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

Step I : To calculate k
`t_(1//2)=2.31xx10^(3)` min
`therefore k=(0.693)/(t_(1//2))=(0.693)/(2.31xx10^3min)`
`=(6.93xx10^(-1-3))/(2.31)min^(-1)`
`=3xx10^(-4)min^(-1)`
Step II : To calulate t
`[R]_0=1("say")[R]=1//5`
`k=3xx10^(-4)min^(-1),t=?`
`therefore t=(2.303)/(k)log([R]_0)/([R])`
`=(2.303)/(3xx10^(-4)min^(-1))log(1)/(1//5)`
`=(2.303log5)/(3xx10^(-4)min^(-1))`
`=(2.303xx0.6990)/(3xx10^(-4))`min
`=0.5366xx10^4` min
`=5.366xx10^3` min