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

Given, `Ararr B` and, `20%` solution of `A` decomposes `25%` in `20` minute at `25^(@)C`.
`:.` Initial amount, `a=20`
`:.` Amount left, `(a-x)=20xx75/100=15`
`:. K_(25)=2.303/20"log"_(10)20/15=0.0144 "minute"^(-1), ( :' t=20 minute)`
Now `2.303 "log"_(10)K_(40)/K_(25)=E_(a)/R([T_(2)-T_(1)])/(T_(1)T_(2))`
`:. 2.303"log"_(10)K_(40)/K_(25)=(70xx10^(3))/(8.314)`
`[(313-298)/(313xx298)]`
`K_(40)/K_(25)=3.872`
`:. K_(40)=0.05575 min^(-1)`
`( :' K_(25)=0.0144)`
Now suppose amount `m` is left in `30%` solution in `20` minute at `40^(@)C`.
`K_(40)=2.303/t"log"_(10)a/((a-x))`
`0.05575=2.303/20"log"_(10)30/m`
`:. m=9.838`
`:. %` decomposition
`=((a-m))/(a)xx100=(30-9.838)/(30)xx100=67.21%`