A first order reaction ArarrBrequires activation energy of 70"kJ.mol"^(-1)When a 20% solutions of A was kept at25^(@)Cfor 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.

A first order reaction ArarrBrequires activation energy of 70"kJ.mol"^

1.	A first order reaction ArarrBrequires activation energy of 70"kJ.mol"^(-1)When a 20% solutions of A was kept at25^(@)Cfor 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.
Answer» Solution :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%`