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`
[Where unit of k is `s^(-1)`]
Calculate `E_(a)` for this reaction and at what temperature will its half-period be 256 minutes?

Explanation

Calculation of `E_(a)`:The reaction is first order.
The decomposition of `H_(2)O_(2)` occurs in reaction .It is given that,
log k=`14.34-1.25xx10^(4)K//T` ….(a)
log k=`log A-(E_(a))/(2.303 RT)` (Arrhenius(b))
Comparing equation (a) and (b), log k and 14.34 =log A is accepted and removing it,
So,`(E_(a))/(2.303RT)=(1.25xx10^(4))/(T)k`
`therefore E_(a)=1.25xx104 kxx2,303xx8.314 JK^(-1) mol^(-1)` Calculation of decomposition constant k on the base of `t_((1)/(2))`:
`k=(0.693)/(t)` Where ,`t_((1)/(2))=256` minute
`=(0.693)/(256xx60s)` Calculation of `t_((1)/(2))` temperature:
Given equation :log k=14.34-`(1.25xx10^(4)K)/(T)`
`therefore log(4.51xx10^(-5))=14.34-(1.25xx10^(4)K)/(T)`
`therefore -4.3458-14.34=-(1.25xx10^(4)K)/(T)`
`therefore-18.6858=-(1.25xx10^(4)K)/(T)`
`therefore T=(1.25xx10^(4)K)/(18.6858)=668.96 K=669K`