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N(2) + 3H(2) rarr 2NH(3), Delta H = - 46...

`N_(2) + 3H_(2) rarr 2NH_(3), Delta H = - 46K`. Cals. From the above reaction, heat of formation of ammonia is

A

46k Cals

B

`-46` kCals

C

`-23` kCals

D

23kCals

Text Solution

Verified by Experts

The correct Answer is:
C
Heat of formation `rArr` for 1 mole
`rArr Delta H_(f) = (-46)/(2) = -23`
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