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Question

For a given reaction, energy of activation for forward reaction (Eaf) is 80kJ.mol1. ΔH=40kJ.mol1 for the reaction. A catalyst lowers Eaf to 20kJ.mol1. The ratio of energy of activation for reverse reaction before and after addition of catalyst is :

A
1
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B
0.5
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C
1.2
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D
2
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Solution

The correct option is D 2
Δ H = Ef Eb -40 = 80 - Eb Eb = 120 kJ/mole,
catalyst lower the Ef To 20 kJ/ mole for forward Rxn then Ef' = 20 kJ/mol
we know catalyst decreases the Activation energy equal amount
in both direction
Eb=(12060) =60kJ/mol
EbEb=12060=2.0

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