A reaction is carried out at two different initial concentrations of a reactant [A], that is [A]0,1 and [A]0,2 and the respective t1/2 values (t1/2) and (t1/2)2 are obtained. The order of the reaction can be calculated as follows:
Example 9:
A first order reaction has k=3 x 10-6 s-1 at 5/3 K. The reaction is allowed to run for five hours. After this time interval what percentage of the initial concentration would have changed into products? What is the half-life of the reaction?
Suggested answer :
In the integrated rate law expression k is given as
On substitution, the expression becomes
= 0.052
Hence, 5.2% of the initial concentration has changed to products.
