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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

The correct Answer is:
A
(a) `Delta H = E_(f) - E_(b)`
`-22 =70 - E_(b)`
`E_(b) = 92 kcal`
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