A first order reaction is 20% complete in the 10 minutes. Calculate the time period for 75% completion of the reaction.

Explanation

`k = (2.303)/(t) "log" ([A]_0)/([A])`
`k = (2.303)/(10) "log" 100/80`
`k = (2.303)/(10) xx 0.09677`
`= 0.0223 "min"^(-1)`
`t = (2.303)/(k) "log" ([A]_0)/([A])`
`t = (2.303)/(0.0223) "log" 100/25`
`t = (2.303)/(0.0223) xx 0.6020`
`=62 ` min.