Questions: Transition State Theory and the Eyring Equation

Question 1

A bimolecular reaction is observed to have a large negative ΔS‡. Which interpretation is most consistent with transition state theory?

• A. The transition state is more disordered than the reactants, releasing entropy.
• B. The two reactants must adopt a precise, constrained geometry to reach the transition state.
• C. The activation enthalpy ΔH‡ is the dominant contribution to the reaction rate.
• D. Quantum tunneling through the barrier is significant, so κ ≪ 1.

Answer: (B)

Entropy measures the number of accessible microstates. A large negative ΔS‡ means the transition state has far fewer accessible configurations than the reactants — both molecules must approach with the correct orientation and geometry. This entropic penalty slows the rate beyond what barrier height alone would predict. It is a hallmark of bimolecular reactions that demand tight steric and electronic alignment.

Question 2

The Arrhenius activation energy E_a and the TST activation free energy ΔG‡ are equivalent quantities that can be used interchangeably.

• T. True
• F. False

Answer: False

They are related but distinct. E_a is an empirical parameter extracted from the temperature dependence of ln k; it approximates the enthalpic barrier. ΔG‡ is the free energy difference between the transition state and reactants: ΔG‡ = ΔH‡ − TΔS‡. Even for a reaction with negligible entropy of activation, E_a ≈ ΔH‡ + RT, not ΔH‡ exactly. When ΔS‡ ≠ 0, the two quantities diverge further. Using E_a where ΔG‡ is required would ignore the entropic contribution to the rate.

Question 3

What physical phenomenon does the transmission coefficient κ account for in the Eyring equation, and why does it become especially important for reactions involving proton transfer?

Model answer: κ accounts for the probability that a trajectory reaching the transition state actually proceeds to products rather than recrossing back to reactants, and for quantum tunneling through the barrier. Proton transfer involves the lightest nucleus (mass ≈ 1 amu), so its de Broglie wavelength is large enough that it has significant probability of tunneling through rather than over the activation barrier, making κ substantially greater than it would be for heavier atom transfers.

Classical TST assumes every trajectory that reaches the top of the barrier proceeds forward — κ = 1. Real dynamics can involve recrossing (κ < 1) or tunneling (effectively κ > 1 relative to the classical rate). Protons tunnel because their small mass gives them a large quantum wavelength at thermal energies; deuterium, being twice as heavy, tunnels less, explaining the kinetic isotope effect often observed in proton-transfer reactions.