At 380°C, the half-life period for the first order decomposition of `H_2 O_2` is 360 min. The energy of activation of the reaction is 200 `"kJ mol"^(-1)`. Calculate the time required for 75% decomposition at `450^(@)` C.

Explanation

Given, `t_(1//2) = 360" min"`
`E_a = 200" kJ mol"^(-1)`
`T_1 = 380^(@)C = 380+373= 653 K`
`T_2 = 450^(@) C = 450 + 273 = 723K`
As, `log"" (k_2 [450^(@) C ] )/( k_1 [380^(@) C] ) = (E_a)/( 2.303 R) [ (1)/( T_1) - (1)/( T_2) ]`
`k_1` at `380^(@) C = (0.693)/( t_(1//2) ) = (0.693)/( 360)`
`therefore log"" (k_2)/( 0.693// 360 ) = (200 xx 10^(3) )/( 2.303 xx 8.314 ) [ (1)/( 653) - (1)/( 723) ]`
`log"" (k_2)/( 0.0019) = 1.5487`
`(k_2)/( 0.0019)=" antilog" 1.5487 (k_2)/( 0.0019) = 35.37`
`rArr k_2 = 0.067" min"^(-1)`
Thus, the time required for 75% decomposition at 450°C is given by
`t = (2.303)/( k) log"" ([A]_0)/([A]) = (2.303)/( 0.067) log"" (100)/( 25)`
`= 20.69` min