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Question

For a given reaction, energy of activation for forward reaction(Eaf) is 80 kJ mol1. Enthalpy change, ΔH=40 kJ mol1 for the reaction. A catalyst lowers Eaf to 20 kJmol1. The ratio of energy of activation for reverse reaction before and after addition of catalyst is:

A
1.0
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B
0.5
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C
1.2
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D
2.0
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Solution

The correct option is D 2.0

Enthalpy change, ΔH=EfEb40=80Eb

Eb=120 kJ/mol
The catalyst lowers the Eafto20kJ/mol for forward reaction then Eaf=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

Eb=(12060)=60 kJ/mol

EbEb=12060=2.0

Hence, (d) is correct.


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