Given : NH_(3)(g)+3Cl_(2)(g) hArr NCl_(3)(g), +3HCl(g), -DeltaH_(1)N_(2)(g) + 3H_(2)(g) hArr 2NH_(3)(g), -DeltaH_(2) H_(2)(g) + Cl_(2)(g) hArr 2HCl(g), DeltaH_(3) The heat of formation of NCl_(3)(g) in terms of Delta H_(1), DeltaH_(2)andDeltaH_(3)is

Given : NH_(3)(g)+3Cl_(2)(g) hArr NCl_(3)(g), +3HCl(g), -DeltaH_(1)N_(

1.	Given : NH_(3)(g)+3Cl_(2)(g) hArr NCl_(3)(g), +3HCl(g), -DeltaH_(1)N_(2)(g) + 3H_(2)(g) hArr 2NH_(3)(g), -DeltaH_(2) H_(2)(g) + Cl_(2)(g) hArr 2HCl(g), DeltaH_(3) The heat of formation of NCl_(3)(g) in terms of Delta H_(1), DeltaH_(2)andDeltaH_(3)is
Answer» `DeltaH_(F) = - DeltaH_(1)+ ( DeltaH_(2))/( 2) - ( 3)/( 2) DeltaH _(3)` `DeltaH_(f) = DeltaH_(1)+ ( DeltaH_(2))/( 2) - ( 3)/( 2) DeltaH _(3)` `DeltaH_(f) = DeltaH_(1)- ( DeltaH_(2))/( 2) - ( 3)/( 2) DeltaH _(3)` NONE of the above Solution :We aim at`: (1)/(2) N_(2)(g) + (3)/(2) Cl_(2)(g)RARR NCl_(3)(g), DeltaH_(f)= ?` `(1)/(2) `Eqn. (II) `+ (3)/(2)`Eqn. (iii) and SUBTRACTING the sum fromEqn. (i) gives the required equation.Hence, `DeltaH _(f)= - DeltaH_(1) - [(- DeltaH_(2))/( 2)+(3)/(2) DeltaH_(3)]` `= - DeltaH_(1) + ( DeltaH_(2))/( 2) - (3)/(2)DeltaH_(3)`