A first reaction takes `40 mi n ` for `30%` decomposition. Calculate `t_(1//2)`.

Answer

`77.7 mi n`

Explanation

Use direct relation,
`(t_(1//2))/(t_(30%))=(3.0)/(log((100)/(100-30)))`
`:.t_(1//2)=40 mi nxx(0.3)/(log((10)/(7)))`
`=(40mi nxx0.3)/((1-0.845))[log7=0.845]`
`~~77.7mi n` Alternatively
`x=30, a =100`
`k=(2.303)/(t)log (a)/(a-x)`
`=(2.303)/(40 mi n ) log((100)/(100-30))`
`=(2.303)/(40)xx0.155mi n^(-1)`
`=8.918xx10^(-3)mi n^(-1)`
`t_(1//2)=(0.693)/(K)=(0.693)/(8.918xx10^(-3))~~77.7mi n`