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

Applying Arrhenius equation,
`k = Ae^(-E_(a)//RT)`
or `logk = log A - (E_(a))/(2.303 RT)`
`:. log.(k_(2))/(k_(1)) = (E_(a))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`
`= (70 xx 10000)/(2.303 xx 8.314)[(1)/(298)-(1)/(313)]`
or `(k_(2))/(k_(1)) = 3.872`
For a first order reaction,
`k_(1)=(2.303)/(t) log.(a)/(a-x) [{:(a= 100%x=25%,,,,),( :. a-x=75%,,,,),(t=20min,,,,):} ]`
`=(2.303)/(20)log.(100)/((100-25))`
`= 0.014386 min^(-)`
Calculation of `%` decompoistion at `40^(@)C`
form Eq. (i), `(k_(2))/(0.014386)= 3.872`
or `k_(2)= 0.05571 min^(-1)`
Again, for a first order reaction,
`k_(2)= (2.303)/(t) log.(a)/(a-x)`
or `0.5571=(2.303)/(20)log.(100)/(100-x)`
or `x= 67.17`
At `313K, 67.17%` decompoistion takes place.