Transition-State Theory

Introduction

The Arrhenius equation describes how rate constants vary with temperature, but its pre-exponential factor is only an empirical number. Transition-state theory, developed by Eyring, Evans and Polanyi in the 1930s, offers a physical interpretation. It treats the top of the energy barrier as a short-lived species, the activated complex, in quasi-equilibrium with reactants. The result expresses rate constants through familiar thermodynamic quantities: a Gibbs energy, an enthalpy and an entropy of activation.

Core explanation

The transition state. On a potential energy surface, reactants and products occupy valleys separated by a pass. The saddle point at the top of the lowest pass is the transition state, marked ‡. It is not an intermediate that can be isolated; it has a lifetime of roughly one molecular vibration, around 10⁻¹³ s.

Key assumptions. 1. Reactants are in quasi-equilibrium with activated complexes, with an equilibrium constant K‡. 2. Every complex that crosses the saddle point towards products goes on to form products; none recross. 3. Motion along the reaction coordinate can be treated separately from other molecular motions.

The Eyring equation. Under these assumptions,

k = (k BT/h) K‡ = (k BT/h) exp(−ΔG‡/RT)

where k B is the Boltzmann constant and h the Planck constant. The factor k BT/h, about 6.2 × 10¹² s⁻¹ at 298 K, is a universal frequency for crossing the barrier. For a unimolecular reaction k has units of s⁻¹; for bimolecular reactions a standard concentration is included.

Splitting ΔG‡. Since ΔG‡ = ΔH‡ − TΔS‡,

k = (k BT/h) exp(ΔS‡/R) exp(−ΔH‡/RT)

Enthalpy of activation, ΔH‡, reflects bonds being stretched or broken in reaching the transition state. For a solution reaction it is related to the Arrhenius activation energy by Eₐ ≈ ΔH‡ + RT.

Entropy of activation, ΔS‡, reflects the change in freedom of motion. A negative ΔS‡ indicates a more ordered transition state than reactants: two molecules combining into one complex, a cyclic arrangement, or solvent molecules becoming organised around developing charges. A positive ΔS‡ suggests loosening, such as a bond breaking to release fragments. ΔS‡ is thus a diagnostic tool for mechanism: bimolecular associative steps typically show ΔS‡ of −50 to −150 J K⁻¹ mol⁻¹, while dissociative steps are often near zero or positive.

Eyring plot. Rearranging gives

ln(k/T) = ln(k B/h) + ΔS‡/R − ΔH‡/(RT)

so ln(k/T) against 1/T has gradient −ΔH‡/R and an intercept from which ΔS‡ is obtained.

Formulae

k = (k BT/h) exp(−ΔG‡/RT); ΔG‡ = ΔH‡ − TΔS‡; Eₐ ≈ ΔH‡ + RT (solution). k B = 1.381 × 10⁻²³ J K⁻¹; h = 6.626 × 10⁻³⁴ J s.

Step-by-step reasoning

1. Measure k at several temperatures. 2. Compute ln(k/T) and 1/T for each point. 3. Fit a straight line; ΔH‡ = −R × gradient. 4. From the intercept, subtract ln(k B/h) = 23.76 and multiply by R to get ΔS‡. 5. Combine to give ΔG‡ at a chosen temperature and interpret the signs.

Visual explanation

Draw a reaction profile: a reactant plateau, a peak labelled ‡, and a product plateau. Label the peak height ΔG‡. Imagine a contour map beneath it in which the route passes over a mountain saddle — lowest along the path, highest across it.

Real-world analogy

Crossing a mountain range, travellers choose the lowest pass. The height of the pass is the enthalpy barrier; the width of the pass is the entropy term. A low but very narrow pass may let fewer travellers through per hour than a slightly higher, broad one.

Real-world example

In organic chemistry, Diels–Alder cycloadditions have strongly negative entropies of activation, around −140 J K⁻¹ mol⁻¹, because two molecules join into a single ordered cyclic transition state. Chemists use such values as evidence for a concerted associative mechanism rather than a stepwise one.

Why?

Why does entropy influence rate? The rate depends on how many ways the system can reach the transition state. If reaching it demands a precise alignment, only a tiny fraction of energetic encounters succeed, which appears as a negative ΔS‡ and a smaller rate constant.

Common misconception

"The transition state is a reaction intermediate." An intermediate sits in an energy minimum and has a measurable lifetime; a transition state sits at an energy maximum along the reaction coordinate and cannot be isolated.

Worked example

Question: A unimolecular reaction has ΔG‡ = 80 kJ mol⁻¹ at 298 K. Estimate k.

Reasoning: k BT/h = 6.21 × 10¹² s⁻¹. ΔG‡/RT = 80 000 ÷ (8.314 × 298) = 32.3. exp(−32.3) ≈ 9.4 × 10⁻¹⁵. k ≈ 6.21 × 10¹² × 9.4 × 10⁻¹⁵.

Answer: k ≈ 5.8 × 10⁻² s⁻¹, a half-life of about 12 s.

Quick check

1. A reaction has a strongly negative entropy of activation. What does this suggest about its transition state? Answer: The transition state is more ordered than the reactants, as in an associative or cyclic step.

Exam focus

Know the Eyring equation, the link ΔG‡ = ΔH‡ − TΔS‡ and the axes of an Eyring plot. Be able to interpret the sign of ΔS‡ in mechanistic terms. Note the useful rule that each 5.7 kJ mol⁻¹ increase in ΔG‡ lowers k tenfold at 298 K.

Advanced insight

Real reactions may recross the barrier, so a transmission coefficient κ, usually slightly below 1, multiplies the Eyring rate. For light-atom transfers, especially hydrogen, quantum tunnelling through the barrier can make κ greater than 1 and produce large kinetic isotope effects, curved Eyring plots and rates exceeding classical predictions.

Summary

Transition-state theory treats the activated complex at the saddle point as in quasi-equilibrium with reactants, giving the Eyring equation k = (k BT/h) exp(−ΔG‡/RT). Splitting ΔG‡ into ΔH‡ and ΔS‡ reveals the energetic and organisational costs of reaching the transition state, and an Eyring plot of ln(k/T) against 1/T measures both.

Practice questions

1. What is the gradient of an Eyring plot? Answer: −ΔH‡/R, obtained from a plot of ln(k/T) against 1/T. 2. Give one structural reason for a positive entropy of activation. Answer: A bond breaks in the transition state, releasing fragments with more freedom of motion, or ordered solvent is released. 3. Two reactions have the same ΔH‡, but reaction P has ΔS‡ = −100 J K⁻¹ mol⁻¹ and reaction Q has ΔS‡ = 0. Which is faster? Answer: Q, because a less negative ΔS‡ gives a smaller ΔG‡ and hence a larger rate constant. 4. Relate Eₐ to ΔH‡ for a reaction in solution at 300 K if ΔH‡ = 60.0 kJ mol⁻¹. Answer: Eₐ ≈ ΔH‡ + RT = 60.0 + 2.5 = 62.5 kJ mol⁻¹.