`70%` of a first order reaction was completed in `70 min`. What is the half life of the reaction?

Options

A. `4.2 min`
B. `42 min`
C. `4.2 hr`
D. `4.2 s`

Answer

B

Explanation

Use direct formula:
`t_(1//2) = 0.3, t_(70%) = log.(100)/(100-70)`
`log.(10)/(3) = [1-0.48] = 0.52 ~~ 0.5`
`(t_(1//2))/(t_(70%)) = "(0.3)/(0.5)`
`t_(1//2) = (0.3 xx 70)/(0.5) = 42 min`