form the following data for the reaction between `A` and `B`,

(a) Calculate the order of the reaction with respect to `A` and with respect to `B`.
(b) Calculate the rate constant at `300 K`.
( c) Calculate the pre-expontential factor.

Explanation

Lat rate expression be `r=K[A]^(m)[B]^(n)`
(i) `:. 5.0xx10^(-4)=K[2.5xx10^(-4)]^(m)[3.0xx10^(-5)]^(n) …(1)`
`4.0xx10^(-3)=K[5.0xx10^(-4)]^(m)[6.0xx10^(-5)]^(n) …(2)`
`1.6xx10^(-2)=K[1.0xx10^(-3)]^(m)[6.0xx10^(-5)]^(n) …(3)`
By eqs. (2) and (3), `m=2`
By eqs. (1) and (2), `n=1`
`:.` Rate `=K[A]^(2)[B]^(1)`
or Order with respect to `A` and `B` are `2` and `1` respectively.
(ii) By eq. (1)
`5.0xx10^(-4) =K[2.5xx10^(-4)]^(2)[3xx10^(-5)]^(1)`
`:. K=2.66xx10^(8) litre^(2) mol^(-2) sec^(-1)` at `300 K`
(iii) `:' 2.303 "log"_(10) K_(2)/K_(1)=E_(a)/R([T_(2)-T_(1)])/(T_(1)T_(2)), [:' r_(2)/r_(1)=K_(2)/K_(1)]`
`:' 2.303"log"_(10)(2.0xx10^(-3))/(5.0xx10^(-4))=E_(a)/8.314 [(320-300)/(300xx320)]`
`E_(a)=55.33 kJ`
(iv) Also `2.303 log_(10) K=2.303 log_(10) A-E_(a)/(RT)`
`2.303 log_(10) 2.66xx10^(8) =2.303 log_(10) A-(55.33xx10^(3))/(8.314xx300)`
`:. A=1.140xx10^(18)`