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For N(2)(g) + 3H(2)(g) rarr 2NH(3)(g) + ...

For `N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g) + 22kcal`, `E_(a)` for the reaction is `70 kcal`. Hence, the activation energy for `2NH_(3)(g) rarr N_(2)(g) + 3H_(2) (g)` is :

A

`70kcal`

B

`92kcal`

C

`48kcal`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
`E_(a(fo rward))-E_(a(backward))=DeltaH`
`E_(a(backward))E_(a)-DeltaH`
`=70-(-22)=92kcal`
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