The decomposition of `H_(2)O_(2)` is a first order reaction . The rate constant of the reaction can be expressed as `logk=14.34-(1.25xx10^(4))/(T)` . Calculate the activation energy of the reaction . At what temperature does the half-life of the reaction become 265min?

Explanation

`(E_(a))/(2.303R)=1.25xx10^(4) therefore E_(a)=239.34 "kJ.mol"^(-1)`
`k=(0.693)/(256xx60)=4.51xx10^(-5)s`
Now, `log k = 14.34-(1.25xx10^(4))/(T)`
or, `log(4.51xx10^(-5))=14.34-(1.25xx10^(4))/(T)`
`therefore T = 668.75K`