 # Explain half-life for zero-order and second-order reactions

The half-life for zero-order and second-order reactions, half-life changes based on the concentration of the reactant. By definition, the half-life of any of the reactions is the amount of time the reactants take to consume half of the starting material. In a second-order reaction, the half-life of the reaction is inversely proportional to the initial concentration of the reactant(A). For each second-order reaction is each half-life is twice as long as the life span of the one before.

To find the equation for the half-life of a second-order reaction, set [A]t=0.5[A]0 in the second-order equation.

1/0.5[A]0=kt1/2+1/[A]0

2/[A]0=kt1/2+1/[A]0

2/[A]0-1/[A]0=kt1/2

t1/2=1/k[A]0 (3)

One important thing to notice in the above reaction is this, the half-life of a second-order reaction depends only on the initial concentration, in contrast to the first-order reactions. Because of this reason, the concept of half-life for a second-order reaction is less useful in comparison to the first-order reaction.

CASE-1 IDENTICAL REACTANTS

When two identical reactants (A) combine in a single elementary step and gives a product P.

A+A→P (4)

2A→P (5)

The reaction rate for this above reaction can be written as-

Rate=−1/2d[A]dt=+d[P]/dt (6

And the rate of loss of reactant A

dA /dt=-k[A][A]

=-k[A]2 (7)

In this 7th reaction, the k is the second-order rate constant and the unit of k is M-1 min-1 or M-1 s-1. Therefore, if the concentration of the reactant will be doubled than the rate of the reaction will increase to four times. In this particular case, another different reactant (B) would be present with the reactant A, however, there is no change in the rate of the reaction due to change in the concentration, that is the reaction order for B is zero and we can express the rate law as

Rate law=k[A]2[B]0

Case-2 DIFFERENT REACTANTS

Taking two reactants A and B combine together to form a product in a single elementary step:

A+B→P (8)

The reaction rate of the above reaction can be written in the following way

Rate=-d[A]/dt=-d[B]/dt=+d[P]/dt (9)

And the rate of loss of the reactant A in the above chemical reaction,

d[A]/dt=-k[A][B] (10)

In the above reaction, the reaction order of both the reactants is 1. From this, we can conclude that when the concentration of the reactant is doubled, and four times the concentration of the reactant will quadruple the rate. If we double the reactant concentration of A and increase the concentration of reactant B four times at the same time, then the rate of the reaction will be increased by the factor of 8. This relationship can hold true for any of the varying concentrations of both the reactant that is A and B. (0) (0)