The energy change accompanying the equilibrium reaction `A hArr B` is `-33.0 kJ mol^(-1)`.

Assuming that pre-exponential factor is same for forward and backward reaction answer the following:
The equilibrium constant `k` for the reaction at `300 K`

Options

A. `5.55 xx 10^(5)`
B. `5.67 xx 10^(3)`
C. `5.55 xx 10^(6)`
D. `5.67xx 10^(2)`

Answer

A

Explanation

`k_(f) = Ae^(-Ef//RT), k_(b) = Ae^(-E_(b)//RT)`
`:. k = (k_(f))/(k_(b)) = (Ae^(-E_(f)//RT))/(Ae^(-E_(b)//RT)) = e^((E_(b)-E_(f))//RT)`
or `log k = (E_(b) - E_(f))/(2.303 RT)`
`= (33000 J mol^(-1))/(2.303 xx 8.314 J K^(-1) mol^(-1) xx 300 K) = 5.745`
`k = "Antilog" (5.745) = 5.55 xx 10^(5)`