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))

The value of Delta_(f) H^( Theta ) for NH_(3) is -91.8 "kJ mol"^(-1).

1.	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))
Answer» SOLUTION :`(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)`.