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.

Answer

(a) `OR` w.r.t. `A = 2`
`OR` w.r.t. `B = 1`
(b) `k = 2.66 xx 10^(8) L^(2) mol^(-2) s^(-1)`
( c) `A = 1.25 xx 10^(18) s^(-1)`

Explanation

a. form careful observation of the given data it can be concluded that doubling the concentration of `A` and `B` together increases the rate of reaction by eight times. Therefore, to overall order with respect to `A` and `B` is `3`.
Doubling the concentration of `A` alone increases the rate of the reaction by four times, therefore, the order with respect to `A` is `2`.
Hence, the order with respect to `B` must be `1`.
Rate `= k[A]^(2)[B]`
Alternatively
Rate `=k[A]^(x)[B]^(y)`
form (ii) and (iii), we get `implies2^(x)=4impliesx=2`
form (i) and (iii), we get `implies2^(x+y)=8impliesx+y=3`
`impliesy=1`
Thus, rate `=k[A]^(2)[B]`
b. Rate `=k_(1)[A]^(2)[B]`
At `300 K`,
`5.0xx10^(-4)=k_(1)[2.5xx10^(-4)]^(2)xx[3.0xx10^(-5)]`
or `k_(2)= 1.07xx10^(9)L^(2)mol^(-2)s^(-1)`
Also,
`k= Ae^(-E_(a)//RT)`
or `log k=log A-(E_(a))/(2.303RT)`
`:. log.(k_(2))/(k_(1))= (E_(a))/(2.303)[(1)/(T_(1))-(1)/(T_(2))]`
or `log.(1.07xx10^(9))/(2.66xx10^(8))=(E)/(2.303xx8.314)[(1)/(300)-(1)/(320)]`
or `E_(a)= 55.56xx10^(3)J`
Calculation of `A` at `300 K`
d.`k=Ae^(-E_(a)//RT)`
or `log k= log A- (E_(a))/(2.303RT)`
or `log.2.66xx10^(8)= log A-(55.56xx10^(3))/(2.303xx8.314xx300)`
or `A= 1.25xx10^(18)s^(-1)`