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) rarr NH_(3) (g), Delta(f) H^(Θ) = - 91.8 kJ mol^(-1)`
(`Delta_(f) H^(Θ)` means enthalpy of formation of 1 mole of `NH_(3)`)
`:.` Enthalpy change for the formation of 2 moles of `NH_(3)`
`N_(2) (g) + 3H_(2) (g) rarr 2NH_(3) (g), Delta_(f) H^(Θ) = 2 xx -91.8 = - 183.6 kJ mol^(-1)`
And for the reverse reaction,
`2NH_(3) (g) rarr N_(2) (g) + 3H_(2) (g), Delta_(f) H^(Θ) = + 183.6 kJ mol^(-1)`
Hence, the value of `Delta_(f) H^(Θ)` for `NH_(3)` is `+ 183.6 kJ mol^(-1)`