rate theory $^{8}$. A typical plot of the energy, say as calculated by the method of Eyring and Polanyi, that the reacting molecules must trace is plotted in Fig. 1. This is Fig. 3 of the forthcoming Bulletin article. In this plot, the energy of the reactants is on the left and the energy of the products is on the right. As the incoming molecule approaches the molecule with which it will react, the energy increases. This energy barrier must be surmounted, rather like a hiker hiking up to and passing over a pass and then descending. The energy state of the products is on the right and this energy state may be greater or lesser than or equal to that of the reactants. In Fig. 1, the products have a lower energy; this is irrelevant to our argument. The height of this barrier is Δ E $^{‡}$. As would a hiker, the constituents of the reaction may hesitate briefly at the pass. Eyring coined the name activated complex for this chemically unstable species at the top of the barrier. It is convenient to use a simpler, perhaps simplistic, nonrigorous derivation than that used by Eyring. In addition to being simpler, this derivation has the advantage of not requiring that the reaction take place in a dilute gas. In a canonical ensemble, the probability, P ( E ), of the system having an energy E is where R is the gas constant and T is the temperature. The denominator is a normalizing factor that ensures that the total probability is one. The probability of the system having enough energy to reach the top of the pass is the integral of P ( E ) from Δ E ‡ to infinity. This gives This result assumes the canonical system, where the volume and number are constant. However, in the reaction it is the pressure and chemical potential that are constant. Hence, it is the Gibbs’ free energy, G , rather than the energy, that should be used. We should consider The mode in the activated complex that takes part in the reaction may be thought of a soft spring. This mode is soft, with a negative spring constant, because the activated P (Δ G $^{‡}$) = exp( − Δ G $^{‡}$/RT ) . (3) P (Δ E $^{‡}$) = ∫ ∞ Δ E ‡ exp( − E/RT ) dE ∞ 0 exp( − E/RT ) dE = exp( − Δ E $^{‡}$/RT ) . (2) P ( E ) = exp( − E/RT ) ∫ ∞ 0 exp( − E/RT ) dE , (1) 5