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Organic Mechanisms. Xiaoping Sun
Читать онлайн.Название Organic Mechanisms
Год выпуска 0
isbn 9781119618867
Автор произведения Xiaoping Sun
Жанр Химия
Издательство John Wiley & Sons Limited
From Equation 1.11, we have ln[A] – ln[A]0 = −kt
Therefore,
Equation 1.12 is the integrated rate law for a unimolecular reaction.
The half‐life (t1/2) of reactant A (the time required for conversion of one‐half of the reactant to the product, i.e., when t = t1/2, [A] = ½ [A]0) can be solved from Equation 1.12 as follows:
Therefore,
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:
where k is the rate constant (with the typical unit of M−1s−1) for the reaction.
Rearranging Equation 1.14 leads to
Integrating Equation 1.15 on both sides and applying the boundary condition t = 0, [A] = [A]0 (initial concentration), we have
From Equation 1.16, we have
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.
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
If the quantities of the two reactants A and B are in stoichiometric ratio ([A] 0 = [B]0), Equation 1.19 becomes
Rearranging Equation 1.20 leads to Equation 1.21.
Integrating Equation 1.21 on both sides and applying the boundary condition t = 0, x = 0, we have
From Equation 1.22, we have
Since [A] = [A]0 − x, Equation 1.23 becomes
If the reactants A and B have different initial concentrations, Equation 1.19 becomes
Integrating Equation 1.24 on both sides and applying the boundary condition t = 0, x = 0, we have
From