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Enthalpy of formation of ammonia is -46....

Enthalpy of formation of ammonia is `-46.0 kJ "mol"^(-1)`.The enthalpy change for the reaction:
`2NH_(3)(g) to N_(2)(g) + 3H_(2)(g)`

A

`46.0 kJ "mol"^(-1)`

B

`-23.0 kJ "mol"^(-1)`

C

`92.0 kJ "mol"^(-1)`

D

`-92.0 kJ "mol"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
(c) Enthalpy of formation for 2 mol of `NH_(3) = 2 xx -46.0 = -92.0 kJ`
Enthalpy change for the reaction = 92.0 kJ (`therefore` It is reverse reaction)
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