The rate constant for the first order decomposition of `H_(2)O_(2)` is given by the
following equation:
`"log"k=14.34-1.25xx10^(4)k//t`
Calculate `E_(a)` for this reaction and at what temperature will its half-period be 256 minutes?

Answer

`a.` `E_(a)=239.339kJ mol^(-1)`
`b. T=669K`

Explanation

`a.` According to Arrhenius equation, `k=Ae^(-Ea//RT)`
`logk=logA-(E_(a))/(2.303RT) ......(i)`
Comparing with the griven equation
`logk=14.34-1.25xx10^(4)K//T ...(ii)`
`:. (E_(a))/(2.303RT)=(1.25xx10^(4)K)/(T)`
`or E_(a)=2.303R xx 1.25xx10^(4)K`
`=2.303xx(8.314J K^(-1) mol^(-1))xx1.25xx10^(4)K`
`=239.34kJ mol^(-1)`
`b.` When `t_(1//2)=256mi n, k=(0.693)/(t_(1//2)mi n)=(0.693)/(256xx60s)`
`=4.51xx10^(-5)s^(-1)`
Substitute the value of `k` in `Eq. (ii)`,
`log(4.51xx10^(-5)s^(-1))=14.34-(1.25xx10^(4)K)/(T)`
`log(-5+0.6542)=14.34-(1.25xx10^(4)K)/(T)`
`or(1.25xx10^(4)K)/(T)=18.6858`
`or T=669K`