The half-life period of a `1^(st)` order reaction is 60 minutes. What percentage will be left over after 240 minutes ?

Options

A. `6.25%`
B. `4.25%`
C. `5%`
D. `6%`

Answer

A

Explanation

`t_(1//2) = (0.693)/(k) implies (0.693)/(t_(1//2) = k implies (0.693)/(60) = k`
k=0.01155` "min"^(-1)`
`k= (2.303)/t log""((a)/(a-x))`
Let the initial amount (a) be 100.
`0.01155"min"^(-1) =(2.303)/(240 "min") log ((100)/(a-x))`
`(0.01155 "min"^(-1)xx40"min")/(2.303) = log ((100)/(a-x))`
`1.204 = log 100 - log(a-x)`
`1.204 = 2- log(a-x)`
`log(a-x) = 2- 1.204`
`log(a-x) = 0.796`
`(a-x) = 6.25%`