For the equilibrium, `A(g)hArrB(g),DeltaH` is `-40kJ//mol`, if the ratio of the activation energies of the forward `(E_(f))` and reverse `(E_(b))` reactions is 2/3, then

Options

A. `E_(f)=80kJ//mol,E_(b)=120kJ//mol`
B. `E_(f)=60kJ//mol,E_(b)=100kJ//mol`
C. `E_(f)=30kJ//mol,E_(b)=70kJ//mol`
D. `E_(f)=70kJ//mol,E_(b)=30kJ//mol`

Answer

A

Explanation

Given, `A(g) hArr B(g)" "DeltaH=-40kJ//mol`
So, `(E_(f))/(E_(b))=(2)/(3)`
We know that,
`E_(f)-E_(b)=DeltaH`
`E_(f)-E_(b)=-40`
`E_(b)xx(2)/(3)-E_(b)=-40`
Hence, `E_(b)=120kJ//mol,E_(f)=80kJ//mol`