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

It does not matter whether you take 20%, 30%, 40%, or 70% of A,
At `25^(@)C, 20%` of A decomposes 25%
`rArr" "kt =2.303 log_(10)""(A_(0))/(A)`
`rArr " "k(40) =2.303 log_(10)""(100)/(75)`
`rArr" "k("at "25^(@)C) =0.0143" min"^(-1)`
Using Arrhenus equation find k at `40^(@)C`.
`log_(10)""(k_(40^(@)))/(k_(25^(@)))=(E_(a))/(2.303 R) ((T_(2)-T_(1))/(T_(1)T_(2)))`
`rArr log_(10) ""(k_(40^(@)))/(0.0143) =(70xx10^(3))/(2.303xx8.31)((40-25)/(298xx313))`
`rArr " "k("at "40^(@)C)=0.055" min"^(-1)`
Now calculate % decomposition at `40^(@)C` using first order kinetics.
`kt =2.303 log_(10)""(A_(0))/(A)`
`rArr" "0.055 xx40=2.303 log_(10)""(100)/(100-x)`
`rArr x=67.1=67.1%" decomposition of A at "40^(@)C`