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`.

Answer

`0.03267 min^(-1)`

Explanation

For reaction `(I)`
`A rarr` Products
`T_(1) = 300 K, T_(2) = 310 K`
`(k_(1))_(A)` i s at `300 K` and `(k_(2))_(A)` is at `310 K`
Given:
`[(k_(2)(310 K))/(k_(1)(300 K))]_(A) = 2` …(i)
Uisng Arrhenius equation for reaction `(I)` :
`[(k_(2)(310 K))/(k_(1)(300 K))]_(A) = (E_(a(A)))/(2.303 R)((T_(2)-T_(1))/(T_(1)T_(2)))` ...(ii)
`log 2 = (E_(a(A)))/(2.303R) ((T_(2)-T_(1))/(T_(1)T_(2)))` ...(iii)
(For reaction `II`)
`B rarr` Products
Since at `310 K , B` decomposes twice as fast as `A`.
`because [k_(2)(310 K)]_(A) = 0.0231 min^(-1)`
`:. [k_(2)(310 K)]_(B) = 2 xx 0.0231 min^(-1)`
`= 0.0462 min^(-1)`
We have to calculate
`[k_(1) (300 K)]_(B) =` ?
Uisng Arrhenius equation for reaction `(II)`
`log[(k_(2)(310 K))/(k_(1)(300 K))]_(B) = (E_(a(B)))/(2.303 R) ((T_(2)-T_(1))/(T_(1)T_(2)))` ...(iv)
Given: `(E_(a))_(B) = (1)/(2)(E_(a))_(A)` ...(v)
Substitute the value of Eq. (v) in Eq. (iv),
`log[(k_(2)(310 K))/(k_(1)(300 K))]_(B) = (1)/(2) xx (E_(a(A)))/(2.303 R)((T_(2)-T_(1))/(T_(1)T_(2)))` (vi)
Comparing Eqs. (iii) and (vi), we get
`log[(k_(2)(310 K))/(k_(1)(300 K))]_(B) = (1)/(2) xx log 2`
`:. (0.0462)/([k_(1)(300 K)]_(B)) = (2)^((1)/(2)) = sqrt(2) = 1.414` ....(vii)
`:. [k_(1) (300 K)]_(B) = (0.0462)/(1.414) = 0.03267 min^(-1)`