/RT). Assume for the moment that κ, ΔH‡° and ΔS‡° are approximately constant over the measured range. Divide by T and take natural logarithms: ln(k/T) = ln(κk B/h) + ΔS‡°/R − (ΔH‡°/R)(1/T). If y = ln(k/T) and x = 1/T, this is y = b + mx, with slope m = −ΔH‡°/R and intercept b = ln(κk B/h) + ΔS‡°/R. Therefore ΔH‡° = −mR. If κ is known or reasonably approximated as one, ΔS‡° = R[b − ln(κk B/h)] under the stated standard-state convention.

The logarithm of a dimensional quantity needs care. Laboratory practice often reports k and T in explicitly stated units and treats ln(k/T) as shorthand for the logarithm of a numerical value relative to those units. A rigorous dimensionless form takes ln[(k/T)/(1 s⁻¹ K⁻¹)] for first-order k, or divides by another declared reference. The slope is unaffected by a constant reference-unit choice, but the numerical intercept changes when units change. For a bimolecular reaction, the standard-concentration factor and the chosen rate-constant units also enter the intercept. This is why a published activation entropy must be accompanied by units and standard-state conventions.

A negative slope usually follows from a positive ΔH‡°. Plotting against 1/T means higher temperature appears to the left, so do not read the graph as if x increased with T. If the fitted slope is −6.0 × 10³ K, the ideal activation enthalpy is −(−6.0 × 10³ K)(8.314 J mol⁻¹ K⁻¹) ≈ 49.9 kJ mol⁻¹. The intercept then estimates entropy after the frequency and transmission terms are handled. A raw intercept is not itself ΔS‡°.

Experimental fitting deserves more care than drawing a line through two points. Each measured rate constant has uncertainty. Taking logs changes the error distribution, and points with different relative errors should not automatically receive equal weight. Temperature uncertainty also changes 1/T. A narrow temperature span can yield an apparently straight plot yet poorly constrain both slope and intercept; extrapolating that line far outside the measured range magnifies error. Report the span and uncertainty, and use residuals to look for systematic curvature.

Curvature can have several causes. Activation heat capacity may make ΔH‡° and ΔS‡° vary with temperature. A change in rate-controlling step or population of reactant conformers may produce a bend. The transmission factor κ may vary with temperature because of recrossing or tunnelling. A diffusion-limited solvent reaction may also break the simple elementary barrier model. One curve cannot uniquely identify which cause applies; additional mechanistic measurements are needed. The IUPAC terminology report provides the conventions for activation quantities, and the IUPAC reaction-rate theory discussion identifies assumptions behind their transition-state interpretation.

Comparing an Eyring plot with an Arrhenius plot is instructive. Arrhenius analysis plots ln k against 1/T and yields an empirical activation energy Ea from a local slope. The Eyring plot uses ln(k/T), accounting for the explicit T factor, and yields an activation enthalpy under its model. For many ideal gas-phase unimolecular treatments over a limited range, Ea ≈ ΔH‡° + RT, but the exact relationship can vary with reaction order and conventions. Do not call the two slopes identical merely because both graphs look straight.

Step-by-step reasoning

1. Measure k at several well-controlled absolute temperatures and state the rate law and units. 2. Calculate x = 1/T and y = ln(k/T) with a declared numerical-unit convention. 3. Fit a straight line and inspect residuals and uncertainties rather than relying only on visual alignment. 4. Compute ΔH‡° = −mR from the fitted slope and convert to suitable energy units. 5. Compute ΔS‡° from the intercept only after stating κ, standard state and rate-constant units. 6. Treat unusual slopes or curvature as evidence to investigate mechanisms or temperature-dependent corrections.

Visual explanation

Draw three temperature points on a graph with 1/T increasing to the right and ln(k/T) increasing upward. Connect them with a descending line. Mark the vertical drop divided by horizontal change as slope −ΔH‡°/R, and label the intercept as “ln(κk B/h) + ΔS‡°/R,” not just “entropy.” Add a second, slightly curved dashed data series to show how systematic departures from a line may signal temperature-dependent activation parameters or a changing mechanism.

Real-world analogy

Suppose a machine's output depends on both how often it cycles and what fraction of cycles succeed. If cycle frequency itself rises with temperature, dividing output by that predictable frequency exposes changes in success probability. The Eyring plot similarly removes the explicit T factor before examining the activation factor. The analogy is limited because the molecular crossing factor comes from statistical mechanics rather than a literal motor.

Real-world example

Researchers may measure an isomerisation rate at several temperatures by spectroscopy. If the first-order decay curves give rate constants and ln(k/T) versus 1/T is linear, they can report an apparent ΔH‡° and ΔS‡° with uncertainty and a temperature interval. Comparing those values with calculations can test a proposed barrier. They should still confirm that the same reaction path operates throughout the interval, because a competing conformer or side reaction can make the fitted parameters misleading.

Why?

Why is ln(k/T), rather than ln k, used for an Eyring plot? The Eyring equation contains a prefactor proportional to T. Dividing by T leaves κ(k B/h) times the activation exponentials. Under approximately constant ΔH‡°, ΔS‡° and κ, taking the logarithm then gives a linear expression in 1/T whose slope is −ΔH‡°/R. Keeping the T factor inside k would add an extra ln T contribution to the plot.

Common misconception

“The Eyring-plot intercept directly equals ΔS‡°.” It also contains ln(κk B/h) and may require a standard-state or unit conversion. A second mistake is treating a straight line as proof of one mechanism: several mechanisms can approximate a line over a narrow range. A third is plotting Celsius temperature or 1/°C. Both the Boltzmann factor and reciprocal temperature require absolute kelvin.

Worked example

A set of first-order data fitted over a modest temperature interval gives slope m = −6.00 × 10³ K for ln(k/T) versus 1/T. The activation enthalpy is ΔH‡° = −mR = 6.00 × 10³ × 8.314 = 49,884 J mol⁻¹, or 49.9 kJ mol⁻¹. To check the slope's meaning, take T₁ = 300 K and T₂ = 310 K. The model predicts ln[(k₂/T₂)/(k₁/T₁)] = (ΔH‡°/R)(1/T₁ − 1/T₂) ≈ 6,000(1/300 − 1/310) ≈ 0.645. Therefore (k₂/T₂)/(k₁/T₁) ≈ e^0.645 ≈ 1.91. The ratio of the unscaled rate constants is also multiplied by 310/300, giving about 1.97.

Quick check

1. What is the horizontal-axis variable in a standard Eyring plot? Answer: Reciprocal absolute temperature, 1/T in K⁻¹.

Exam focus

Derive the line equation from the Eyring equation before reading parameters. Label the axes ln(k/T) and 1/T; keep all temperatures in kelvin. Use slope −ΔH‡°/R, not the Arrhenius formula blindly. Treat the intercept as a combination of transmission, frequency and entropy contributions, and state the conditions under which the parameters are meaningful.

Advanced insight

If κ varies with temperature, the observed slope includes d ln κ/d(1/T) as well as the activation enthalpy term. If ΔH‡° varies because ΔC p‡ is nonzero, a curved Eyring plot can be modeled with an explicit temperature-dependent activation free energy. A piecewise straight fit may be tempting, but distinguishing a mechanism switch from smooth heat-capacity curvature requires independent data. Bayesian or weighted nonlinear fitting can propagate measurement uncertainties and compare these alternative models more honestly than estimating a slope by eye.

Summary

An Eyring plot graphs ln(k/T) against 1/T. Under constant activation parameters and transmission factor, its slope is −ΔH‡°/R and its intercept contains ΔS‡°/R together with the universal frequency and transmission terms. Standard-state and unit conventions are necessary to interpret the intercept. Good analysis uses several temperatures, uncertainty-aware fitting and residual checks; curvature suggests additional physical effects rather than a single automatic diagnosis.

Practice questions

1. A line has slope −5.00 × 10³ K. Find ΔH‡°. Answer: ΔH‡° = −mR = 5.00 × 10³ × 8.314 ≈ 41.6 kJ mol⁻¹.

2. Why must Celsius values not be placed on the reciprocal-temperature axis? Answer: The Boltzmann factor and thermodynamic equations use absolute temperature, so 1/T must use kelvin.

3. What other quantity, besides ΔS‡°, contributes to a unimolecular Eyring intercept when κ is constant? Answer: ln(κk B/h), expressed using a stated convention for logarithms and units.

4. Name two physical explanations for a curved Eyring plot. Answer: Temperature-dependent activation heat capacity and a change in mechanism are possibilities; temperature-dependent tunnelling or recrossing can also contribute.

5. A fitted slope is −4,000 K. Is the ideal activation enthalpy positive or negative? Answer: Positive: ΔH‡° = −mR ≈ 33.3 kJ mol⁻¹.