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Question

For a given reaction, energy of activation for forward reaction (Eaf)is80kJ.mol1.ΔH=40kJ.mol1 for the reaction. A catalyst lowers Eafto20kJ.mol1 . Find out the ratio of energy of activation for reverse reaction before and after addition of catalyst.

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Solution

ΔH=EfEb40=80Eb, Eb=120kJ/mole

Catalyst lower the Eaf to 20kJ/mole for forward Rxn then Ef=20kJ/mol

We know catalyst decreases the activation energy equal amount in both directions:

Eb=(12060)=60kj/mol

EbEb=12060=2.0

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