Q2(a)(ii). Theory of absolute reaction rates; how a catalyst works

1. Activated complex. As A and B approach, the energy rises to a maximum (saddle point on the potential-energy surface). The arrangement at the top, with bonds partly broken and partly formed, is the activated complex [AB]°.
2. Quasi-equilibrium. The complex is taken to be in equilibrium with the reactants: [AB°]=K°[A][B]
3. Crossing the top. One vibration of the complex, along the reaction coordinate, has no restoring force. Its frequency is ν=k_BT/h, so rate =κ ν [AB°], where the transmission coefficient κ≈1.
4. Eyring equation. With Δ G°=-RTln K°=Δ H°-TΔ S°: 3pt] k=κ (k_BT)/(h) K°=κ (k_BT)/(h) e^Δ S°/R e^-Δ H°/RT
5. Why “absolute”: k_BT/h is a universal frequency, 6.2×10¹² s⁻¹ at 298 K, so k can in principle be calculated from the properties of the reactants and the complex alone (through partition functions), with no adjustable steric factor. The entropy of activation Δ S° replaces the steric factor P of collision theory; Eₐ=Δ H°+RT in solution.

How a catalyst makes the reaction faster

1. The catalyst combines with a reactant to form a different activated complex on a new path (adsorbed complex, enzyme–substrate complex, intermediate compound). Its Δ G° is lower.
2. Since k∝ e^-Δ G°/RT, a small fall in the barrier gives a large rise in rate. Decomposition of H₂O₂: Eₐ 76 kJ mol⁻¹, 57 with iodide ion, 8 with the enzyme catalase. The iodide path alone gives e^(19 000/(8.314×298))=2.1×10³ times faster reaction at 298 K.
3. Reactants and products are untouched, so Δ G° and K do not change; the forward and reverse barriers fall by the same amount, and equilibrium is only reached sooner.