Thermodynamics of Activation

Why does adding a bulky substituent slow a reaction more than temperature alone can explain? The Arrhenius equation gives an activation energy and a prefactor, but it cannot tell you whether the slowdown is enthalpic or entropic. Transition-state theory answers that question directly by attaching thermodynamic state functions to the activated complex. This matters because real reactions are shaped by molecular organization, not just bond energies. A sluggish bimolecular reaction between two flexible molecules often suffers from a large negative entropy of activation, not a large enthalpy barrier. Distinguishing the two lets chemists diagnose mechanism from kinetic data alone. This article builds the thermodynamic formulation of transition-state theory from the Eyring equation. You will derive $\Delta^{\ddagger}G$ , $\Delta^{\ddagger}H$ , and $\Delta^{\ddagger}S$ . Then you will connect them rigorously to the Arrhenius $E_a$ and $A$ . Two worked examples show how to extract these quantities from real rate data.

Key takeaways

The Eyring equation writes the rate constant as $k = (k_BT/h)K^{\ddagger}$ , treating the transition state as a quasi-equilibrium species.
$K^{\ddagger}$ is recast as a Gibbs energy of activation, $\Delta^{\ddagger}G = \Delta^{\ddagger}H - T\Delta^{\ddagger}S$ .
$\Delta^{\ddagger}H$ closely tracks the Arrhenius $E_a$ for reactions in solution; for gas-phase unimolecular or bimolecular reactions the relation carries an extra $RT$ or $2RT$ term.
$\Delta^{\ddagger}S$ measures the loss (or gain) of order in forming the transition state and controls the size of the Arrhenius pre-exponential factor $A$ .
An Eyring plot of $\ln(k/T)$ against $1/T$ gives $\Delta^{\ddagger}H$ from the slope and $\Delta^{\ddagger}S$ from the intercept.
Key idea: splitting the activation barrier into enthalpy and entropy turns a single empirical number, $E_a$ , into a diagnosis of why a reaction is fast or slow.

The core insight

Because the activated complex is treated as being in quasi-equilibrium with reactants, its formation obeys ordinary equilibrium thermodynamics, so the empirical Arrhenius barrier can be decomposed into an enthalpy of activation (bond distortion, electrostatic reorganization) and an entropy of activation (loss of translational, rotational, and conformational freedom), each carrying distinct mechanistic information that $E_a$ alone cannot provide.

The physical picture

Transition-state (Eyring) theory pictures reactants climbing a free-energy surface toward a saddle point, the activated complex $\ddagger$ . That species is assumed to be in a quasi-equilibrium with reactants, characterized by an equilibrium constant $K^{\ddagger}$ . Because $K^{\ddagger}$ is an equilibrium constant, it can be written in terms of a standard Gibbs energy change, exactly as for any chemical equilibrium. This is the conceptual leap beyond the plain Arrhenius picture. Arrhenius theory treats $E_a$ as a single empirical energy barrier with no internal structure. The thermodynamic formulation of Eyring theory instead asks: how much of the barrier is enthalpic (bond stretching, charge separation, solvent reorganization) and how much is entropic (loss of rotational and translational freedom, ordering of solvent molecules)? A reaction can have a modest enthalpic barrier but still be slow because the transition state is highly ordered, giving a very negative $\Delta^{\ddagger}S$ . The sign and magnitude of $\Delta^{\ddagger}S$ is itself mechanistic evidence. Bimolecular reactions that must bring two separate molecules into a single ordered complex typically show substantially negative $\Delta^{\ddagger}S$ , often $-80$ to $-150\ \mathrm{J\,K^{-1}\,mol^{-1}}$ in solution. Unimolecular ring-opening or dissociation reactions, where the transition state is looser than the reactant, can show small positive $\Delta^{\ddagger}S$ .

Figure

Reaction profile in enthalpy–entropy language: reactants, transition state, and products

The second figure shows how the two thermodynamic contributions separately govern the pre-exponential factor and the exponential term once the Eyring equation is written out.

Figure

Eyring plot: ln(k/T) versus 1/T separates ΔH‡ and ΔS‡

The core relations

k = \frac{k_BT}{h}K^{\ddagger} = \frac{k_BT}{h}\exp\!\left(-\frac{\Delta^{\ddagger}G^{\circ}}{RT}\right)

where $k_B$ is Boltzmann's constant, $h$ Planck's constant, and $\Delta^{\ddagger}G^{\circ}$ the standard Gibbs energy of activation.

\Delta^{\ddagger}G^{\circ} = \Delta^{\ddagger}H^{\circ} - T\Delta^{\ddagger}S^{\circ}

where $\Delta^{\ddagger}H^{\circ}$ and $\Delta^{\ddagger}S^{\circ}$ are the standard 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 the first exponential sets the effective pre-exponential factor and the second is the Boltzmann-like enthalpy term.

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

where plotting $\ln(k/T)$ against $1/T$ (the Eyring plot) gives slope $-\Delta^{\ddagger}H^{\circ}/R$ and the bracketed intercept.

E_a = \Delta^{\ddagger}H^{\circ} + RT \quad(\text{unimolecular, or any reaction in solution treated as first order in the TS step})

where $E_a$ is the Arrhenius activation energy from $k = A e^{-E_a/RT}$ .

A = \frac{k_BT}{h}\,e\,\exp\!\left(\frac{\Delta^{\ddagger}S^{\circ}}{R}\right)

where $e = 2.71828\ldots$ arises from the $RT$ correction between $E_a$ and $\Delta^{\ddagger}H^{\circ}$ .

Derivation AFrom the Eyring equation to Δ‡H° and Δ‡S°, and their link to Ea and A

Goal. Starting from the Eyring rate law, obtain the Eyring-plot form and derive the exact relations between $\Delta^{\ddagger}H^{\circ}$ , $\Delta^{\ddagger}S^{\circ}$ , and the Arrhenius parameters $E_a$ , $A$ .

Step 1 — Write the Eyring rate constant. Transition-state theory (assumed known) gives

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

with transmission coefficient $\kappa \approx 1$ . Treat $K^{\ddagger}$ as a true equilibrium constant between reactants and the activated complex.

Step 2 — Introduce activation thermodynamics. Since $K^{\ddagger}$ is an equilibrium constant,

-RT\ln K^{\ddagger} = \Delta^{\ddagger}G^{\circ} = \Delta^{\ddagger}H^{\circ} - T\Delta^{\ddagger}S^{\circ}.

Substituting,

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

Step 3 — Linearize for the Eyring plot. Divide by $T$ and take logarithms:

\ln\frac{k}{T} = \ln\frac{k_B}{h} + \frac{\Delta^{\ddagger}S^{\circ}}{R} - \frac{\Delta^{\ddagger}H^{\circ}}{R}\cdot\frac{1}{T}.

A plot of $\ln(k/T)$ against $1/T$ is linear (assuming $\Delta^{\ddagger}H^{\circ}$ and $\Delta^{\ddagger}S^{\circ}$ are roughly $T$ -independent over the range studied), with slope $-\Delta^{\ddagger}H^{\circ}/R$ and intercept $\ln(k_B/h) + \Delta^{\ddagger}S^{\circ}/R$ .

Step 4 — Compare with the Arrhenius definition. Arrhenius defines $E_a$ through

E_a \equiv RT^2\frac{d\ln k}{dT}.

From Step 2, $\ln k = \ln(k_B/h) + \ln T + \Delta^{\ddagger}S^{\circ}/R - \Delta^{\ddagger}H^{\circ}/(RT)$ . Differentiate term by term:

\frac{d\ln k}{dT} = \frac{1}{T} + \frac{\Delta^{\ddagger}H^{\circ}}{RT^2} + \frac{1}{R}\frac{d\Delta^{\ddagger}H^{\circ}}{dT}\cdot\frac{1}{T}

Neglecting the (usually small) temperature dependence of $\Delta^{\ddagger}H^{\circ}$ itself (constant-heat-capacity-of-activation approximation, $\Delta^{\ddagger}C_p \approx 0$ ), this reduces to

\frac{d\ln k}{dT} \approx \frac{1}{T} + \frac{\Delta^{\ddagger}H^{\circ}}{RT^2}.

Multiplying by $RT^2$ ,

E_a = RT + \Delta^{\ddagger}H^{\circ}.

Step 5 — Recover A. Since $k = Ae^{-E_a/RT}$ and $k = (k_BT/h)e^{\Delta^{\ddagger}S^{\circ}/R}e^{-\Delta^{\ddagger}H^{\circ}/RT}$ , substitute $\Delta^{\ddagger}H^{\circ} = E_a - RT$ :

k = \frac{k_BT}{h}e^{\Delta^{\ddagger}S^{\circ}/R}\,e\,e^{-E_a/RT}

so that, matching to the Arrhenius form,

\boxed{A = e\,\frac{k_BT}{h}\exp\!\left(\frac{\Delta^{\ddagger}S^{\circ}}{R}\right), \qquad E_a = \Delta^{\ddagger}H^{\circ} + RT}

Caveat. For a bimolecular gas-phase reaction, the standard-state conversion between concentration and pressure-based equilibrium constants introduces an extra factor of $RT$ in $\Delta^{\ddagger}H^{\circ}$ versus $E_a$ , giving $E_a = \Delta^{\ddagger}H^{\circ} + 2RT$ instead. The single- $RT$ result above applies to unimolecular reactions and to solution-phase reactions analyzed with concentration standard states, which is the common textbook case.

Worked examples

Example 1 — Foundational: extracting Δ‡H°, Δ‡S°, Δ‡G° from a single rate constant. The base hydrolysis of a certain ester is first order overall, with rate constant $k = 3.15\times10^{-3}\ \mathrm{s^{-1}}$ at $T = 298\ \mathrm{K}$ , and independently measured Arrhenius activation energy $E_a = 55.0\ \mathrm{kJ\,mol^{-1}}$ . Step 1: Get $\Delta^{\ddagger}H^{\circ}$ . Treating this as a solution-phase, effectively unimolecular activation step,

\Delta^{\ddagger}H^{\circ} = E_a - RT = 55{,}000\ \mathrm{J\,mol^{-1}} - (8.314\ \mathrm{J\,K^{-1}mol^{-1}})(298\ \mathrm{K})

= 55{,}000 - 2478 = 52{,}522\ \mathrm{J\,mol^{-1}} = 52.5\ \mathrm{kJ\,mol^{-1}}.

Step 2: Get $\Delta^{\ddagger}G^{\circ}$ directly from $k$ . Rearranging the Eyring equation,

\Delta^{\ddagger}G^{\circ} = -RT\ln\!\left(\frac{kh}{k_BT}\right).

Compute the dimensionless ratio:

\frac{kh}{k_BT} = \frac{(3.15\times10^{-3}\ \mathrm{s^{-1}})(6.626\times10^{-34}\ \mathrm{J\,s})}{(1.381\times10^{-23}\ \mathrm{J\,K^{-1}})(298\ \mathrm{K})}

= \frac{2.087\times10^{-36}}{4.116\times10^{-21}} = 5.07\times10^{-16}.

Then

\Delta^{\ddagger}G^{\circ} = -(8.314)(298)\ln(5.07\times10^{-16}) = -(2478)(-35.22) = 87{,}300\ \mathrm{J\,mol^{-1}} = 87.3\ \mathrm{kJ\,mol^{-1}}.

Step 3: Get $\Delta^{\ddagger}S^{\circ}$ . From $\Delta^{\ddagger}G^{\circ} = \Delta^{\ddagger}H^{\circ} - T\Delta^{\ddagger}S^{\circ}$ ,

\Delta^{\ddagger}S^{\circ} = \frac{\Delta^{\ddagger}H^{\circ} - \Delta^{\ddagger}G^{\circ}}{T} = \frac{52{,}522 - 87{,}300}{298} = \frac{-34{,}778}{298} = -116.7\ \mathrm{J\,K^{-1}mol^{-1}}.

Sanity check. A strongly negative $\Delta^{\ddagger}S^{\circ}$ of about $-117\ \mathrm{J\,K^{-1}mol^{-1}}$ is entirely consistent with hydroxide attacking an ester carbonyl: two separate species (ester + OH⁻) must come together and freeze out relative translation and rotation to form a single ordered, partially bonded transition state. The large positive $\Delta^{\ddagger}G^{\circ}$ (87.3 kJ/mol) versus the smaller $\Delta^{\ddagger}H^{\circ}$ (52.5 kJ/mol) shows the reaction is slowed mainly by this entropic penalty, not by enthalpy.

Example 2 — Applied: comparing two mechanisms from Eyring-plot data. A kineticist measures rate constants for a unimolecular isomerization at two temperatures: $k_1 = 4.0\times10^{-4}\ \mathrm{s^{-1}}$ at $T_1 = 300\ \mathrm{K}$ , and $k_2 = 3.6\times10^{-3}\ \mathrm{s^{-1}}$ at $T_2 = 320\ \mathrm{K}$ . Find $\Delta^{\ddagger}H^{\circ}$ , $\Delta^{\ddagger}S^{\circ}$ , and $E_a$ , and decide whether the transition state is more ordered or looser than the reactant. Step 1: Two-point Eyring slope. Using

\ln\frac{k_2/T_2}{k_1/T_1} = -\frac{\Delta^{\ddagger}H^{\circ}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)

compute the left side. $k_1/T_1 = 4.0\times10^{-4}/300 = 1.333\times10^{-6}\ \mathrm{K^{-1}s^{-1}}$ ; $k_2/T_2 = 3.6\times10^{-3}/320 = 1.125\times10^{-5}\ \mathrm{K^{-1}s^{-1}}$ .

\ln\frac{1.125\times10^{-5}}{1.333\times10^{-6}} = \ln(8.438) = 2.133.

The temperature term:

\frac{1}{320}-\frac{1}{300} = 3.125\times10^{-3} - 3.333\times10^{-3} = -2.083\times10^{-4}\ \mathrm{K^{-1}}.

2.133 = -\frac{\Delta^{\ddagger}H^{\circ}}{8.314}\times(-2.083\times10^{-4})

\Delta^{\ddagger}H^{\circ} = \frac{2.133\times8.314}{2.083\times10^{-4}} = \frac{17.74}{2.083\times10^{-4}} = 85{,}170\ \mathrm{J\,mol^{-1}} = 85.2\ \mathrm{kJ\,mol^{-1}}.

Step 2: Arrhenius $E_a$ (unimolecular, so single $RT$ correction) at the mean temperature $T \approx 310\ \mathrm{K}$ :

E_a = \Delta^{\ddagger}H^{\circ} + RT = 85{,}170 + (8.314)(310) = 85{,}170 + 2577 = 87{,}750\ \mathrm{J\,mol^{-1}} = 87.7\ \mathrm{kJ\,mol^{-1}}.

Step 3: Get $\Delta^{\ddagger}S^{\circ}$ using the data point at $T_1 = 300\ \mathrm{K}$ . First find $\Delta^{\ddagger}G^{\circ}_1$ :

\frac{k_1h}{k_BT_1} = \frac{(4.0\times10^{-4})(6.626\times10^{-34})}{(1.381\times10^{-23})(300)} = \frac{2.650\times10^{-37}}{4.143\times10^{-21}} = 6.398\times10^{-17}.

\Delta^{\ddagger}G^{\circ}_1 = -(8.314)(300)\ln(6.398\times10^{-17}) = -(2494)(-37.29) = 93{,}000\ \mathrm{J\,mol^{-1}}.

Then

\Delta^{\ddagger}S^{\circ} = \frac{\Delta^{\ddagger}H^{\circ} - \Delta^{\ddagger}G^{\circ}_1}{T_1} = \frac{85{,}170 - 93{,}000}{300} = \frac{-7830}{300} = -26.1\ \mathrm{J\,K^{-1}mol^{-1}}.

Sanity check and interpretation. $\Delta^{\ddagger}S^{\circ} \approx -26\ \mathrm{J\,K^{-1}mol^{-1}}$ is modestly negative, much smaller in magnitude than the $-117\ \mathrm{J\,K^{-1}mol^{-1}}$ found for the bimolecular ester hydrolysis in Example 1. That is exactly the expected pattern: a unimolecular isomerization only has to reorganize internal geometry (e.g., partial bond rotation or ring puckering), not corral a second molecule from solution, so it loses much less entropy on the way to the transition state. The modest negative value suggests a somewhat rigid, ordered transition state (e.g., a cyclic or partially bonded arrangement) rather than a loose, bond-breaking one.

Limiting cases & common traps

Situation	What to watch for
Bimolecular gas-phase reaction	Use $E_a = \Delta^{\ddagger}H^{\circ} + 2RT$ , not $+RT$ , because of the extra concentration/pressure standard-state factor.
Very large negative $\Delta^{\ddagger}S^{\circ}$	Signals a highly ordered, associative transition state (e.g., $S_N2$ , cycloadditions); pre-exponential factor $A$ falls well below the "normal" $10^{13}\ \mathrm{s^{-1}}$ collision estimate.
Positive $\Delta^{\ddagger}S^{\circ}$	Common in unimolecular fragmentations/eliminations where the transition state is looser than the reactant; $A$ exceeds $10^{13}\ \mathrm{s^{-1}}$ .
Nonlinear Eyring plot	Indicates $\Delta^{\ddagger}C_p^{\circ} \neq 0$ or a change in rate-determining step with temperature; a single $\Delta^{\ddagger}H^{\circ}$ is no longer meaningful over the whole range.
Solvent reorganization entropy	In polar solvents, ionic or charge-separating transition states can show large $\Delta^{\ddagger}S^{\circ}$ contributions from ordering solvent dipoles, on top of any intrinsic loss of freedom.

The chemist behind it

Henry Eyring formulated the statistical-thermodynamic treatment of reaction rates in 1935, building on transition-state ideas developed alongside Meredith Gwynne Evans and Michael Polanyi. Eyring's innovation was to treat the activated complex as a genuine thermodynamic species in quasi-equilibrium with reactants, so that all the machinery of classical thermodynamics — Gibbs energy, enthalpy, entropy — could be brought to bear on reaction rates, not just on equilibria. This reframing let chemists connect measured rate constants directly to molecular structure and solvation effects, well before computational methods could estimate barriers from first principles.

Origins · 1 minHenry Eyring and how this was discovered

Henry Eyring

1901–1981 · Formulated transition-state theory and its thermodynamic (activated-complex) formulation in 1935

Ask the tutor: How does adding a catalyst or changing solvent polarity typically shift Δ‡H° versus Δ‡S° for a given reaction?

Is Δ‡G° the same as the Arrhenius activation energy Ea?

No. $\Delta^{\ddagger}G^{\circ}$ is a full Gibbs energy of activation, combining enthalpy and entropy contributions, and it directly sets the rate constant through $k=(k_BT/h)e^{-\Delta^{\ddagger}G^{\circ}/RT}$ . $E_a$ is defined purely from the temperature dependence of $k$ via $E_a = RT^2\,d\ln k/dT$ , and turns out to equal $\Delta^{\ddagger}H^{\circ}$ plus a small correction of one or two $RT$ terms, depending on molecularity. They are related but conceptually distinct quantities.

Why is Δ‡S° often large and negative for bimolecular reactions?

Two separate reactant molecules, each with three translational and (if nonlinear) three rotational degrees of freedom, must merge into a single activated complex with only one such set. That loss of independent translational and rotational freedom is entropically costly, typically contributing $-80$ to $-150\ \mathrm{J\,K^{-1}mol^{-1}}$ . Solvent ordering around a developing charge can add further negative contributions in polar solvents, especially for ionic transition states.

Can Δ‡H° be negative?

Yes, in principle, though it is unusual. A negative $\Delta^{\ddagger}H^{\circ}$ would imply the transition state is enthalpically more stable than the separated reactants, which can occur in some strongly exothermic, barrierless-looking associative processes or diffusion-controlled reactions where weak, favorable interactions dominate the approach to the transition state. In practice, most chemically interesting reactions have positive $\Delta^{\ddagger}H^{\circ}$ reflecting real bond reorganization.

How many data points do I need to get reliable Δ‡H° and Δ‡S° values?

Strictly, an Eyring plot needs several rate constants across a reasonably wide temperature range, at least four to six points, to fit a reliable slope and intercept with regression. A two-point calculation, as shown in Example 2, gives a quick estimate but assumes $\Delta^{\ddagger}H^{\circ}$ and $\Delta^{\ddagger}C_p^{\circ}$ are constant over that interval, and it carries no information about linearity or curvature.

Does this framework require the transition state to be a real, isolable species?

No. The activated complex is a mathematical construct at the saddle point of the free-energy surface, with a lifetime on the order of a bond vibration, roughly $10^{-13}$ seconds. Treating it as being in quasi-equilibrium with reactants is a modeling assumption, justified because most trajectories crossing the saddle point proceed to products without recrossing, not a claim that it can be isolated or observed directly.