For a given reaction, energy of activation for forward reaction `(E_(af)` is 80 kJ `mol^(-1).DeltaH=-40"kJ mol"^(-1)` for the reaction. A catalyst lowers `E_(af)` to 20 kJ `mol^(-1)`. The ratio of energy of activation for reverse reaction before and after addition of catalyst is :

Options

A. `1.0`
B. `0.5`
C. `1.2`
D. `2.0`

Answer

D

Explanation

`Delta H = E_(r) = E_(k) rArr - 40 = 80 - E_(a) " "E_(k) = 120 "kJ/mol"`,
Catalyst lower the `E_(t)` to 20 kJ/mol for forward reaction then `E_(r) =20` kJ/mol
We know catalyst decreases the Activation energy equal amount in both direction
`E_a = (120 - 60) =60` kJ/mol
`(E_b)/( E_b^1) = (120)/( 60 ) = 2.0`