Question

A first order reaction is `20%` complete in 10 minutes. Calculate the time taken for the reaction to go to `80%` completion.

Answer 1

For a first order reaction,`t=(2.303)/k log a/(a-x)`
Ist case: `a=100%, (a-x)=80%, t=10min`
`10"min" = 2.303/k log100/80= 2.303/k log1.25`……………….(i)
II nd case:
`a=100%, (a-x)=20%`
`t=(2.303)/(k) log 100/20 = 2.303/k log 5`………..(ii)
Dividing (ii) by eqn (i),
`(t/10"min")= (log5)/(log 1.25) = (0.69897)/(0.969)` or `t=(0.69897)/(0.0969) xx (10 "min") = 72.13`