40% of a first order reaction is completed in 50 minutes. How much time will it take for the completion of 80% of this reaction?

Explanation

For a first order reaction :
Time required is given by,
`t= (1)/(k_(1))ln [(a)/(a-x)]`
where a is the initial concentration.
Hence,
`(t_(40))/(t_(80))=((1)/(k_(1))ln((100)/(100-40)))/((1)/(k_(1))ln[(100)/(100-80)])`
`(t_(40))/(t_(80))=(ln[(100)/((100-40))])/(ln[(100)/((100-80))])`
`(50)/(t_(80))=(ln[(100)/(60)])/(ln[(100)/(20)])`
`(50)/(t_(80))=(ln(1.66))/(ln(5))`
`(50)/(t_(80))=(0.50)/(1.61)=161`min.
Hence, time required for 80% completion would be 161 min.