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In a second-order reaction, the sum of the exponents in the rate law is equal to two. The two most common forms of second-order reactions will be discussed in detail in this section.
Reaction Rate
Integration of the second-order rate law
d[A]dt=−k[A]2. (1.1)
yields
1[A]=1[A]0+kt. (1.2)
which is easily rearranged into a form of the equation for a straight line and yields plots similar to the one shown on the left below.
The half-life is given by
t1/2=1k[Ao]. (1.3)
Notice that the half-life of a second-order reaction depends on the initial concentration, in contrast to first-order reactions. For this reason, the concept of half-life for a second-order reaction is far less useful. Reaction rates are discussed in more detail here. Reaction orders are defined here.Here are explanations of zero and first order reactions.
Case 1: Identical Reactants
Two of the same reactant (A) combine in a single elementary step.
A+A⟶P. (1.4)
2A⟶P. (1.5)
The reaction rate for this step can be written as
Rate=−12d[A]dt=+d[P]dt. (1.6)
and the rate of loss of reactant A
dAdt=−k[A][A]=−k[A]2. (1.7)
where k is a second order rate constant with units of M-1 min-1or M-1 s-1. Therefore, doubling the concentration of reactant A will quadruple the rate of the reaction. In this particular case, another reactant (B) could be present with A. however, its concentration does not affect the rate of the reaction, i.e., the reaction order with respect to B is zero, and we can express the rate law as v=k[A]2[B]0.
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