For a given reaction, energy of activation for forward reaction(Eaf) is 80 kJ mol−1. Enthalpy change, ΔH=−40 kJ mol−1 for the reaction. A catalyst lowers Eaf to 20 kJmol−1. The ratio of energy of activation for reverse reaction before and after addition of catalyst is:
Enthalpy change, ΔH=Ef−Eb⇒−40=80−Eb
Eb=120 kJ/mol
The catalyst lowers the Eafto20kJ/mol for forward reaction then E′af=20kJ/mol
As we know that a catalyst decreases the activation energy by equal amount in both direction, the decrease in both the directions is by 60 kJ/mol
E′b=(120−60)=60 kJ/mol
EbE′b=12060=2.0
Hence, (d) is correct.