The energy of activation of a first order reaction is `187.06 kJ mol^(-1)` at `750 K` and the value of pre-exponential factor A is `1.97xx10^(12)s^(-1)`. Calculate the rate constant and half life. `(e^(-30)= 9.35xx10^(-14))`

Options

A. k=0.184`s^(-1)` and `t_(1//2)`=3.76s
B. k=0.154`s^(-1)` and `t_(1//2)`=3.76s
C. k=0.184`s^(-1)` and `t_(1//2)`=2.76s
D. none of the above

Answer

A

Explanation

Given `E_(a)= 187.06 KJ mol^(-1)`
`T= 750 K`
`R= 8.314 JK^(-1)mol^(-1)`
We known `K=Ae^(-Ea//RT)`
`:. (E_(a))/(RT)=(187.06xx10^(3))/(8.314xx750)~~30`
`:. k= Ae^(-30)= 1.97xx10^(12)xx9.35xx10^(-14)`
`k= 0.184s^(-1)`
Half life `(t_(1//2))=(0.693)/(0.184)=3.76s`