At `380^(@)C` 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

For the first order reaction half-life `=(0.693)/(k)`
or, `360=(0.693)/(k) thereforek_(380^(@)C)=(0.693)/(360)`
We know , `log.(k_(2))/(k_(1))=(E_(a))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`
`thereforelog.k_(450^(@)C)/((0.693)/(360))=(200xx1000)/(2.303xx8.314)[(1)/(653)-(1)/(723)]`
or, `k_(450^(@)C)=6.81xx10^(-2)"min"^(-1)`
For 75% decomposition at 723K `(450^(@)C)`,
`k_(450^(@)C)=(2.303)/(t)log.(a)/(a-x)`
or, `6.81xx10^(-2)=(2.303)/(t)log.(100)/((100-75))" or, "t=20.358"min"`