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

(I) `A rarr "Products"`
(II) `B rarr "Products"`
`:' t_(1//2)` for (I) `310 K=30 "minute"`
`:. K_((I)) at 310 K =0.693/30=0.0231 min^(-1) …(1)`
`:. Rate =K [ ]` and both reactions are of I order
also given, `(K_(1) at 310)/(K_(1) at 300)=2 …(2)`
Also given, `(K_(II) at 310)/(K_(1) at 310)=2 ...(3)`
Also we have, `E_(a_(II))/E_(a_(I))=1/2 ...(4)`
For `I: 2.303 "log"_(10) (K_(1) at 310)/(K_(1) at 300)=E_(a)/R[(310-300)/(310xx300)]...(5)`
For `II: 2.303"log"_(10)(K_(II) at 310)/(K_(II) at 300)=E_(a_(II))/R[(310-300)/(310xx300)] ...(6)`
Dividing eqs. (5) by (6),
`:. ("log"_(10)(K_(I) at 310)/(K_(I) at 300))/("log"_(10)(K_(II) at 310)/(K_(II) at 300))=E_(a_(I))/E_(a_(II))=2, By eq. (4) ....(7)`
or `"log"_(10)(K_(1) at 310)/(K_(1) at 300)=2 "log"_(10) [(K_(II) at 310)/(K_(II) at 300)]`
or `(K_(1) at 310)/(K_(1) at 300)=[(K_(II) at 310)/(K_(II) at 300)]...(8)`
By eqs. (2) and (8),
or `[(K_(II) at 310)/(K_(II) at 300)]^(2)=2`
or `K_(II) at 310 K=sqrt(2) K_(II) at 300K ...(9)`
By eqs. (3) and (9),
`2xxK_(1) at 310 K =sqrt(2) (K_(II) at 300 K)...(10)`
`K_(II) at 300K =(2xxK_(1) at 310)/(sqrt(2))`
By eqs. (1) and (10),
`K_(II) at 300 K= sqrt(2)xx0.0231`
`K_(II) at 300K =3.27xx10^(-2) min^(-1)`
`=0.0327 min^(-1)`