Activation Parameters — Eyring vs Arrhenius

By Aniket Bhardwaj · 4 September 2026 · Advanced Chemistry

Both equations describe how a rate constant grows with temperature, and both are fitted from the same experimental data. The difference is what they claim. Arrhenius is an empirical fit with two adjustable parameters; Eyring comes from transition state theory and its parameters are real thermodynamic quantities belonging to a real (if fleeting) species. That is why mechanistic papers report ΔH‡ and ΔS‡ rather than E_a and A — the entropy of activation is a statement about the shape of the transition state.

The two equations

Arrhenius k = A e^−E_a/RT
Linear form: ln k = ln A − (E_a/R)(1/T) → plot ln k vs 1/T

Eyring k = κ (k_BT/h) e^−ΔG‡/RT = κ (k_BT/h) e^ΔS‡/R e^−ΔH‡/RT
Linear form: ln(k/T) = −(ΔH‡/R)(1/T) + ln(k_B/h) + ΔS‡/R → plot ln(k/T) vs 1/T

κ is the transmission coefficient, the fraction of systems that cross the barrier and stay crossed; it is normally taken as 1. The prefactor k_BT/h is a universal frequency — 6.25 × 10¹² s⁻¹ at 300 K — and it is the same for every reaction. All chemical individuality sits in ΔG‡.

Notice the structural difference in the plots. Arrhenius plots ln k; Eyring plots ln(k/T). The slopes therefore differ slightly, and so do the quantities extracted: the Arrhenius slope gives −E_a/R while the Eyring slope gives −ΔH‡/R. The two are related but not equal.

The bridge between them

In solution (any molecularity) and for gas-phase unimolecular reactions:
E_a = ΔH‡ + RT and A = e · (k_BT/h) · e^ΔS‡/R

For a gas-phase bimolecular reaction:
E_a = ΔH‡ + 2RT and A = e² · (k_BT/h) · e^ΔS‡/R

General gas-phase form: E_a = ΔH‡ + (1 − Δn‡)RT, where Δn‡ is the change in the number of molecules on forming the activated complex.

Rearranging the A relation gives the working formula for the entropy of activation:

ΔS‡ = R [ ln( A h / k_BT ) − 1 ] (unimolecular / solution case)

Worked example — one data set, both treatments

Q. A first-order reaction in solution has k = 2.0 × 10⁻⁴ s⁻¹ at 300 K and k = 3.6 × 10⁻³ s⁻¹ at 330 K. Find E_a, A, then ΔH‡, ΔS‡ and ΔG‡ at 300 K.

Step 1 — activation energy from two points.

ln(k₂/k₁) = (E_a/R)(1/T₁ − 1/T₂)
k₂/k₁ = 3.6 × 10⁻³ / 2.0 × 10⁻⁴ = 18.0, so ln(18.0) = 2.8904
1/300 − 1/330 = 0.00333333 − 0.00303030 = 3.0303 × 10⁻⁴ K⁻¹
E_a = R × 2.8904 / 3.0303 × 10⁻⁴ = 8.314 × 2.8904 / 3.0303 × 10⁻⁴
= 24.030 / 3.0303 × 10⁻⁴ = 79 300 J mol⁻¹ = 79.3 kJ mol⁻¹

Step 2 — pre-exponential factor.

ln A = ln k₁ + E_a/(RT₁) = ln(2.0 × 10⁻⁴) + 79 300/(8.314 × 300)
= −8.5172 + 79 300/2494.2 = −8.5172 + 31.793 = 23.276
A = e^23.276 = 1.28 × 10¹⁰ s⁻¹

Step 3 — activation enthalpy. A first-order reaction in solution, so E_a = ΔH‡ + RT:

ΔH‡ = 79.30 − (8.314 × 300)/1000 = 79.30 − 2.49 = 76.81 kJ mol⁻¹

Step 4 — activation entropy. First evaluate the universal prefactor:

k_BT/h = (1.380649 × 10⁻²³ × 300)/(6.62607 × 10⁻³⁴) = 6.251 × 10¹² s⁻¹
A/(k_BT/h) = 1.28 × 10¹⁰ / 6.251 × 10¹² = 2.054 × 10⁻³
ΔS‡ = R[ln(2.054 × 10⁻³) − 1] = 8.314 × (−6.188 − 1) = 8.314 × (−7.188)
= −59.8 J K⁻¹ mol⁻¹

Step 5 — free energy of activation at 300 K.

ΔG‡ = ΔH‡ − TΔS‡ = 76.81 − 300 × (−0.0598) = 76.81 + 17.93 = 94.74 kJ mol⁻¹

Check. Substituting back into the Eyring equation must return the original rate constant:
k = 6.251 × 10¹² × e^{−94 740/2494.2} = 6.251 × 10¹² × e^−37.98 = 6.251 × 10¹² × 3.19 × 10⁻¹⁷ = 2.0 × 10⁻⁴ s⁻¹ ✓

That closing check is worth doing in an exam whenever time allows — it catches a dropped factor of 1000 or a sign error immediately.

What the sign of ΔS‡ tells you

This is the payoff, and the reason the extra algebra is worth doing. E_a tells you how high the barrier is; ΔS‡ tells you what the top of the barrier looks like.

ΔS‡	Transition state	Typical mechanisms
Strongly negative (−50 to −200 J K⁻¹ mol⁻¹)	More ordered than reactants — two species bound together, rotations frozen	S_N2, Diels–Alder and other cycloadditions, associative ligand substitution, bimolecular addition
Near zero (−20 to +20)	Little change in order; often a simple bond stretch	Isomerisations, some radical reactions, interchange mechanisms
Positive (+10 to +60)	Looser than reactants — a bond breaking, fragments gaining freedom	S_N1 ionisation, dissociative ligand substitution, unimolecular fragmentation, retro-cycloadditions

Our worked example gave ΔS‡ = −59.8 J K⁻¹ mol⁻¹, which — despite first-order kinetics — points to a highly ordered transition state, for instance one in which solvent molecules are strongly organised around a developing charge. A solvolysis showing first-order kinetics with a positive ΔS‡ would instead support clean S_N1 ionisation. Kinetics alone cannot distinguish these; activation parameters can. That is the whole reason mechanistic chemistry bothers with Eyring analysis.

Common mistakes

Using E_a = ΔH‡ + RT for a gas-phase bimolecular reaction. That case needs 2RT. Decide the phase and molecularity before converting.
Ignoring the units of A when computing ΔS‡. For a second-order reaction A carries concentration units, so ΔS‡ depends on the standard state you choose (mol dm⁻³ versus mol m⁻³ shifts it by R ln 1000 ≈ 57 J K⁻¹ mol⁻¹). Always state the standard state.
Plotting ln k when you meant Eyring. Eyring requires ln(k/T). Using the Arrhenius plot and calling the slope ΔH‡/R introduces an error of about RT.
Treating E_a as temperature-independent over a wide range. Both equations assume it; over 100 K or more, curvature in the plot is real physics (temperature-dependent ΔC_p‡), not scatter.
Reading ΔG‡ as a reaction ΔG. ΔG‡ refers to the activated complex, is always positive, and says nothing about whether the reaction is thermodynamically favourable.
Two-point fits presented as precise. A 30 K interval with 5 % error in k already puts several kJ mol⁻¹ of uncertainty on E_a. Use five or more temperatures when the number matters.

Summary

	Arrhenius	Eyring
Origin	Empirical	Transition state theory
Plot	ln k vs 1/T	ln(k/T) vs 1/T
From the slope	−E_a/R	−ΔH‡/R
From the intercept	ln A	ln(k_B/h) + ΔS‡/R
Mechanistic content	Barrier height only	Barrier height and ordering of the transition state

Fit and check your rate data in seconds. The Arrhenius calculator takes two rate constants and two temperatures and returns E_a and A directly — the starting point for every Eyring conversion above.

Open the Arrhenius Equation Calculator →

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