F(v) = 4π 3 /^2
B( 2 πmk T) v^2 e−mv^2 /^2 kBT = 4 π (m/2πkBT)3/2 v^2 exp(–mv^2 /2kBT)
k: Boltzmann’s constant = R/N 0 = 1.38 × 10–23 J K–1
e is 2.718, the base of natural logarithmsFigure 6.4 The distribution of energies at two temperatures, T 1 and T 2 (T 1 > T 2 ).
The number of collisions with energies greater than the free energy of
activation is indicated by the appropriately shaded area under each
curve.
3) Because of the way energies are distributed at different temperature, increasing
the temperature by only a small amount causes a large increase in the number of
collisions with larger energies.
- The relationship between the rate constant (k) and ∆G‡:
k = k 0 e−∆G‡/RT
1) k 0 is the absolute rate constant, which equals the rate at which all transition states
proceed to products. At 25 °C, k 0 = 6.2 × 10^12 s–1.
2) A reaction with a lower free energy of activation will occur very much faster
than a reaction with a higher one.- If a reaction has a ∆G‡ less than 84 kJ mol–1 (20 kcal mol–1), it will take place
readily at room temperature or below. If ∆G‡ is greater than 84 kJ mol–1, heating