The value of `Delta_(f) H^(Θ)` for `NH_(3)` is `-91.8 kJ mol^(-1)`. Calculate enthalpy change for the following reaction.
`2NH_(3) (g) rarr N_(2) (g) + 3H_(2) (g)`

Explanation

Given `1/2 N_(2)(g) + 3/2 H_(2)(g) to NH_(3)(g), Delta_(f)H^(@) = -91.8 kJ "mol"^(-1)`
Multiplying by 2
`N_(2)(g) + 3H_(2)(g) to 2NH_(3)(g)`
`DeltaH = -183.6 kJ`
or `2NH_(3)(g) to N_(2)(g) + 3H_(2)(g)`
`DeltaH = 183.6 kJ`.