For the equilibrium, `A(g)rarrB(g), triangleH` is - 40 kJ `mol^(-1)` . If the ratio^f the activation energies of the forward `(E_(f))` and reverse `(E_(b))` reaction is (2/3) then:

Options

A. `E_(f)=60 kj mol^(-1),E_(b)=100 kj mol^(-1)`
B. `E_(f)=30 kj mol^(-1),E_(b)=70 kj mol^(-1)`
C. `E_(f)=80 kj mol^(-1),E_(b)=120 kj mol^(-1)`
D. `E_(f)=70 kj mol^(-1),E_(b)=30 kj mol^(-1)`

Answer

C

Explanation

`A(g) rarrB(g) triangle H=-40 kj`
`(3x)/(5)-(2x)/(5)=+40 rarr (x)/(5)=40 rarr x=200`
Therefore `E_(b)=(3x)/(5)=(3xx200)/(5)=120 kj mol^(-1)`
`E_(f)=(2x)/(5)=(2xx200)/(5)=80 kj mol^(-1)`