The rate constant of a first order reaction follows the equation `logk(s^(-1))=22.3-(12.16xx10^(3))/(T)K`
At what temperature will the half-life of the reaction be 115.5 min ?

Explanation

For a first order reaction,
`k=(0.693)/(t_(1//2))`
If `t_(1//2)=115.5"min"=115.5xx60=6930,` then
`k=(0.693)/(6930)=10^(-4)s^(-1)`
Putting `k=10^(-4)s^(-1)` into the given equation,
We have `log(10^(-4))=22.3-(12.16xx10^(3))/(T)`
or, `-4=22.3-(12.16xx10^(3))/(T) " or, "T=462.3K`
Therefore , the half-life of the reaction will be 115.5 min at 462.3K.