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# Reaction Half-life

Author: John Hutchinson

Once we know the rate law for a reaction, we should be able to predict how fast a reaction will proceed. From this, we should also be able to predict how much reactant remains or how much product has been produced at any given time in the reaction. We will focus on the reactions with a single reactant to illustrate these ideas.

Consider a first order reaction like A products, for which the rate law must be

 Rate = -d[A]/dt = k[A] [10]

From Calculus, it is possible to use equation 10 to find the function [A](t) which tells us the concentration [A] as a function of time. The result is

 [A]=[A]0 e-kt [11]

or equivalently

 ln[A] = ln[A]0 - kt [12]

Equation 12 reveals that, if a reaction is first order, we can plot ln[A] versus time and get a straight line with slope equal to -k. Moreover, if we know the rate constant and the initial concentration, we can predict the concentration at any time during the reaction.

An interesting point in the reaction is the time at which exactly half of the original concentration of A has been consumed. We call this time the half life of the reaction and denote it as t1/2. At that time, [A] = 1/2[A]0. From equation 12 and using the properties of logarithms, we find that, for a first order reaction

 t1/2 = ln2/k [13]

This equation tells us that the half-life of a first order reaction does not depend on how much material we start with. It takes exactly the same amount of time for the reaction to proceed from all of the starting material to half of the starting material as it does to proceed from half of the starting material to one-fourth of the starting material. In each case, we halve the remaining material in a time equal to the constant half-life in equation 13.

These conclusions are only valid for first order reactions. Consider then a second order reaction, such as the butadiene dimerization discussed above. The general second order reaction A products has the rate law

 Rate = -d[A]/dt = k[A]2 [14]

Again, we can use Calculus to find the function [A](t) from equation 14. The result is most easily written as

 1/[A] = 1/[A]0 + k(t) [15]

Note that, as t increases, 1/[A] increases, so [A] decreases. Equation 15 reveals that, for a reaction which is second order in the reactant A, we can plot 1/[A] as a function of time to get a straight line with slope equal to k. Again, if we know the rate constant and the initial concentration, we can find the concentration [A] at any time of interest during the reaction.

The half-life of a second order reaction differs from the half-life of a first order reaction. From equation 15, if we take [A] = 1/2 [A]0, we get

 t1/2 = 1 / k[A]0 [16]

This shows that, unlike a first order reaction, the half-life for a second order reaction depends on how much material we start with. From equation 16, the more concentrated the reactant is, the shorter the half-life.

Last Update: 2011-02-16