A first order reaction is `20%` complete in 10 minutes. Calculate the time for the completion of `75%` of the reaction.

Explanation

For the first order reaction,
`t= 2.303/k log a/(a-x)`
Ist case, a=`100%`, x=`20%`, (a-x)=100-20 = 80%, t=10 min
`t_(20%) = 2.303/k log (100%)/(80%)`
10 min `= 2.303/k log 100/80 or 10 min = 2.303/k log 1.25`
Iind case, `a=100%, x=75%, (a-x) = 100-75=25%`
`t_(75%) = 2.303/k log 100/25 or t_(75%) = 2.303/k log4`
Divide eqn. (ii) by eqn. (i),
`t_(75%)/(10 min) = (log4)/(log 1.25) = (0.6021)/(0.0969)` or `t_(75%) = 0.6021/0.0969 xx 10 min = 62.14 min`