The rate constant of a reaction at 500 K and 700 K are 0.02 `s^(-1)` and 0.07 `s^(-1)` respectively. The value of `E_a` is

Options

A. 20.24 kJ `mol^(-1)`
B. 19.6 kJ `mol^(-1)`
C. 18.23 kJ `mol^(-1)`
D. 17.42 kJ `mol^(-1)`

Answer

C

Explanation

`k_1=0.02s^(-1),k_2=0.07s^(-1),T_1=500 K,T_2=700 K`
`log(k_2)/(k_1)=(E_a)/(2.303R)((1)/(T_1)-(1)/(T_2))`
`implies log(0.07)/(0.02)=(E_a)/(2.303xx8.314)((1)/(500)-(1)/(700))`
`E_a=2.303xx8.314((500xx700)/(700-500))log(0.07)/(0.02)`
=18230.36 J`//`mol or 18.32 kJ`//`mol