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

Explanation

`(1)/(2) N_(2(g)) + (3)/(2) H_(2(g)) to NH_(3(g)) , Delta_(f) H^( Theta ) = -91.8 "kJ mol"^(-1)`
(`Delta_(f) H^( Theta )` means enthalpy of formation of 1 mole of `NH_(3)` )
`therefore` Enthalpy change for the formation of 2 moles of `NH_(3)`
`N_(2(g)) + 3H_(2(g)) to 2NH_(3(g)) , Delta_(f) H^( Theta )= 2 xx -91.8 =-183.6 "kJ mol"^(-1)`
for the reverse reaction,
`2NH_(3(g)) to N_(2(g)) + 3H_(2(g)) , Delta_(f) H^( Theta ) = + 183.6 "kJ mol"^(-1)`
Hence, the value of `Delta_(f) H^( Theta ) ` for `NH_(3)` is `+183.6 "kJ mol"^(-1)`.