The rate constant for the first order decomposition of `H_(2) O_(2)` is given by the following equation,
`log k = 14.34 - (1.25 xx 10^(4) K)/( T)`
Calculate `E_a` for this reaction and at what temperature will its half period be 256 min?

Explanation

(i) According to Arrhenius equation,
`k= .Ae^(- E_(a) // RT)`
`log k = log A - (E_a)/( 2.303 RT) " "...(i)`
`log k = 14.34 - (1.25 xx 10^(4) K)/( T) " "...(ii)`
On comparing both equations,
`(E_a)/( 2.303 RT) - (1.25 xx 10^(4) K)/( T)`
`E_a = 1.25 xx 10^(4) K xx 2.303 xx 8.314" JK"^(-1)" mol"^(-1)`
`= 23.93 xx 10^(4)"J mol"^(-1) = 239.3" kJ mol"^(-1)`
(ii) If `t_(1//2) = 256` min
For first order reaction,
`k = (0.693)/( t_(1//2) ) = (0.693)/((256" min")) = (0.693)/( 256 xx 60" s" ) = 4.51 xx 10^(-5)" s"^(-1)`
According to Arrhenius theory,
`log k = 14.34 - (1.25 xx 10^(4) K)/( T)`
`log (4.51 xx 10^(-5) ) = 14.34 - (1.25 xx 10^(4) K)/( T)`
`(1.25 xx 10^(4) )/( T) = 14.34 + 4.35 = 18.69`
`T = (1.25 xx 10^(4) )/( 18.69) = 669K`