Question

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

Answer 1

For a first order reaction ,
`t = (2.303)/(k) "log" ([R]_(0))/([R])`
`k = (2.303)/( 40 "min") "log" (100)/(100 -30)`
= `(2.303)/(40 "min") "log" (10)/(7)`
= `8.918 xx 10^(-3) "min"^(-1)`
Therefore , `t_(1//2)` of the decomposition reaction is
`t_((1)/(2)) = (0.693)/(k)`
= `(0.693)/(8.918 xx 10^(-3)) ` min
= `77.7` min (approximately)