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For the equilibrium A(g)B(g),ΔH is 40kJ/mol, If the ratio of activation energies of the forward Ef and reverse Eb is 2/3, then

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By Expert Faculty of Sri Chaitanya
a
Ef=60kJ/mol;Eb=100kJ/mol
b
Ef=30kJ/mol;Eb=70kJ/mol
c
Ef=80kJ/mol;Eb=120kJ/mol
d
Ef=70kJ/mol:Eb=30kJ/mol
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detailed solution

Correct option is C

 . 

The given reaction is,

A(g)B(g)H=40KJ/mol&EfEb=23Ef=23Eb

From graph we have,

H=EbEf40=Eb23Eb40=13EbEb=120KJ/mol&Ef=EbΔH=120(40)80KJ/mol


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