The half life period for a reaction of first order is `2.31xx10^3` min. How long will it take for `1/5^(th)` of the reactants to be left behind.

Explanation

`t_(1//2)=2.31xx10^(3)` min
`therefore` k=`(0.693)/(t_(1//2))=(0.693)/(2.31xx10^(3)min)`
`[R]_0`=1 (say) [R]=`1//5`,
`k=3xx10^(-4)min^(-1)`
t=`(2.303)/(k)log([R]_0)/([R])`
`=(2.303)/(3xx10^(-4)min^(-1))log(1)/(1//5)`
`=(2.303)/(3xx10^(-4)min^(-1))log5`
`=(2.303xx0.6990)/(3xx10^(-4))` min
`=(2.303xx6.99)/(3)xx10^3` min
`=5.348xx10^3` min