At `380^(@)C` , the half-life periof for the first order decompoistion 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%` decompoistion at `450^(@)C`.

Answer

`20.358 min`

Explanation

The first order reaction
Half life `=(0.693)/(k)`
or `360=(0.693)/(k)`
`k_(380^(@)C)=(0.693)/(360)`
`k=Ae^(-E_(a)//RT)`
or `log k = log A-(E_(a))/(2.303RT)`
`:. log.(k_(2))/(k_(1))= (E_(a))/(2.303TR)[(1)/(T_(1))-(1)/(T_(2))]`
or `log.(k_(450^(@)C))/((0.693)/(360))=(200xx1000)/(2.303xx8.314)[(1)/(653)-(1)/(723)]`
or `k_(450^(@)C)=6.18xx10^(2-)min^(-1)`
For `75%` decompoistion at `723 K`
`k_(450^(@)C)= (2.303)/(t) log.(a)/(a-x)`
or `6.81xx10^(-2)=(2.303)/(t)log.(100)/(25)`
or `t= 20.358min`