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)

Enthalpy of formation of ammonia is -46.0 kJ"mol"^(-1).The enthalpy ch

1.	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)
Answer» `46.0 KJ"mol"^(-1)` `-23.0 kJ "mol"^(-1)` `92.0 kJ"mol"^(-1)` `-92.0 kJ"mol"^(-1)` Solution :(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)