Question

The reaction A → products was found to be second order order and have a rate constant, k, of 0.707 M-1 s-1. If the initial concentration of the reaction was 0.737 M, what is the half life for the reaction?

Answer 1

Half-life for the reaction is 1.92s

Step-by-step explanation:

Integrated rate equation for the given second order reaction is-

`(1)/([A]_(t))=kt+(1)/([A]_(0))`

Where
`[A]_(t)` is concentration of A after "t" time and
`[A]_(0)` is initial concentration of A

At half-life,
`[A]_(t)=([A]_(0))/(2)`

Here
`[A]_(0)=0.737M` and
`k=0.707M^(-1)s^(-1)`

Plug-in all the values in the above equation-

`(1)/(([A]_(0))/(2))=(0.707M^(-1)s^(-1)* t)+(1)/([A]_(0))`

or,
`(1)/([A]_(0))=0.707M^(-1)s^(-1)* t`

or,
`t=(1)/((0.737M* 0.707M^(-1)s^(-1)))`

or,
`t=1.92s`

So, half-life for the reaction is 1.92s