`60%` of a first order reaction was completed in `60 min`. The time taken for reactants to decompose to half of their original amount will be

Options

A. `~~ 30 min`
B. `~~ 45 min`
C. `~~ 20 min`
D. `~~ 40 min`

Answer

B

Explanation

`t_(1//2) = (0.69)/(k) = (0.3 xx 2.3)/(k)`
`t_(60%) = (2.3)/(k)log.(100)/(100-60) = (2.3)/(k)log.(10)/(4)`
`k = (2.3)/(60)log.(10)/(4)`
`(0.3 xx 2.3)/(t_(1//2)) = (2.3)/(60)log.(10)/(4)`
`= (1)/(60)[1-2 log2]=(1)/(60)[1-0.6]`
`(0.3)/(t_(1//2)) = (0.4)/(60)`
`:. t_(1//2) = (60 xx 0.3)/(0.4) = 45 min`
Use direct relation
`t_(1//2) = 0.3, t_(x%) = (log.(100)/(100-x))`
`t_(60%) = log.(10)/(4) = (0.4)`
`(t_(1//2))/(t_(60%)) = (0.3)/(0.4)`
`:. t_(1//2) = t_(60%) xx (0.3)/(0.4) xx (60 xx3)/(4) = 45 min`