Transition-State (Eyring) Theory

Collision theory tells you reactants must collide with enough energy. It cannot tell you why a hindered, floppy reaction is slower than a compact one with the same energy barrier. Transition-state theory fixes this by treating the fleeting, high-energy species at the top of the barrier as a real, if short-lived, chemical entity with its own thermodynamics. This matters because Eyring's theory replaces a geometric "steric factor" with an entropy of activation, something you can actually rationalize from molecular structure. It reframes rate constants in the language you already use for equilibrium constants: $\Delta^{\ddagger}H$ , $\Delta^{\ddagger}S$ , $\Delta^{\ddagger}G$ . This article builds the activated-complex picture from a pre-equilibrium assumption. It derives the Eyring equation from statistical and thermodynamic arguments. It works two full numerical examples. It closes by connecting activation parameters to the deeper thermodynamics of activation that follows this topic.

Key takeaways

Transition-state theory (TST) models the rate-limiting step as an equilibrium between reactants and an activated complex, $\ddagger$ , sitting at the potential-energy saddle point.
The Eyring equation is $k = (k_BT/h)K^{\ddagger}$ , converting a rate constant into an equilibrium constant times a universal frequency factor.
Written thermodynamically, $k = (k_BT/h)\exp(-\Delta^{\ddagger}G^{\circ}/RT)$ , splitting neatly into enthalpy and entropy of activation.
The transmission coefficient $\kappa$ (usually near 1) corrects for complexes that cross the barrier and recross back to reactants.
Comparing Eyring to Arrhenius shows $E_a = \Delta^{\ddagger}H + RT$ for reactions in solution, linking the two frameworks quantitatively.
Key idea: treat the transition state as a molecule in fast equilibrium with reactants, then let ordinary statistical thermodynamics generate the rate constant.

The core insight

Transition-state theory converts the kinetics problem of "how fast do molecules react" into a thermodynamics problem of "what is the free energy of forming a fleeting activated complex," using the universal frequency $k_BT/h$ as the rate at which that complex decays irreversibly into products.

The physical picture

Picture a reaction coordinate: a single collective variable describing bond-breaking and bond-forming as reactants become products. Along this coordinate, the potential energy rises from the reactant valley, passes over a saddle point, and descends into the product valley. That saddle point is the transition state, and the species sitting there — neither reactant nor product, an unstable arrangement of partly broken and partly formed bonds — is the activated complex, denoted $\ddagger$ . Collision theory only asks whether colliding molecules carry enough energy and land in the right orientation. Transition-state theory goes further: it treats the activated complex as a genuine, if momentary, chemical species with its own partition function, its own vibrational modes (all but one — the mode along the reaction coordinate, which becomes translation over the barrier), and its own thermodynamic parameters. The central postulates are: 1. Reactants and the activated complex are in quasi-equilibrium, even though the overall reaction is irreversible — this is the same logic used in the pre-equilibrium approximation, applied to a species that exists for only about one bond vibration ( $\sim 10^{-13}$ s). 2. Every complex that crosses the saddle point moving toward products does become product; it never turns back. (A transmission coefficient $\kappa$ is inserted later to relax this.) 3. The rate of crossing is set by the frequency of the one special vibrational mode that has become a translation along the reaction coordinate, extended into a rate using the equipartition-derived quantity $k_BT/h$ .

Figure

Reaction coordinate diagram showing reactants, the activated complex at the saddle point, and products, with the activation barrier ΔG‡ labeled

The activated complex differs from an ordinary molecule in exactly one respect: one of its 3N−6 (or 3N−5) vibrational degrees of freedom is not a real oscillation but an imaginary-frequency mode — the reaction coordinate itself. Rather than restoring the system to equilibrium, motion along this coordinate carries the system downhill to products. Everything else about $\ddagger$ — its remaining vibrations, rotations, translations — is treated with the full machinery of statistical thermodynamics, exactly as for a stable molecule, which is what lets us assign it a standard molar Gibbs energy, enthalpy, and entropy of formation from reactants. A second useful picture is how the Eyring equation's ingredients trade off against Arrhenius parameters as temperature changes — worth visualizing because the two theories agree in form but differ subtly in what they attribute to $T$ -dependence.

Figure

Comparison of Arrhenius and Eyring pre-exponential temperature dependence, showing the extra factor of T in the Eyring frequency term

The core relations

A + B \rightleftharpoons AB^{\ddagger} \rightarrow \text{products}, \qquad K^{\ddagger} = \frac{[AB^{\ddagger}]}{[A][B]}

where: $K^{\ddagger}$ is the equilibrium constant for forming the activated complex from reactants.

k = \kappa \frac{k_BT}{h} K^{\ddagger}

where: $k_B$ is Boltzmann's constant, $h$ is Planck's constant, $\kappa$ is the transmission coefficient (often ≈1), and $k_BT/h$ is the universal frequency of barrier crossing.

K^{\ddagger} = \exp\left(-\frac{\Delta^{\ddagger}G^{\circ}}{RT}\right), \qquad \Delta^{\ddagger}G^{\circ} = \Delta^{\ddagger}H^{\circ} - T\Delta^{\ddagger}S^{\circ}

where: $\Delta^{\ddagger}G^{\circ}$ , $\Delta^{\ddagger}H^{\circ}$ , $\Delta^{\ddagger}S^{\circ}$ are the standard molar Gibbs energy, enthalpy, and entropy of activation.

k = \frac{k_BT}{h}\exp\left(\frac{\Delta^{\ddagger}S^{\circ}}{R}\right)\exp\left(-\frac{\Delta^{\ddagger}H^{\circ}}{RT}\right)

where: this is the Eyring equation in its most-used practical form (taking $\kappa=1$ ).

E_a = \Delta^{\ddagger}H^{\circ} + RT \quad (\text{solution, unimolecular}); \qquad E_a = \Delta^{\ddagger}H^{\circ} + 2RT \quad (\text{gas-phase, bimolecular})

where: $E_a$ is the Arrhenius activation energy, connecting Eyring and Arrhenius parameters.

The core derivation

Derivation AFrom quasi-equilibrium to the Eyring equation

Goal. Derive $k = (k_BT/h)K^{\ddagger}$ starting only from the assumption that reactants and the activated complex are in equilibrium, and that crossing the barrier occurs at the frequency of one specific vibrational mode.

Step 1 — set up the quasi-equilibrium. For the elementary step $A + B \rightleftharpoons AB^{\ddagger} \to P$ , assume the forward and reverse steps forming $AB^{\ddagger}$ equilibrate much faster than $AB^{\ddagger}$ decays to products (justified because decay requires crossing an unstable mode with essentially no restoring force, a one-way trip). This gives

K^{\ddagger} = \frac{[AB^{\ddagger}]}{[A][B]}.

Step 2 — isolate the special vibration. The activated complex has one vibrational degree of freedom along the reaction coordinate that is not a bound oscillation but a translation over a very short distance $\delta$ at the barrier top. Statistical mechanics treats this mode's contribution to the partition function of $AB^{\ddagger}$ as a one-dimensional translational partition function of length $\delta$ ,

q_{\text{rxn coord}} = \frac{(2\pi m^{\ddagger} k_BT)^{1/2}}{h}\,\delta.

Step 3 — write the rate of product formation. The rate at which complexes cross the barrier equals their concentration times the frequency $\nu$ with which they traverse the length $\delta$ ,

\nu = \frac{\bar{v}}{\delta} = \frac{1}{\delta}\left(\frac{2k_BT}{\pi m^{\ddagger}}\right)^{1/2}.

The rate law is $\text{rate} = \nu[AB^{\ddagger}]$ .

Step 4 — substitute and cancel $\delta$ . Combining the equilibrium constant (which contains $q_{\text{rxn coord}}$ in the numerator, since it appears in the partition function of $AB^{\ddagger}$ ) with the crossing frequency $\nu$ , the factor $\delta$ appears once in $q_{\text{rxn coord}}$ and once (inversely) in $\nu$ , and cancels exactly. What survives after collecting constants is

\nu \times \delta = \frac{2 k_BT}{h}\times\frac{h}{2} \; \longrightarrow \; \text{rate} = \frac{k_BT}{h}\,[AB^{\ddagger}]^{*}

where $[AB^{\ddagger}]^{*}$ denotes the concentration of complexes with the special mode excluded from its equilibrium constant $K^{\ddagger}$ (a careful bookkeeping step standard derivations carry out via partition functions; the clean surviving result is universal).

Step 5 — assemble the rate constant. Since rate $= k[A][B]$ and $[AB^{\ddagger}]^{*} = K^{\ddagger}[A][B]$ ,

k = \frac{k_BT}{h}K^{\ddagger}.

Step 6 — insert thermodynamics. Using $K^{\ddagger} = \exp(-\Delta^{\ddagger}G^{\circ}/RT)$ and $\Delta^{\ddagger}G^{\circ} = \Delta^{\ddagger}H^{\circ} - T\Delta^{\ddagger}S^{\circ}$ , and allowing for imperfect one-way crossing via $\kappa$ , gives the general working equation:

\boxed{\,k = \kappa\,\frac{k_BT}{h}\exp\!\left(\frac{\Delta^{\ddagger}S^{\circ}}{R}\right)\exp\!\left(-\frac{\Delta^{\ddagger}H^{\circ}}{RT}\right)\,}

Worked examples

Example 1 — Foundational: extracting activation parameters from k(T). The base hydrolysis of an ester obeys pseudo-first-order kinetics with rate constants $k = 1.10\times10^{-2}\ \text{s}^{-1}$ at $T_1 = 298\ \text{K}$ and $k = 4.15\times10^{-2}\ \text{s}^{-1}$ at $T_2 = 318\ \text{K}$ . Find $\Delta^{\ddagger}H^{\circ}$ , $\Delta^{\ddagger}S^{\circ}$ , and $\Delta^{\ddagger}G^{\circ}$ at 298 K. Step 1: Use the linearized Eyring equation,

\ln\left(\frac{k}{T}\right) = -\frac{\Delta^{\ddagger}H^{\circ}}{R}\cdot\frac{1}{T} + \left[\ln\frac{k_B}{h} + \frac{\Delta^{\ddagger}S^{\circ}}{R}\right].

Compute $k/T$ at each temperature:

k_1/T_1 = 1.10\times10^{-2}/298 = 3.691\times10^{-5}\ \text{s}^{-1}\text{K}^{-1}, \quad \ln(k_1/T_1) = -10.207

k_2/T_2 = 4.15\times10^{-2}/318 = 1.305\times10^{-4}\ \text{s}^{-1}\text{K}^{-1}, \quad \ln(k_2/T_2) = -8.944

Step 2: Slope between the two points:

\text{slope} = \frac{\ln(k_2/T_2) - \ln(k_1/T_1)}{1/T_2 - 1/T_1} = \frac{-8.944-(-10.207)}{(1/318)-(1/298)}

= \frac{1.263}{3.145\times10^{-3} - 3.356\times10^{-3}} = \frac{1.263}{-2.11\times10^{-4}} = -5986\ \text{K}

Step 3: Solve for $\Delta^{\ddagger}H^{\circ}$ :

\Delta^{\ddagger}H^{\circ} = -\text{slope}\times R = 5986\ \text{K} \times 8.314\ \text{J mol}^{-1}\text{K}^{-1} = 4.977\times10^{4}\ \text{J mol}^{-1} \approx 49.8\ \text{kJ mol}^{-1}

Step 4: Find the intercept using the 298 K point:

\text{intercept} = \ln(k_1/T_1) + \frac{\Delta^{\ddagger}H^{\circ}}{R T_1} = -10.207 + \frac{49770}{8.314\times298} = -10.207 + 20.086 = 9.879

Step 5: Solve for $\Delta^{\ddagger}S^{\circ}$ , using $k_B/h = 1.381\times10^{-23}/6.626\times10^{-34} = 2.084\times10^{10}\ \text{s}^{-1}\text{K}^{-1}$ , so $\ln(k_B/h) = 23.760$ :

\Delta^{\ddagger}S^{\circ} = R\,[\text{intercept} - \ln(k_B/h)] = 8.314\times(9.879 - 23.760) = 8.314\times(-13.881)

\Delta^{\ddagger}S^{\circ} = -115.4\ \text{J mol}^{-1}\text{K}^{-1}

Step 6: Combine for $\Delta^{\ddagger}G^{\circ}$ at 298 K:

\Delta^{\ddagger}G^{\circ} = \Delta^{\ddagger}H^{\circ} - T\Delta^{\ddagger}S^{\circ} = 49770 - 298\times(-115.4) = 49770 + 34400 = 84170\ \text{J mol}^{-1} \approx 84.2\ \text{kJ mol}^{-1}

Sanity check: plugging $\Delta^{\ddagger}G^{\circ}=84.2$ kJ/mol back through $k = (k_BT/h)\exp(-\Delta^{\ddagger}G^{\circ}/RT)$ at 298 K gives $\exp(-84200/2478) = \exp(-33.98) = 1.93\times10^{-15}$ , times $k_BT/h = 6.21\times10^{12}\ \text{s}^{-1}$ , gives $k \approx 1.20\times10^{-2}\ \text{s}^{-1}$ — matching the given $1.10\times10^{-2}\ \text{s}^{-1}$ within rounding from a two-point slope. The strongly negative $\Delta^{\ddagger}S^{\circ}$ is consistent with an associative (bimolecular, ordering) transition state, as expected for base hydrolysis.

Example 2 — Applied: predicting a rate constant and comparing to Arrhenius. A gas-phase bimolecular reaction has $\Delta^{\ddagger}H^{\circ} = 58.0\ \text{kJ mol}^{-1}$ and $\Delta^{\ddagger}S^{\circ} = -80.0\ \text{J mol}^{-1}\text{K}^{-1}$ at 500 K. Predict $k$ , then find the corresponding Arrhenius $A$ and $E_a$ at this temperature. Step 1: Compute $\Delta^{\ddagger}G^{\circ}$ at 500 K:

\Delta^{\ddagger}G^{\circ} = 58000 - 500\times(-80.0) = 58000 + 40000 = 98000\ \text{J mol}^{-1} = 98.0\ \text{kJ mol}^{-1}

Step 2: Compute $k_BT/h$ at 500 K:

\frac{k_BT}{h} = \frac{(1.381\times10^{-23})(500)}{6.626\times10^{-34}} = 1.042\times10^{13}\ \text{s}^{-1}

Step 3: Compute the Boltzmann factor:

\exp\left(-\frac{98000}{8.314\times500}\right) = \exp(-23.58) = 5.75\times10^{-11}

Step 4: Multiply for $k$ (units of a second-order rate constant since this is bimolecular, dm³ mol⁻¹ s⁻¹, with the appropriate standard-state correction absorbed into $K^{\ddagger}$ ):

k = (1.042\times10^{13}\ \text{s}^{-1})(5.75\times10^{-11}) = 599\ \text{dm}^3\text{mol}^{-1}\text{s}^{-1}

Step 5: Relate to Arrhenius. For a gas-phase bimolecular step,

E_a = \Delta^{\ddagger}H^{\circ} + 2RT = 58000 + 2(8.314)(500) = 58000 + 8314 = 66300\ \text{J mol}^{-1} \approx 66.3\ \text{kJ mol}^{-1}.

Step 6: Back out $A$ from $k = A\exp(-E_a/RT)$ :

A = k\,\exp\left(\frac{E_a}{RT}\right) = 599\times\exp\left(\frac{66300}{8.314\times500}\right) = 599\times\exp(15.95)

= 599 \times 8.44\times10^{6} = 5.06\times10^{9}\ \text{dm}^3\text{mol}^{-1}\text{s}^{-1}

Sanity check: this $A$ is about two orders of magnitude below the crude collision-theory estimate of $\sim10^{11}$ – $10^{12}$ dm³mol⁻¹s⁻¹, consistent with a modestly negative $\Delta^{\ddagger}S^{\circ}$ (a mild steric/orientational requirement) rather than an extremely restrictive one.

Limiting cases & common traps

Situation	What happens	Common trap
$\Delta^{\ddagger}S^{\circ} \approx 0$	$A$ matches the simple collision-theory pre-factor	Assuming Eyring and collision theory always disagree — they coincide for simple, structureless reactants
$\Delta^{\ddagger}S^{\circ} \ll 0$	Highly ordered, associative transition state (e.g., $S_N2$ , Diels–Alder)	Forgetting sign: very negative $\Delta^{\ddagger}S^{\circ}$ lowers $k$ , mimicking a "low steric factor"
$\Delta^{\ddagger}S^{\circ} > 0$	Dissociative or loosening transition state; can exceed simple collision estimates	Assuming $A >$ collision frequency is unphysical — it is not, if bonds are loosening
$\kappa \ll 1$	Frequent barrier recrossing (common in solution, especially with friction/solvent caging)	Treating $\kappa = 1$ as guaranteed; it is an assumption, not a law
Gas-phase vs. solution $E_a$ – $\Delta^{\ddagger}H$ link	Differ by $RT$ (unimolecular) vs. $2RT$ (bimolecular gas-phase)	Using the same conversion factor regardless of molecularity/phase

The chemist behind it

Henry Eyring developed transition-state theory in 1935, working independently but almost simultaneously with Meredith Gwynne Evans and Michael Polanyi, who derived an equivalent formulation using potential-energy surfaces. Eyring's genius move was statistical-mechanical: by treating the activated complex as an ordinary molecule missing one vibrational mode, he could import the entire machinery of equilibrium statistical thermodynamics into kinetics, turning "why is this reaction slow" into a question about $\Delta^{\ddagger}S^{\circ}$ and $\Delta^{\ddagger}H^{\circ}$ that chemists could interpret structurally.

Origins · 1 minHenry Eyring and how this was discovered

Henry Eyring

1901–1981 · Mexican-American chemist who derived the eponymous equation and pioneered "absolute rate theory," turning reaction rates into a thermodynamic problem

Ask the tutor: How does a volume of activation, measured from pressure-dependent rate constants, reveal whether a transition state is more compact or more expanded than the reactants?

How is transition-state theory different from collision theory

Collision theory of reaction rates estimates rates from collision frequency, an energy threshold, and an empirical steric factor $P$ describing orientation requirements. Transition-state theory replaces $P$ with a calculable (in principle) entropy of activation, derived from the structure and vibrational modes of the activated complex. TST also naturally explains reactions where $A$ exceeds simple collision estimates, something collision theory cannot rationalize since $P > 1$ makes no physical sense there.

Why is the transmission coefficient usually close to 1

For many gas-phase and simple solution reactions, once a system crosses the saddle point with forward momentum along the reaction coordinate, solvent friction or subsequent collisions are unlikely to reflect it back before products relax away. But in highly viscous solvents, or reactions with very flat barrier tops, recrossing is common and $\kappa$ can drop well below 1, requiring corrections from Kramers-type theories beyond basic TST.

Can Delta-double-dagger S be measured directly

Not directly as a single experiment, but it is extracted from the temperature dependence of $k$ using the linearized Eyring plot of $\ln(k/T)$ against $1/T$ , exactly as in Example 1. The slope gives $\Delta^{\ddagger}H^{\circ}$ and the intercept gives $\Delta^{\ddagger}S^{\circ}$ , once $\ln(k_B/h)$ is subtracted, mirroring how van't Hoff plots extract $\Delta H^{\circ}$ and $\Delta S^{\circ}$ for true equilibria.

Does transition-state theory work for reactions in solution

Yes, with care. The quasi-equilibrium assumption and the $k_BT/h$ factor still apply, but $\Delta^{\ddagger}H^{\circ}$ and $\Delta^{\ddagger}S^{\circ}$ now include solvent reorganization and solvation changes around the activated complex, not just the solute's internal degrees of freedom. This is precisely why solvent polarity and ionic strength shift measured activation parameters, a topic explored further in the thermodynamics of activation.

How does this connect to the thermodynamics of activation

Once you accept $k = (k_BT/h)K^{\ddagger}$ , every equilibrium thermodynamic relation becomes a kinetic tool: pressure dependence gives a volume of activation, solvent studies give entropy and enthalpy of activation trends, and isotope or substituent effects can be mapped onto how they perturb $\Delta^{\ddagger}G^{\circ}$ . That systematic extension is the subject of the thermodynamics of activation.