The energies of activation for forward and reverse reaction for `A_(2)+B_(2) hArr 2AB` are `180 kJ "mol"^(-1)` and `200 kJ "mol"^(-1)` respectively. The presence of catalyst lowers the activation energy of both (forward and reverse) reactions by `100 kJ "mol"^(-1)`. The enthalpy change of the reaction `(A_(2) + B_(2) rarr 2 AB)` in the presence of catalyst will be (in `kJ "mol"^(-1)`):

Options

A. 300
B. 120
C. 280
D. -20

Answer

D

Explanation

`A_(2)+B_(2)rarr2AB`
In the presence of Catalyst,
`E_(a)("Forward")=180-100=80 kJ"mol"^(-1)`
`E_(a)(Backward)=200-100=100kJ"mol"^(-1)`
`DeltaH=E_(a)("forward")-E_(a)(Backward)`
`=80-100=-20kJ"mol"^(-1)`