Question

The time required for `10%` completion of a first order reaction at `298 K` is equal to that required for its `25%` completion at `308 K`. If the pre-exponential factor for the reaction is `3.56 xx 10^(9) s^(-1)`, calculate its rate constant at `318 K` and also the energy of activation.

Answer 1

Correct Answer - `k = 9.3 xx 10^(-4) s^(-1)`
`E_(a) = 76.6 kJ mol^(-1)`
`t_(1) = (2.3)/(k_(1))log((100)/(100-25))`
`t_(2) = (2.3)/(k_(2))log((100)/(100-25))`
As `t_(1) = t_(2)`,
` (2.3)/(k_(1))log.((10)/(9)) = (2.3)/(k_(2))log.((4)/(3))`
`:. (k_(2))/(k_(2)) = (log((4)/(3)))/(log((10)/(9))) = 2.73`
Also, `log.(k_(2))/(k_(1)) = (E_(a))/(2.3 R)((T_(2)-T_(1))/(T_(1)T_(2)))`
`log(2.73) = (E_(a))/(2.3 xx 8.314 J K^(-1) mol^(-1)) xx((10)/(298 xx 398))`
`:. E_(a) = 76.6 xx 10^(3) J mol^(-1) = 76.6 kJ mol^(-1)`
Further, `k = Ae^(-E_(a)//RT)`.
or `ln k = ln A-(E_(a))/(RT)`
`ln k = ln(3.56 xx 10^(9)s^(-1))`
`- (76.6 K J mol^(-1))/(8.314 xx 10^(-3) K J mol^(-1) K^(-1) xx 318)`
`:. k = 9.3 xx 10^(-4) s^(-1)`

Answer 2

Correct Answer - `k = 9.3 xx 10^(-4) s^(-1)`
`E_(a) = 76.6 kJ mol^(-1)`
`t_(1) = (2.3)/(k_(1))log((100)/(100-25))`
`t_(2) = (2.3)/(k_(2))log((100)/(100-25))`
As `t_(1) = t_(2)`,
` (2.3)/(k_(1))log.((10)/(9)) = (2.3)/(k_(2))log.((4)/(3))`
`:. (k_(2))/(k_(2)) = (log((4)/(3)))/(log((10)/(9))) = 2.73`
Also, `log.(k_(2))/(k_(1)) = (E_(a))/(2.3 R)((T_(2)-T_(1))/(T_(1)T_(2)))`
`log(2.73) = (E_(a))/(2.3 xx 8.314 J K^(-1) mol^(-1)) xx((10)/(298 xx 398))`
`:. E_(a) = 76.6 xx 10^(3) J mol^(-1) = 76.6 kJ mol^(-1)`
Further, `k = Ae^(-E_(a)//RT)`.
or `ln k = ln A-(E_(a))/(RT)`
`ln k = ln(3.56 xx 10^(9)s^(-1))`
`- (76.6 K J mol^(-1))/(8.314 xx 10^(-3) K J mol^(-1) K^(-1) xx 318)`
`:. k = 9.3 xx 10^(-4) s^(-1)`