A first order reaction `A rarr B` requires activation energy of `70 kJ mol^(-1)`. When a `20%` solution of `A` was kept at `25^(@)C` for `20 min`, `25%` decomposition took place. What will be the percentage decomposition in the same time in a `30%` solution maintained at `40^(@)C` ? (Assume that activation energy remains constant in this range of temperature)

Explanation

In `{(k(40^(@)C))/(k(25^(@)C))}=(E_(a))/(R) ((15)/(298xx313))=(70xx10000)/(8.314) xx(15)/(298xx313)=1.35`
`rArr (k(40^(@)C))/(k(25^(@)C))=3.87`
Also `k(25^(@)C) =(1)/(20)" In "(100)/(75) =(1)/(20)" In "(4)/(3)`
`rArr k(40^(@)C) =3.87xxk(25^(@)C) =3.87xx(1)/(20)" In "(4)/(3)=55.66xx10^(-3)" m in"^(-1)`
Now `k(40^(@)C) xx20 =" In "(100)/(100-x)`
`rArr 55.66xx10^(-3) xx20=" In "(100)/(100-x) rArr x=67%`