Two reaction, `(I)A rarr` Products and `(II) B rarr` Products, follow first order kinetics. The rate of reaction `(I)` is doubled when the temperature is raised form `300 K` to `310K`. The half life for this reaction at `310K` is `30 min`. At the same temperature `B` decomposes twice as fast as `A`. If the energy of activation for reaction `(II)` is twice that of reaction `(I)`, (a) calculate the rate of constant of reaction `(II)` at `300 K`.

Explanation

`t_(1//2) = 21.2" m in, "k =3.26xx10^(-2)" m in"^(-1)`
(i) `A overset(k_(A)) rarr` Product
(ii) `B overset(k_(B)) rarr` Product
For (i) `(E_(a))/(R ) ((10)/(300xx310))=" In "2`
`rArr E_(a)" (i) =9300 In "2=53.6 kJ " rArr E_(a) (iii) =(E_(a)(ii))/(2) =26.8 kJ`
At 310 `k t_(1//2) (i) =30 min`
`because" Rate of (ii) =2 rate of (i) " rArr t_(1//2) (ii) =15` m in
Now for reaction (ii) :
`"In "{(k_(B) (310))/(k_(B) (300))}=" In "{(t_(1//2) (300))/(t_(1//2)(310))}=(E_(a)(ii))/(R) ((10)/(300xx310))`
`rArr" In "{(t_(1//2)(300))/(15)}=("In 2")/(2) rArr t_(1//2) (300)=21.2` m in
`rArr k_(B) (300) =("In "2)/(t_(1//2))=(0.693)/(21.2) =3.26xx10^(-2)" m in"^(-1)`