Organic Mechanisms. Xiaoping Sun

(1.11)

From Equation 1.11, we have ln[A] – ln[A]₀ = −kt

      Therefore,

(1.12)

Equation 1.12 is the integrated rate law for a unimolecular reaction.

      The half‐life (t_1/2) of reactant A (the time required for conversion of one‐half of the reactant to the product, i.e., when t = t_1/2, [A] = ½ [A]₀) can be solved from Equation 1.12 as follows:

Therefore,

(1.13)

Equation 1.13 shows that the half‐life of a substance that undergoes first‐order decay is inversely proportional to the rate constant and independent of the initial concentration.

       Bimolecular reactions

      A bimolecular reaction that involves two reactant molecules of the same compound (Eq. 1.4: 2A ➔ P) follows the second‐order rate law as shown below:

(1.14)

      where k is the rate constant (with the typical unit of M⁻¹s⁻¹) for the reaction.

Rearranging Equation 1.14 leads to

(1.15)

Integrating Equation 1.15 on both sides and applying the boundary condition t = 0, [A] = [A]₀ (initial concentration), we have

(1.16)

From Equation 1.16, we have

(1.17)

Equation 1.17 is the integrated rate law for a bimolecular reaction involving two molecules from the same compound.

A bimolecular reaction that involves two reactant molecules of different compounds (Eq. 1.5: A + B ➔ P) also follows the second‐order rate law (first‐order in each of the reactants) as shown in Equation 1.18.

(1.18)

Assume that at a given time t, the molar concentration of the product P is x. Therefore, the molar concentrations of reactants A and B are [A] = [A]₀ − x and [B] = [B]₀ − x, respectively. [A]₀ and [B]₀ are initial concentrations of reactants A and B, respectively.

      From Equation 1.18, we have

(1.19)

If the quantities of the two reactants A and B are in stoichiometric ratio ([A] ₀ = [B]₀), Equation 1.19 becomes

(1.20)

Rearranging Equation 1.20 leads to Equation 1.21.

(1.21)

      Integrating Equation 1.21 on both sides and applying the boundary condition t = 0, x = 0, we have

(1.22)

From Equation 1.22, we have

(1.23)

Since [A] = [A]₀ − x, Equation 1.23 becomes

      If the reactants A and B have different initial concentrations, Equation 1.19 becomes

(1.24)

Integrating Equation 1.24 on both sides and applying the boundary condition t = 0, x = 0, we have

(1.25)

From