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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

92 Kcal

B

70 Kcal

C

48 Kcal

D

22 Kcal

Text Solution

Verified by Experts

`DeltaH=E_(f)-E_(b)`
`-22=70-E_(b)`
`E_(b)=92 kcal`
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