A first order reaction `ArarrB` requires activation energy of `70"kJ.mol"^(-1)` When a 20% solutions 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

From the Arrhenius equation,
`logk=logA-(E_(a))/(2.303RT)`
or, `log.(k_(2))/(k_(1))=(E_(a))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`
`thereforelog.(k_(2))/(k_(1))=(70xx10^(3))/(2.303xx8.314)[(1)/(298)-(1)/(313)] " or, " (k_(2))/(k_(1))=3.872" " ...[1]`
For a first order reaction, `k=(2.303)/(t)log.(a)/(a-x)`
Given x = 0.25a,
`thereforea-x=(a-0.25a)=0.75a " at " t=20min`
`thereforek_(1)=(2.303)/(20)log.(a)/(0.75a)=0.014386"min"^(-1)" "...[2]`
Putting the value of `k_(1)` into equation [1] , we have
`k_(2)=3.872xx0.014386=0.05571"min"^(-1)`
For a first order reaction, `k_(2)=(2.303)/(t)log.(a)/(a-x)`
or, `0.05571=(2.303)/(20)log.(a)/(a-x)`
[x = decrease in concentration of the reactant ]
or, `log.(a)/(a-x)=(0.05571xx20)/(2.303)`

or, `log.(a)/(a-x)=0.48381 " or, " x=0.6717a`
Hence , the percentage decomposition `=(0.6717a)/(a)xx100`
`=67.17%`